Class 9 Mathematics

Chapter 13 Construction Exercise 13

Question 1. We draw a line segment PQ of length 5 cm. We take an external point A on the line segment. Let us draw a parallel line through point A to line segment PQ. [Let us draw three alternative processes].

Solution:
First method:

 

 

1. First draw a 5 cm long straight line PQ with the help of scale and outside PQ take any point.
2. On PQ straight line take any point R. A and R joined by scale, consequently is formed ∠ARQ. Now on point A on the opposite side of ∠ARQ ∠RAS joining is formed S and A by the scale and extending on both sides, got straight line XY

∴ ∠SAR = ∠ARQ, Alternate angles. XY II PQ
∴ XY is the required line through A and parallel to PQ.

Read and Learn More WBBSE Solutions For Class 9 Maths

Second Method:

 

 

1. First draw a 5 cm long straight line PQ with help of scale and outside PQ take any point.
2. On PQ straight line takes any point D.
3. D and A are joined by scale.

DA is extended to point E, consequently, ∠ADQ is formed.

4. On the same side of AD, equal to ∠ADQ on straight line EA at point A, ∠EAB is formed.
5. BA straight line is extended to point R.

∴ ∠EAB = ∠ADQ (corresponding angles). ∵RB II PQ
RB is the required straight line through A and parallel to PQ.

Third Method:

 

 

1. First draw a 5 cm long straight line PQ with the help of scale and outside PQ take any point
2. On PQ straight line take any point B. B and A are joined.
3. From BQ a segment BO is cut off.
4. Taking O as the centre, and taking a radius equal to BA, an arc is drawn.

5. Now taking A as a centre and taking a radius equal to BO another arc is drawn which cuts the former at point R. Joining A and R by the scale and extending on both sides we get MN straight line.

∴MN is the required straight line through A and parallel to PQ.

Question 2. We draw a triangle with lengths of sides 5 cm, 8 cm and 11 cm and draw a parallelogram equal in area to that triangle and having an angle of 60°. [Let us write instruction process and proof].

Solution:
1. By scale, an 11 cm straight line BC is drawn. Taking B and C as centres, and taking radii equal to 8 cm and 5 cm respectively, two arcs are drawn which cut each other at point A. A B and AC are joined. ABC is a triangle whose sides AB = 8 cm, AC 5 cm and BC= 11 cm.
2. In the ΔABC side BC is bisected at point D.
3. In ΔABC from A to BC a parallel straight line PQ is drawn.
4. In ΔABC on side BC at point D an angle of 60°, ∠GDC is drawn which cuts PQ at E.

5. From EQ, DC equal to EF is cut and a joining C, F, and CDEF parallelogram is formed.

 

 

 

 

6. Proof: A, and D are joined.

In □EDCF, DC II EF (By construction) and DC EF (By construction)
∴ EDCF is a parattelogram.

ΔADC and parallelogram □EDCF are on the same base DC and between parallel lines DC and EF.

∴ ΔADC = \(\frac{1}{2}\) parallelogram □EDCF …..(1)

Again, in □ABC, AD is the median

∴ ΔADC = \(\frac{1}{2}\) ΔABC

From (1) and (2), ΔABC = parallelogram □EDCF
∴ Equal to ΔAABC in the area a parallelogram EDCF is formed whose one angle is 60°.

Question 3. We draw a triangle in which AB = 6cm, BC =9 cm, ∠ABC= 55°, let us draw a parallelogram equal in area to that triangle having an angle of 60° and length of one side is \(\frac{1}{2}\) of AC.

Solution:
1. A straight line AP is drawn in which segment AB equal to 6 cm is cut off. On point angle, ∠ABQ= 55° is drawn. From segment BQ, BC equal to 9 cm is cut off. A and C are joined; consequently ∠ABC is formed whose side AB = 6 cm, BC = 9 cm and <ABC = 55°

 

 

2. In ΔABC side AC is bisected at point D.
3. In ΔABC from point B parallel to AC an st. line XY is drawn.
4. In ΔABC on side AC at point D equal to 60° angle GDC is drawn which cuts XY at point E.
5. DC equal to EF is cut off from EY and joining C and F a parallelogram CDEF is formed.

Proof: B and D are joined.

In □CDEF, DC II EF (By construction) and DC = EF (By construction)
∴ CDEF is a parallelogram.

ΔBDC and parallelogram CDEF lie on same base DC and in between parallel lines DC and XY.

∴ ΔBDC = \(\frac{1}{2}\) parallelogram □CDEF

Again, in ΔABC, BD is the median

∴ ΔBDC = \(\frac{1}{2}\) ΔABC

∴ \(\frac{1}{2}\) ΔΑΒC = \(\frac{1}{2}\) parallelogram □CDEF

∴ ΔABC = parallelogram □ CDEF

Question 4. In ΔPQR, ∠PQR = 30°, ∠PRQ = 75° and QR = 8 cm. Let us draw a rectangle equal in area to that triangle.

Solution:
1. PQR is a triangle whose ∠PQR = 30°, ∠PRQ = 75° and QR = 8 cm.
2. In ΔPQR from point P parallel to RQ a straight line MN is drawn.
3. RQ is bisected at S which cuts MN at T.
4. From TN equal to SQ segment TO is cut and O, and Q are joined. TSQO is the required rectangle.

 

 

 

 

 

Proof: P and S are joined.
In □TSQO, SQ II TO (By construction)
and SQ = TO (By construction)
and ∠TSQ = 90°

∴TSQO is a rectangle.
ΔPSQ and rectangle TSQO lie on the same base SQ and in between parallel lines SQ and PO.

∴ ΔPSQ= \(\frac{1}{2}\) rectangle TSQO

Again, in ΔPRQ, PS is the median

∴ ΔPSQ =\(\frac{1}{2}\) ΔPRQ

∴ \(\frac{1}{2}\) ΔPRQ= \(\frac{1}{2}\) rectangle TSQO

∴ PRQ rectangle TSQO Proved

Question 5. Draw an equilateral triangle with a length of each side of 6.5 cm and let us draw a parallelogram equal in area to that triangle and having an angle of 45°.

Solution:
Construction process:
1. An equilateral triangle AABC having a side 6 cm is drawn.
2. In ΔABC side BC is bisected at point D and from point A parallel to BC a straight line PQ is drawn.
3. In ΔABC at point D on side BC angle ∠EDC equal to 45° is drawn which cuts side PQ at point E.
4. Equal to DC EF is cut from the EQ segment. Joining C and F, the parallelogram CDEF is formed.

 

 

Proof: In □CDEF, DC II EF (By construction) and CD = EF (By construction)
∴ CDEF is a parallelogram.

ΔADC and parallelogram CDEF lie on the same base CD and in between parallel lines CD and AF.

∴ ΔADC = \(\frac{1}{2}\) parallelogram □CDEF

Again, in ΔABC, AD is the median

∴ΔADC = \(\frac{1}{2}\) ΔABC

∴ \(\frac{1}{2}\) ΔABC= \(\frac{1}{2}\) parallelogram □CDEF

∴ ΔABC = parallelogram □CDEF Proved

Question 6. The length of each equal side of an isosceles triangle is 8 cm and the length of the base is 5 cm. Let us draw a parallelogram equal in area to that circle and having one angle of a parallelogram is equal to one of the equal angles of an isosceles triangle and one side is \(\frac{1}{2}\) of equal side.

Solution: ABC, an isosceles triangle is formed whose base BC = 5 cm and equal sides AB = AC = 8 cm.

 

 

Equal to this triangle in area, a parallelogram ADEF is formed whose ∠D is equal to ∠C of the isosceles triangle.

Question 7. Construct an isosceles triangle whose equal sides are each equal to 8 cm and the vertical angle is 30°. Construct a parallelogram equal in area to this triangle.

Solution: ΔABC, an isosceles triangle is formed whose sides AB = AC = 8 cm and ΔBAC = 30°.

 

Equal to this triangle in the area is the parallelogram BEGH is formed.

 

Chapter 12 Theorems On Area Exercise 12

Question 1. P and Q are the midpoints of sides AB and DC of parallelogram ABCD, let’s prove that the area of quadrilateral field = \(\frac{1}{2}\) x area of parallelogram field.

Solution:

 

 

Join: P, Q.

Proof: ∵ P is the midpoint of AB

∴ AP = \(\frac{1}{2}\) AB

= \(\frac{1}{2}\) DC [∵ opposite sides of parallelogram]

= QC (∵ Q is the midpoint of CD)

Read and Learn More WBBSE Solutions For Class 9 Maths

In □PCQ, AP = QC & AP II QC (∵ AB II DC)
∴ APCQ is a parallelogram.

In □PQD, AP II DQ & AP = DC
∴  APQD is a parallelogram.

In □PBCQ, PB = QC & PB II QC
∴ PBCQ is a parallelogram.
∴ Area of quadrilateral APCQ = ΔAPQ+ΔPCQ

= \(\frac{1}{2}\) parallelogram □APQD + \(\frac{1}{2}\) parallelogram □PBCQ

= \(\frac{1}{2}\) (parallelogram □APQD+ parallelogram □PBCQ)

= \(\frac{1}{2}\) Area of parallelogram ABCD  Proved

Question 2. The distance between two sides AB and DC of a rhombus ABCD is PQ and the distance between sides AD and BC is RS; let’s prove that PQ = RS.

Solution:

 

 

To prove PQ = RS.

Proof: ∵  Area of rhombus = base x height
∴ Area of rhombus ABCD = AB.PQ

Again, the area of rhombus ABCD = BC.RS
∴ AB.PQ=BC.RS
∴ As AB = BC (Side of the rhombus)
∴ PQ = RS Proved

Question 3. P and Q are the mid-points of sides AB and DC of parallelogram ABCD respectively. Let’s prove that PBQD is a parallelogram and ΔPBC= \(\frac{1}{2}\) parallelogram PBQD.

Solution:

 

 

 

To prove
1. PBQD is a parallelogram

2. ΔPBC =  \(\frac{1}{2}\) Parallelogam PBQD

Join: P & C.

Proof: ∵ ABCD is a parallelogram.
∴ AB = DC

or, \(\frac{1}{2}\) AB= \(\frac{1}{2}\) DC

or, PB = DQ (∵ P & Q are the mid-points of the sides AB & DC respectively)

In parallelogram □PBQD, PB = DQ & PB II DQ (∵ AB II DC)
∴ PBQD is a parallelogram  …… (1)

∴ ΔPBC & parallelogram PBQD are on the same base PB & between the same parallel straight lines PB & DC.

∴ ΔPBC = \(\frac{1}{2}\) parallelogram □PBQD ….(2) Proved

Question 4. In an isosceles triangle, ABC, AB = AC, and P is any point on produced side BC. PQ and PR are perpendicular on sides AB and AC from the point P respectively. BS is perpendicular on side AC from point B; let’s prove that PQ-PR = BS.

Solution:

 

 

 

To prove, PQ-PR = BS
Join: P, Q.

Proof: Area ΔABP = area of ΔABC+ area of ΔAPC

or, \(\frac{1}{2}\) . AB.CQ = \(\frac{1}{2}\) AC.BS+ \(\frac{1}{2}\) .AC.PR

,or, AB.PQ = AC(BS +PR)
or, AC. PQ AC (BS+ PR) (∵ AB = AC)

or, PQ=BS + PR
or, PQ-PR = BS Proved

Question 5. O is any point outside the equilateral triangle ABC and within the angular region on ABC; OP, OQ and OR are the perpendicular distances of AB, BC, and CA respectively from the point O. Let us prove that the altitude of the triangle = OP + OQ – OR.

Solution:

 

 

To prove the Height of the triangle = OP + OQ – OR.

Join: O, A; O, B and O, C & AS is the perpendicular from A on BC.

Proof : ΔABC = ΔAOB + ΔBOC – ΔAOC

or, \(\frac{1}{2}\).BC.AS = \(\frac{1}{2}\).AB.OP+\(\frac{1}{2}\).BC.OQ – \(\frac{1}{2}\) AC.OR

or, BC.AS = BC. OP+ BC. OQ – BC. OR [∵ AB = BC = AC]
or, AS OP + OQ – OR
or, Height of the triangle = OP+OQ-OR Proved

Question 6. A straight line parallel to AB of parallelogram ABCD intersects sides AB, AC, and BC or their produced parts at the points E, F, and G respectively. Let’s prove that ΔAEG = ΔAFD.

Solution:

 

 

To prove  ΔAEG = ΔAFD

Join: A, G & D, F.

Proof: ΔACG & parallelogram CDEG are on the same base CG & in between two parallelogram lines CG & AD.

∴ ΔACG = \(\frac{1}{2}\) parallelogram □CDEG

Again, ΔCDF & parallelogram □CDEG are on the same base & in between two parallel lines CD & EG.

∴ ΔCDF = \(\frac{1}{2}\) parallelogram □CDEG

∴ ΔACG = ΔCDF …..(1)

∵ AC is the diagonal of parallelogram ABCD.
∴ ΔABC = ΔADC ……(2)

Subtracting (1) from (2), ΔABC – ΔACG = ΔADC – ΔCDF
or, ΔABG = ΔAFD
or, ΔAEG = ΔAFD ( ∵ AG is the diagonal of the parallelogam ABGE)
∴ ΔABG = ΔAEG Proved

Question 7. E is any point on side DC of parallelogram ABCD, produced AE meets produced BC at the point F. D, F are joined. Let’s prove that (1) ΔADF= ΔABE; (2)ΔDEF = ΔBEC.

Solution:

 

 

Join: D, F. To prove (1) ΔADF = ΔABE (2) ΔDEF = ΔBEC.

Proof: ΔADF & parallelogram ABCD are on the same base AD & in between two parallelograms AD & BF

∴ ΔADF = \(\frac{1}{2}\) parallelogram □ABCD

Again, ΔABE & parallelogram □ABCD are on the same base AB & between two parallel straight lines AB & DC.

∴ ΔABE = \(\frac{1}{2}\) parallelogram □ABCD

∴ ΔADF = ΔABE
∴ ΔADE+ ΔBEC = \(\frac{1}{2}\) □ABCD

or, ΔADE+ ΔBEC = ΔADF
or, ΔADE + ΔBEC = ΔADE + ΔDEF
or, ΔBEC = ΔDEF
or, ΔDEF = ΔBEC

Question 8. Two triangles ABC and ABD with equal area stand on the opposite side of AB. Let’s prove that AB bisects CD.

Solution:

 

 

To prove AB bisects CD.
CP & DQ are two perpendiculars drawn on AB from points C & D respectively. CD cuts AB at O.

Proof: ∵ΔABC= ΔABD (given)

∴ \(\frac{1}{2}\).AB.OP = \(\frac{1}{2}\).AB.DQ

or, CP = DQ
In ΔCOP & ΔDOQ,
CP = DQ (proved)

∠CPO = ∠DQO (= 90°)
∠COP = vertically opposite ∠DOQ

∴ ΔCPO = ΔDOQ (A-A-S condition)
∴ CO = OD
∴ AB bisects CD. Proved

Question 9. D is midpoint of side BC of triangle ABC. Parallelogram CDEF stands between side BC and parallels to BC through point A. Let’s prove that ΔABC = parallelogram □CDEF.
Solve: To prove AABC = parallelogram CDEF.

Solution:

 

 

 

Join: A, D.

Proof: ΔADC & parallelogram CDEF are on the same base CD & in between two parallel straight lines CD & AF.

∴ ΔADC = \(\frac{1}{2}\) parallelogram □CDEF

∵ AD is the median of ΔABC

∴ ΔADC = \(\frac{1}{2}\) ΔABC

∴ \(\frac{1}{2}\) ABC = parallelogram □CDEF

or, ΔABC = parallelogram □CDEF Proved

Question 10. P is any a point on diagonal BD of parallelogram ABCD. Let’s prove that ΔAPD = ΔCPD.

Solution:

 

 

To prove ΔAPD = ΔCPD
AE & CF are two perpendiculars on BD from points A & C respectively.

Proof: In ΔADE ΔBCF
∠ADE = alternate ∠CBF
∠AED = ∠CFB
& AD = BC

∴ ΔADE ≅ ΔBCF
∴ AE = CF

(∵ AD II BC & BD is the transversal)
[each angle is 90°]
(∵ opposite side of parallelogram)
(A-A-S condition)

As ΔAPD & ΔCPD are on the same base DP
∴ Their heights are equal.
∴ AE = CF
∴ ΔAPD = ΔCPD Proved

Question 11. AD and BE are the medians of triangle ABC. Let’s prove that ΔACD = ΔBCE

Solution:

 

 

In ΔABC, AD & BE are the medians.

To prove: ΔACD = ΔBCE

Proof: In ΔABC is the median.

∴ ΔACD = ΔABD = \(\frac{1}{2}\) ΔABC

Again, In ΔABC, BE is the median.

∴ ΔBCE = \(\frac{1}{2}\) ΔABC

∴ ΔACD = ΔBCE Proved

Question 12. A line parallel to BC of triangle ABC intersects sides AB and AC at the points P and Q respectively. CQ and BQ intersect each other at the point X. Let’s prove that:

1. ΔBPQ= ΔCPQ
2. ΔBCP = ΔBCQ
3. ΔACP= ΔABQ
4. ΔBXP = ΔCXQ

 

 

Solution: To prove

1. ΔBPQ= ΔCPQ
2. ΔBCP = ΔBCQ
3. ΔACP= ΔABQ
4. ΔBXP = ΔCXQ

Proof: ΔBPQ & ΔCPQ are on the same base PQ & in between two parallel lines PQ & BC.
∴ ΔBPQ = ΔCPQ ….(1)

ΔBCP & ΔBCQ are on the same base BC & in between two parallel lines PQ & BC.
∴ ΔBCP = ΔBCQ …(2)

∵ ΔBPQ = ΔCPQ (Proved)
∴ ΔBPQ+ΔAPQ = ΔCPQ + ΔAPQ

∴ ΔABQ = ΔACP
∴ ΔACP = ΔABQ

∵ ΔBPQ = ΔCPQ
∴ ΔBPQ – ΔPQX = ΔCPQ = ΔPQX

∴ ΔBXP = ΔCXQ …..(4) Proved

Question 13. D is the mid-point of BC of triangle ABC and P is any point on BC. Join P,A. Through the point D a straight line parallel to line segment PA meets AB at point Q. Let’s prove that:

1. ΔADQ= ΔPDQ
2. ΔBPQ= \(\frac{1}{2}\) ΔABC.

 

 

Solution: D is the mid-point of BC of triangle ABC and P is any point on BC. P and A are joined. Through the point D a straight line parallel to line segment PA meets AB at point Q.

To prove :
1. ΔADQ= ΔPDQ
2. ΔBPQ= \(\frac{1}{2}\) ΔABC.

Proof: ΔADQ and ΔPDQ lie on the same base DQ and between the same parallel straight lines DQ and AP.

∴ ΔADQ ΔPDQ
∴ ΔADQ = ΔPDQ ….(1)

∴ ΔADQ+ΔBDQ = ΔPDQ + ΔBDQ
∴ ΔABD = ΔBPQ

∵ AD is the median of ΔABC.

∴ ΔΑΒD = \(\frac{1}{2}\) ΔΑΒC

∴ ΔBPQ = \(\frac{1}{2}\) ΔABC ….(2)

Question 14. In triangle ABC of which AB = AC; perpendiculars through the points B and Con sides AC and AB intersect sides AC and AB at the points E and F. Let’s prove that FE II BC.

 

 

Solution: To prove FE II BC
Join E, F.

Proof: In ΔABC, AB = AC

∴ ∠ABC = ∠ACB
∴ ∠FBC= ∠ECB

In ΔBCF & ΔBCE
∠FBC=∠ECB Proved
∠BFC = ∠BEC (= 90°)& BC is the common base

∴ ΔBCF ≅ ΔBCE (A-A-S condition)
∴ Area of ΔBCF = Area of ΔBCE

∴ ΔBCF & ΔBCE are on the same base BC & their areas are equal.
∴ The triangles are in between two parallel lines.
∴ FE II BC Proved

Question 15. In triangle ABC, ∠ABC= ∠ACB; bisectors of an angle ∠ABC and ∠ACB intersect the sides AC and AB at the points E and F respectively. Let’s prove that FE II BC.

 

 

Solution: To prove FE II BC

Join: E, F

Proof: In ΔABC, ΔABC ΔACB

∴ \(\frac{1}{2}\) ∠ABC = \(\frac{1}{2}\) ∠ACB

∴ ∠EBC = ∠FCB

In ΔBCF & ΔBCE
∠FBC =∠ECB  (∵∠ABC = ∠ACB)
∠FCB = ∠EBC (Proved) & BC is the common side.

∴ ΔBCF ≅ ΔBCE (A-A-S Condition)
∴ Area of ΔBCF = Area ΔBCE

∴ ΔBCF & ΔBCE on the same BC & their areas are equal.
∴ The triangles are in between two parallel lines.
∴ FE II BC

Question 16. The shape of two parallelograms ABCD and AEFG, of which ∠A is common, are equal in area and E lies on AB. Let’s prove that DE II FC.

 

 

 

Solution: To prove DE II FC.

Join: D, E; E, C & C, F.

Proof: ΔDEC and parallelogram ABCD are on the same base & in between two parallel lines AB & CD.

∴ ΔDEC = \(\frac{1}{2}\) parallelogram □ABCD

Again, ΔDEF & parallelogram AEFG are on the same base & in between two parallel lines EF & AG.

∴ ΔDEF = \(\frac{1}{2}\) parallelogram □AEFG

∵ parallelogram □ABCD = parallelogram □AEFG (given)

∴ \(\frac{1}{2}\) parallelogram □ABCD = \(\frac{1}{2}\) parallelogram □AEFG

∴ ΔDEC = ΔDEF
ΔDEC & ΔDEF are on the same base DE & their areas are equal.
∴ DE II FC Proved

Question 17. ABCD is a parallelogram and ABCE is a quadrilateral. Diagonal AC divides the quadrilateral field ABCE into two equal parts. Let’s prove that AC II DE.

 

 

Solution: To prove  AC II DE.
Join D, E.

Proof: Diagonal AC bisects quadrilateral ABCD in two equal parts.
∴ Area of AABC = Area of AACE

Again, in the triangle ABC,
∴ Area of ΔABC = Area of ΔADC
∴ Area of ΔADC = Area of ΔACE

∴ ΔADC & ΔACE are of equal area & on the same base.
∴ Triangles are in between two parallel lines.
∴ AC II DC Proved

Question 18. D is the mid-point of side BC of triangle ABC; P and Q lie on sides BC and BA in such a way that ΔBPQ = \(\frac{1}{2}\) ΔABC. Let’s prove that, DQ II PA.

 

 

Solution: To prove  DQ II PA.
Join A, D.

Proof: In ΔABC, D is the mid-point of BC.
∴ AD is the median.

∴ ΔABD = \(\frac{1}{2}\) ΔABC

But ΔBPQ = \(\frac{1}{2}\) ΔABC (given)

∴  ΔABD = ΔBPQ
∴  ΔABD-ΔBDQ= ΔBPQ-ΔBDQ
∴ ΔADQ = ΔDPQ

∴  ΔADQ & ΔDPQ are of equal area and on the same base DQ.
∴  DQ II PA Proved

Question 19. Parallelogram ABCD, of which mid-points of sides AB, BC, CD are E, F, G and H, and DA respectively. Let’s prove that:

1. EFGH is a parallelogram.
2. Area of the shape of parallelogram EFGH is half of the area of the shape of parallelogram ABCD.

 

 

Solution: To prove EFGH is a parallelogram.

Join: A, C

Proof:

1. In ΔABC, E & F are the mid-points of AB & AC.

∴ EF II AC & EF = \(\frac{1}{2}\) AC …….(1)

Again, in ΔADC, the mid-points of AD & CD are H & G.

∴ HG II AC & HG=\(\frac{1}{2}\) AC …….(2)

From (1) & (2), EF II HG & EF = HG
∴ EFGH is a parallelogram.

2. O is the midpoint of AC. Join H, O & G, O. HE & GF cut AC at P & Q. In ΔDAC joining the mid-points of the sides AD, AC & CD, AHOG is formed.

∴ ΔHOG = \(\frac{1}{4}\) ΔABC

∵ HG II PQ

∴ ΔHOG = \(\frac{1}{2}\) parallelogram □HPQG

[ ∵ ΔHOG & parallelogram □HPQG are on the same base HG and in between two parallel lines HG & PQ]

∴ parallelogram □HPQG = 2ΔHOG = 2 x \(\frac{1}{4}\) ΔACD = \(\frac{1}{2}\) ΔACD

Similarly, parallelogram □EPQF = \(\frac{1}{2}\) ΔABC

∴ parallelogram □HPQG + parallelogram □EPQF = \(\frac{1}{2}\) ΔACD + \(\frac{1}{2}\) ΔABC

∴ parallelogram □EFGH = \(\frac{1}{2}\) ( ΔACD+ ΔABC) = \(\frac{1}{2}\) parallelogram □ABCD ….(2) Proved

Question 20. In of a trapezium ABCD, AB II DC and E is mid-point of BC. Let’s prove that area of triangular field AED = \(\frac{1}{2}\) Χ area of the shape of the trapezium field ABCD.

 

 

Solution: In trapezium ABCD, AB II CD and E is the mid-point of BC.

Let’s prove that in the triangular field = area \(\frac{1}{2}\)  x ABCD of the trapezium field.

Construction: From point E parallel to AD a straight line is drawn which cuts DC at P and extended AB at Q. A, and E are joined.

Proof: ΔAED & parallelogram □ADPQ are on the same base AD & in between two parallel lines AD and PQ.

∴ Area of ΔAED = \(\frac{1}{2}\) parallelogram □ADPQ

In ΔPEC & ΔBEQ,
∠PCE = alternate ∠EBQ

∠PEC (vertically opposite angle) ∠BEQ & CE = BE (∵ E is the midpoint of BC)

∴ ΔPEC ≅ ΔBEQ (A-A-S Condition)
∴ Area of ΔPEC = Area ΔBEQ

Area of ABCD trapezium = Area ADPEB + Area ΔPEC
= Area of ADPEB + Area of ABEQ
= Area □ADPQ
= Area 2x ΔAED

∴ 2 x area ΔAED = Area of ABCD trapezium

∴ Area ΔAED = \(\frac{1}{2}\) Χ Area of ABCD trapezium

Question 21. Multiple choice questions

1. D, E, and F are mid-points of sides BC, CA, and AB respectively of a triangle ABC. If AABC= 16 sq. cm, then the area of the shape of trapezium FBCE is

 

 

1. 40 sq. cm
2. 8 sq. cm
3. 12 sq. cm
4. 100 sq. cm

Solution: Join B, E.

∵ BE is the median of ΔABC.

∴ Area ΔABE = Area ΔBEC = Area \(\frac{1}{2}\) ΔABC

= \(\frac{1}{2}\) x 16 sq cm= 8 sq cm

∵ EF is the median of ΔABE

∴ Area ΔBEF= Area \(\frac{1}{2}\) ΔABE

= \(\frac{1}{2}\) x 8 sq cm = 4 sq cm
Area FBCE = Area ΔBEF + Area ΔBCE = (4+8) sq. cm = 12 sq. cm

Solution: 3.12 sq. cm

2.  A, B, C, and D are the mid points of sides PQ, QR, RS, and SP respectively of parallelogram PQRS. If the area of the shape of the parallelogram PQRS = 36 sq. cm then the area of the ABCD field is

 

 

1. 24 sq. cm
2. 18 sq. cm
3. 30 sq. cm
4. 36 sq. cm

Solution: We know the area of the parallelogram formed by joining the midpoints of a parallelogram is half of that parallelogram.

∴ Area of ABCD = \(\frac{1}{2}\) Χ Area of parallelogram □PQRS

= \(\frac{1}{2}\) x 36 sq. cm = 18 sq. cm

Solution: 2. 18 sq. cm

3. O is any point inside parallelogram ABCD. If AAOB +ACOD = 16 sq. cm, then area of the shape of parallelogram ABCD is

 

 

1. 8 sq. cm
2. 4 sq. cm
3. 32 sq. cm
4. 64 sq. cm

Solution: The straight line through O & parallel to AB intersects AD & BC at R & S points respectively.

∴ ΔΑΟΒ = \(\frac{1}{2}\) parallelogram □ARSB

∴ ΔCOD = \(\frac{1}{2}\) parallelogram □CSRD

∴ ΔAOB + ΔCOD = \(\frac{1}{2}\) (parallelogram □ARSB+ parallelogram □CSRD)

∴ ΔAOB+ ΔCOD = \(\frac{1}{2}\) Χ parallelogram □ABCD

or, 16 sq. cm = \(\frac{1}{2}\) parallelogram □ABCD
∴ Area of parallelogram ABCD = 2 x 16 sqcm = 32 sqcm

Solution: 3. 32 sq. cm

4. D is the mid-point of side BC of triangle ABC. E is the mid-point of side BD and O is the mid-point of AE; the area of triangular field BOE is

 

 

1. \(\frac{1}{3}\) Χ Area of ΔABC

2. \(\frac{1}{4}\) Χ Area of ΔABC

3. \(\frac{1}{6}\) Χ Area of ΔABC

4. \(\frac{1}{8}\) Χ Area of ΔABC

Solution:

Area of the triangular area BOE = \(\frac{1}{2}\)  x area ΔABE [ ∵BO is the median of ΔABE]

\(\frac{1}{2}\) x \(\frac{1}{2}\) x area ABC (∵ In ΔABE, BO is the median)

\(\frac{1}{2}\) x \(\frac{1}{2}\) x area ABD (∵ In ΔABD, AE is the median)

=\(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) x area ABC (∵ In ΔABC, AD is the median)

= \(\frac{1}{8}\) X ABC

Solution: 4. \(\frac{1}{8}\) X Area of ΔABC

5. A parallelogram, a rectangle, and a triangle stand on the same base and between the same parallel, and if their areas are P, Q, and T respectively then

 

 

1. P=R=2T
2. P=R= \(\frac{T}{2}{/latex]
3. 2P = 2R = T
4. P=R=T

Solve: Let parallelogram ABCD, rectangle ABEF and AABG are situated on same base AB and in between parallel straight lines AB and FC.

Area of parallelogram □ABCD = Area of rectangle ABEF =2 x area ΔAGB
∴ P = R = 2T

Solution: 1. P=R=2T

Question 22. Short answer type questions :

1. DE is perpendicular on side AB from point D of parallelogram ABCD and BF is perpendicular on side AD from the point B; if AB = 10 cm, AD = 8 cm, and DE = 6 cm, let us write how much is length of BF.

 

 

Solution: Area of parallelogram ABCD = Base x height = 10 x 6 sq. cm
= 60 sq. cm

∴ BF is the height of parallelogram □ABCD
∴ ADX BF = Area of parallelogram □ABCD
or, 8 x BF 60 cm

or, BF = [latex]\frac{60}{8}\) sq. cm

or, BF = 7.5 cm
∴ Length of BF = 7.5 cm

2. The area of the shape of parallelogram ABCD is 100 sq units. P is the mid-point of side BC; let us write how much is the area of the triangular field ABP.

 

 

Solution: Join, A, P, and A, C.

Area of ΔABP = Area of \(\frac{1}{2}\)× ΔABC (: AP is the median of ΔABC)

\(\frac{1}{2}\) x \(\frac{1}{2}\) Χ ABCD Area of parallelogram

= \(\frac{1}{4}\) x 100 sq.cm = 25 sq.cm

3. AD is the median of triangle ABC and P is any point on side AC in such a way that area of ΔADP area of ΔABD = 2: 3. Let us write the ratio, area of ΔPDC area of ΔABC.

 

 

Solution: Let the area of ΔADP & ΔABD be 2x sq. unit & 3x sq. unit respectively.

∵ AD is the median.

∴ Area of ΔACD = Area of ΔABD = 3x sq. cm
∴  Area of ΔPDC = Area of ΔACD = Area of ΔADP
=(3x-2x) sq.cm = x sq. cm

Area of ΔABC = Area of ΔABD + Area of ΔACD =(3x + 3x) sq.cm = 6x sq. cm
∴ Area of ΔPDC: Area of ΔABC =x:6x = 1:6

4. ABDE is a parallelogram. F is the midpoint of side ED. If the area of the triangular field ABD is 20 sq. unit, then let us write how much is an area of the triangular field AEF.

 

 

Solution: ∵ AD is the diagonal of the parallelogram ABDE.
∴ Area of ΔAED = Area of ΔABD = 20 cm

∴ In ΔAED, is the median AF,

∴  Area of ΔAEF = Area of \(\frac{1}{2}\) ΔAED

= \(\frac{1}{2}\) x 20 sq.cm = 10 sq.cm.

5. PQRS is parallelogram. X and Y are the mid-points of sides PQ and SR respectively. Construct diagonal SQ. Let us write the area of the shape of parallelogram field XQRY: area of triangular field QSR.

 

 

Solution: X & Y are the mid-points of PQ & SR respectively.

∴ Area of parallelogram □XQRY = \(\frac{1}{2}\) of Area of parallelogram PQRS

∵ SQ is the diagonal of the parallelogram PQRS.

∴ Area of ΔQSR = \(\frac{1}{2}\) x Area of the parallelogram PQRS

∴ Area of parallelogram □XQRY = Area of ΔQSR
∴ Area of the parallelogram XQRY: Area of ΔQRS = 1:1

 

 

Class IX Maths Solutions WBBSE Chapter 4 Co-ordinate Geometry: Distance Formula Let Us Work Out

Question 1. I measure the length of the straight line joining the following pairs of points

1. (18, 0); (8, 0)

Solution: Let P and Q be two points whose coordinates are (18,0) and (8,0). Points P and Q are situated in the positive direction of the x-axis.

∴ \(\overline{\mathrm{PQ}}=(\overline{\mathrm{OP}}-\overline{\mathrm{OQ}})\) = (18 – 8) unit = 10 units.

Read and Learn More WBBSE Solutions For Class 9 Maths

 

 

∴ The length of the straight line \(\overline{\mathrm{PQ}}\) = 10 units.

∴ Length of the line segment by joining two points is 10 units.

2. (0, 15); (0, 4)

Soltion: Let P and Q are two points whose coordinates are (0, 15) and (0, 4). P and Q are situated in the positive direction of the y-axis.

∴ \(\overline{\mathrm{OP}}\) =15 unit

∴ \(\overline{\mathrm{OQ}}\)= 4 unit

 

 

∴ \(\overline{\mathrm{PQ}}=(\overline{\mathrm{OP}}-\overline{\mathrm{OQ}}))\) = (15 – 4) units = 11 units.

∴ The length of the straight line PQ is 11 units.
∴ The length of the line segment formed by joining the two points is 11 units.

3. (-7, 0) (-2, 0)

Solution: Let P and Q are two points whose coordinates are (-7, 0) and (-2, 0). Points P and Q are situated in the negative direction of the x-axis.

∴ \(\overline{\mathrm{OP}}\) = 7 units

∴\(\overline{\mathrm{OQ}}\)= 2 units

 

 

∴ \(\overline{\mathrm{PQ}}=\overline{\mathrm{QP}}=\overline{\mathrm{OP}}-\overline{\mathrm{OQ}}\) = (7-2) units = 5 units

∴ Length of straight line PQ is 5 units.
∴ The length of the straight line formed by joining the two points is 5 units.

Class 9 Mathematics West Bengal Board

4. (0, -10) (0, -3)

Solution: Let P and Q are two points whose coordinates are: (0, -10) and (0, -3).

∴ \(\overline{\mathrm{OP}}\) = 10 units

∴ \(\overline{\mathrm{OQ}}OQ\) = 3 units

 

 

 

\(\overline{\dot{\mathrm{PQ}}}=\overline{\mathrm{QP}}=\overline{\mathrm{OP}}-\overline{\mathrm{OQ}}\) = (10 – 3) units = 7 units.
∴ Length of straight line PQ is 7 units.
∴ The length of the line segment formed by joining the two points is 7 units.

5. (6, 0) (-2, 0)

Solution: Let P and Q are two points whose coordinates are: (6, 0) and (-2, 0). Point P is situated in the positive direction and point Q is situated in the negative direction of x-axis.

∴ \(\overline{\mathrm{OP}}\) = 6 units

∴ \(\overline{\mathrm{OQ}}\) = 2 units

 

 

\(\overline{\mathrm{PQ}}=\overline{\mathrm{QP}}=(\overline{\mathrm{OP}}+\overline{\mathrm{OQ}})\) = (6 + 2) units = 8 units.

∴ Length of straight line PQ is 8 units.
∴ The length of the line segment formed by joining the two points is 8 units.

Class 9 Mathematics West Bengal Board

6. (0,-5) (0, 9)

Solution: Let P and Q are two points whose co-ordinates are: (0, -5) and (0, 9). Point P is situated in the negative direction and point Q is situated in the positive direction of y-axis.

∴ \(\overline{\mathrm{OP}}\) = 5 units

∴ \(\overline{\mathrm{OQ}}OQ\)=9 units

 

 

∴ \(\overline{\mathrm{PQ}}=\overline{\mathrm{QP}}=(\overline{\mathrm{OP}}+\overline{\mathrm{OQ}})\) = (5+9) units = 14 units.

∴ Length of the straight line PQ is 14 units.
∴ The length of the line segment formed by joining the two points is 14 units.

7. (5, 0): (0, 10)

Solution: Let P and Q are two points whose coordinates are (5, 0) and (0, 10)
∴OP = 5 units
OQ = 10 units

 

 

In right angled ΔPOQ from Pythagoras’ theorem,

PQ² = OP² + OQ²
or, PQ² = (5)² + (10)² sq.units
or, PQ² = 25+ 100 sq.units
or, PQ² = 125 sq.units

∴ PQ= √125 units = 5√5 units
∴ Distance between points P and Q is 5√5 units.
∴ The length of the line segment formed by joining the two points is 5√5 units.

Class 9 Mathematics West Bengal Board

8. (3,0): (0, 4)

Solution: Let P and Q be two points whose co-ordinates are (3, 0) and (0, 4).

∴ OP = 3 units
OQ = 4 units

 

 

In right-angled APOQ we get by Pythagoras’ theorem,

PQ² = OP²+OQ²
or, PQ² = (3)² + (4)² sq.units
or, PQ² = 9 + 16 sq.units
or, PQ² = 25 sq.units

∴ PQ= √25 units = 5 units
∴ Distance between points P and Q is 5 units.
Length of the straight line formed by joining the two points is 5 units.

Class 9 Maths WB Board

9. (4,3); (2, 1)

Solution: Let A and B are two points whose co-ordinates are (2, 1) and (4, 3).
In Image, OM = 2 units and AM = 1 units.
ON 4 units and BN = 3 units.

∴ AP = MN = ON-OM = (4-2) units = 2 units
Again, BP= BN-PN = (3-1) units = 2 units

 

 

In right angled Δ APB we get by Pythagoras’ theorem,

AB² = AP² + BP²
or, AB² = (2)² + (2)² sq.units
or, AB² = 4+4 sq.units
or, AB² = 8 sq.units

∴ AB = √8 units = 2√2 units
∴ Distance between the points A and B, i.e., length of the straight line formed by lining the two points is 2√2 units.

Class 9 Maths WB Board

10. (-2,-2) (2, 2)

Solution: Let A and B are two points whose co-ordinates are (-2, -2) and (2, 2).
∴ A point lies in third quadrant. B lies in first quadrant. From points A and B on axis respectively AP and BQ perpendiculars are drawn which cut x-axis at points p and Q respectively.
∴ OP 2 units and OQ = 2 units.
BQ 2 units and AP 2 units

 

 

From point A on extended-BQ a perpendicular is drawn which cuts extended BQ at point D.
∴ AD = PQ = OP + OQ
(2+2) units = 4 units
BD = BQ + QD = BQ + PA (2+2) units = 4 units

In right angled AADB we get by Pythagoras’ theorem, AB2

AD²+ BD²
or, AB² = (4)²+(4)² sq.units
or, AB² = 16+ 16 sq.units
or, AB² = 32 sq.units

∴ AB = √32 units = 4√√2 units
∴ Length of the straight line AB is 4√2 units, i.e., the length of the straight line formed by joining the two points is 4√2 units.

 

Class 9 Mathematics West Bengal Board Chapter 4 Co-ordinate Geometry: Distance Formula Exercise 4.1

 

Question 1. Let us calculate the distances of the following points from the origin

Here, distance formula = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)unit.

1. (7, -24)

Solution: The distance of the point (7,-24) from the origin is

\(\begin{aligned}
D & =\sqrt{(7-0)^2+(-24-0)^2} \text { unit } \\
& =\sqrt{49+576} \text { unit } \\
& =\sqrt{625} \text { unit } \\
& =25 \text { unit }
\end{aligned}\)

 

2. (3,-4)

Solution:The distance of the point (3, 4) from the origin (0,0) is

\(\begin{aligned}
D & =\sqrt{(3-0)^2+(-4-0)^2} \text { unit } \\
& =\sqrt{9+16} \text { unit } \\
& =\sqrt{25} \text { unit } \\
& =5 \text { unit }
\end{aligned}\)

Maths WBBSE Class 10 Solutions

3. (a+b, a-b)

Solution: The distance of the point (a+b, a-b) from the origin (0,0) is

\(\begin{aligned}
& D=\sqrt{(a+b-0)^2+(a-b-0)^2} \text { unit } \\
& =\sqrt{(a+b)^2+(a-b)^2} \text { unit } \\
& =\sqrt{2\left(a^2+b^2\right)} \text { unit }
\end{aligned}\)

 

Question 2. Let us calculate the distances between the pairs of points given below:

1. (5, 7) and (8, 3)

Solution: The distance between (5, 7) and (8, 3) is

\(\begin{aligned}
& =\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2} \text { unit } \\
& D=\sqrt{(5-8)^2+(7-3)^2} \text { unit } \\
& =\sqrt{-3^2+4^2} \text { unit } \\
& =\sqrt{9+16} \text { unit } \\
& =\sqrt{25} \text { unit } \\
& =5 \text { unit }
\end{aligned}\)

Maths WBBSE Class 10 Solutions

2. (7,0) and (2,-12)

Solution: Distance between (7, 0) and (2,-12) is

\(\begin{aligned}
& =\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2} \text { unit } \\
& D=\sqrt{(5-8)^2+(7-3)^2} \text { unit } \\
& =\sqrt{-3^2+4^2} \text { unit } \\
& =\sqrt{9+16} \text { unit } \\
& =\sqrt{25} \text { unit } \\
& =5 \text { unit }
\end{aligned}\)

 

3. \(\left(-\frac{3}{2}, 0\right)\) and (0, -2)

Solution: Distance between \(\left(-\frac{3}{2}, 0\right)\) and (0, -2)

 

\(=\sqrt{\left(\frac{-3}{2}-0\right)^2+(0+2)^2} \text { unit }\)

 

\(\begin{aligned}
& =\sqrt{\frac{9}{4}+4} \text { unit } \\
& =\sqrt{\frac{9+16}{4}} \text { unit } \\
& =\sqrt{\frac{25}{4}} \text { unit } \\
& =\frac{5}{2} \text { unit } \\
& =2.5 \text { unit }
\end{aligned}\)

Maths WBBSE Class 10 Solutions

4. (3, 6) and (-2, -6)

Solution: Distance between (3, 6) and (-2, -6) is

\(\begin{aligned}
& D=\sqrt{(3+2)^2+(6+6)^2} \text { unit } \\
& =\sqrt{25+144} \text { unit } \\
& =\sqrt{169} \text { unit } \\
& =13 \text { unit }
\end{aligned}\)

 

5. (1,-3) and (8,3)

Solution: Distance between (1,-3) and (8,3) is

\(\begin{aligned}
& =\sqrt{(1-8)^2+(-3-3)^2} \text { unit } \\
& =\sqrt{7^2+6^2} \text { unit } \\
& =\sqrt{49+36} \text { unit } \\
& =\sqrt{85} \text { unit }
\end{aligned}\)

Maths WBBSE Class 10 Solutions

6. (5,7) and (8,3)

Solution: The distance between the points (5, 7) and (8, 3) is

\(\begin{aligned}
D & =\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2} \text { unit } \\
& =\sqrt{(5-8)^2+(7-3)^2} \text { unit } \\
& =\sqrt{3^2+4^2} \text { unit } \\
& =\sqrt{9+16} \text { unit } \\
& =\sqrt{25} \text { unit } \\
& =5 \text { unit }
\end{aligned}\)

 

Question 3. Let us prove that the point (-2, -11) is equidistant from the two points (3,7) and (4,6).

Solution: The distance (D₁) between the points (-2, -11) and (-3, 7) is

\(\begin{aligned}
& =\sqrt{(-2+3)^2+(-11-7)^2} \text { unit } \\
& =\sqrt{1^2+(-18)^2} \text { unit } \\
& =\sqrt{1+324} \text { unit } \\
& =\sqrt{325} \text { unit }
\end{aligned}\)

Maths WBBSE Class 10 Solutions

And the distance (D) between the points (-2, -11) and (4, 6) is

\(\begin{aligned}
& =\sqrt{(-2-4)^2+(-11-6)^2} \text { unit } \\
& =\sqrt{(-6)^2+(-17)^2} \text { unit } \\
& =\sqrt{36+289} \text { unit } \\
& =\sqrt{325} \text { unit }
\end{aligned}\)

 

Here, D₁ = D₂
∴ The point (-2, -11) is equidistant to the two points (-3, 7) and (4, 6). Proved

Question 4. By calculation let us show that the points (7, 9), (3, -7), and C (-3, 3) are the vertices of a right-angled triangle.

Solution: Let the points be A = (7, 9), B = (3, -7), and C = (-3, 3).

 

 

\(\begin{aligned}
\text { Distance } A B & =\sqrt{(7-3)^2+(9+7)^2} \text { unit } \\
& =\sqrt{16+256} \text { unit } \\
& =\sqrt{272} \text { unit }
\end{aligned}\)

WBBSE Solution Class 10 Maths

\(\begin{aligned}
\text { Distance } B C & =\sqrt{(3+3)^2+(-7-3)^2} \text { unit } \\
& =\sqrt{36+100} \text { unit } \\
& =\sqrt{136} \text { unit }
\end{aligned}\)

 

\(\begin{aligned}
\text { Distance } A C & =\sqrt{(7+3)^2+(9-3)^2} \text { unit } \\
& =\sqrt{100+36} \text { unit } \\
& =\sqrt{136} \text { unit }
\end{aligned}\)

WBBSE Solution Class 10 Maths

Here, BC²+ CA² = 136+ 136 = 272 & AB² = 272
∴ BC²+ CA²= AB²
Δ ABC is a right angled triangle, right angle being at ∠C.

Question 5. Let us prove that in both of the following cases, the three points are the vertices of an isosceles triangle:

1. (1, 4), (4, 1), and (8, 8)
   
   Solution: Let A = (1,4); B = (4,1) & C = (8,8)

\(\begin{aligned}
\text { Distance } A B & =\sqrt{(1-4)^2+(4-1)^2} \text { unit } \\
& =\sqrt{9+9} \text { unit } \\
& =\sqrt{18} \text { unit }
\end{aligned}\)

 

\(\begin{aligned}
\text { Distance AC } & =\sqrt{(1-8)^2+(4-8)^2} \text { unit } \\
& =\sqrt{49+16} \text { unit } \\
& =\sqrt{65} \text { unit }
\end{aligned}\)

WBBSE Solution Class 10 Maths

\(\begin{aligned}
\text { Distance } B C & =\sqrt{(4-8)^2+(1-8)^2} \text { unit } \\
& =\sqrt{16+49} \text { unit }
\end{aligned}\)

∴ In ΔABC AC= BC
∴ ABC is an isosceles triangle. Proved

2. (-2,-2), (2, 2) and (4,-4)

Solution:Let P= (-2,-2), Q=2, 2) and R = (4,-4)

\(\begin{aligned}
\text { Distance } P Q & =\sqrt{(-2-2)^2+(-2-2)^2} \text { unit } \\
& =\sqrt{4^2+4^2} \text { unit } \\
& =\sqrt{16+16} \text { unit } \\
& =\sqrt{32} \text { unit }
\end{aligned}\)

 

\(\begin{aligned}
\text { Distance } Q R & =\sqrt{(2-4)^2+(2+4)^2} \text { unit } \\
& =\sqrt{(2)^2+(6)^2} \text { unit } \\
& =\sqrt{4+36} \text { unit } \\
& =\sqrt{40} \text { unit }
\end{aligned}\)

 

\(\begin{aligned}
\text { Distance PR } & =\sqrt{(-2-4)^2+(-2+4)^2} \text { unit } \\
& =\sqrt{(-6)^2+(2)^2} \text { unit } \\
& =\sqrt{36+4} \text { unit } \\
& =\sqrt{40} \text { unit }
\end{aligned}\)

WBBSE Solution Class 10 Maths

∴ In triangle PQR, QR = PR= √40
∴ ΔPQR is an isosceles triangle. Proved

Question 6. Let us prove that the three points A (3, 3), B (8,-2), and C (-2,-2) are the vertices of a right-angled triangle. Let us calculate the length of the hypotenuse of ΔABC.

Solution: Given, A (3,3); B = (8,-2); C = (-2,-2)

AB² = (38)² + (3 + 2)² = 5² +5² = 25+ 25 = 50

AC² = (3 + 2)² + (3 + 2)² = 5² +5² = 25+ 25 =50

BC² = (8+ 2)² + (-2 + 2)² = 10² +0 =100+ 0 = 100

∴ AB²+ AC² = 50+50 = 100∴ Δ ABC is right angled triangle and Hypotenuse BC= √100 = 10 units.

Question 7. Let us show by calculation that the points (2, 1), (0, 0), (-1, 2), and (1, 3) are the angular points of a square.

Solution: A = (2, 1), B = (0, 0), C (-1, 2), and D = (1, 3).

 

 

\(Distance A B=\sqrt{(2-0)^2+(1-0)^2}  units =\sqrt{4+1} units =\sqrt{5} units\)

 

\(Distance B C=\sqrt{(0+1)^2+(0-2)^2}  units =\sqrt{1+4} units =\sqrt{5} units\)

Class 10 Math Solution WBBSE Chapter 4

\(\begin{aligned}
\text { Distance CD } & =\sqrt{(-1-1)^2+(2-3)^2} \text { units } \\
& =\sqrt{4+1} \text { units } \\
& =\sqrt{5} \text { units }
\end{aligned}\)

 

\(\begin{aligned}
\text { Distance DA } & =\sqrt{(1-2)^2+(3-1)^2} \text { units } \\
& =\sqrt{1+4} \text { units } \\
& =\sqrt{5} \text { units }
\end{aligned}\)

 

\(Again, Diagonal A C=\sqrt{(2+1)^2+(1-2)^2} units =\sqrt{9+1} units =\sqrt{10} units\)

Class 10 Math Solution WBBSE Chapter 4

\(\begin{aligned}
\text { and Diagonal BD } & =\sqrt{(0-1)^2+(0-3)^2} \text { units } \\
& =\sqrt{1+9} \text { units } \\
& =\sqrt{10} \text { units }
\end{aligned}\)

 

∴ In the quadrilateral ABCD all the sides are equal and diagonals AC = BD.
∴ ABCD is a square. Proved

Question 8. Let us calculate and see that for what value of y, the distance between the two points (2,y) and (10, -9) will be 10 units.

Solution: Distance of the points (2, y) from the point (10, -9)= 10 units.

∴ \(\begin{aligned}
& \sqrt{(2-10)^2+(y+9)^2}=10 \text { (given) } \\
& \sqrt{(-8)^2+(y+9)^2}=10
\end{aligned}\)

Class 10 Math Solution WBBSE Chapter 4

or, 64 +(y+9)²= 100 (Squaring both sides)
or, (y+9) = 100 -64 =36
∴ y+9 = ±6

(1) y+9=6
y=6-9=-3 and

(2) y+9=-6
∴ y-6-9=-15

Question 9. Let us find a point on x-axis which is equidistant from the two points (3,5) and (1,3).

Hints the required point on x-axis is (x,0) = (x-3)² + (0-5)² = (x-1)² + (0-3)²

Solution: Let the point on the x-axis is (x, 0).
∴ Distance of the point (3, 5) from the point (x,0) is

\(\begin{aligned}
& =\sqrt{(x-3)^2+(0-5)^2} \text { unit } \\
& =\sqrt{(x-3)^2+5^2} \text { unit }
\end{aligned}\)

Class 10 Math Solution WBBSE Chapter 4

and the distance of the point (1,3) from the point (x, 0) is

\(\begin{aligned}
& =\sqrt{(x-1)^2+(0-3)^2} \text { unit } \\
& =\sqrt{(x-1)^2+3^2} \text { unit }
\end{aligned}\)

According to the condition, \(\sqrt{(x-3)^2+25}=\sqrt{(x-1)^2+9}\)

or,(x -3)² + 25 = (x -1 )² +9
or, (3 – x)² – (1 – x) ² = 9 – 25
or, (3-x+1-x) (3-x-1+x)=-16
or, (4-2x) x 2 = -16
or, 2(2 – x) x2 = -16
or, 2 – x = -4
or, 2 + 4 = x
∴ X = 6
∴ The required point is (6,0).

Class 10 Math Solution WBBSE Chapter 4

Question 10. Let us write by calculation whether the three points O(0, 0), A(4, 3), and B (8, 6) are collinear.

Hints: If OA + AB = OB, then they are collinear.

Solution:

 

 

\(\begin{aligned}
\text { Diștance } & O A=\sqrt{(0-4)^2+(0-3)^2} \text { unit } \\
= & \sqrt{16+9} \text { unit } \\
= & \sqrt{25} \text { unit } \\
= & 5 \text { unit }
\end{aligned}\)

 

\(\begin{aligned}
\text { Distance } O B & =\sqrt{(0-8)^2+(0-6)^2} \text { unit } \\
& =\sqrt{64+36} \text { unit } \\
& =\sqrt{100} \text { unit } \\
& =10 \text { unit }
\end{aligned}\)

Class 9 Maths WB Board

\(\begin{aligned}
\text { Distance } A B & =\sqrt{(4-8)^2+(3-6)^2} \cdot \text { unit } \\
& =\sqrt{16+9} \text { unit } \\
& =\sqrt{25} \text { unit } \\
& =5 \text { unit }
\end{aligned}\)

Here OA+ AB OB, hence they are collinear.

Question 11. Let us show that the three points (2,2), (-2,-2), and (-2√3,2√3) vertices of an equilateral triangle.

Solution: Let A (2,2); B (-2,-2) and C (-2√3, 2√3).

\(\begin{aligned}
Distance A B & =\sqrt{(2+2)^2+(2+2)^2} \text { unit } \\
& =\sqrt{16+16} \text { unit } \\
& =\sqrt{32} \text { unit }
\end{aligned}\)

 

\(\text { Distance } B C=\sqrt{(-2+2 \sqrt{3})^2+(-2-2 \sqrt{3})^2} \text { unit }\)

 

\(\begin{aligned}
& =\sqrt{(2 \sqrt{3}-2)^2+(2+2 \sqrt{3})^2} \text { unit } \\
& =\sqrt{2\left\{(2)^2+(2 \sqrt{3})^2\right\}} \text { unit } \\
& =\sqrt{2(4+12)} \text { unit } \\
& =\sqrt{32} \text { unit }
\end{aligned}\)

 

\(\text { Distance } B C=\sqrt{(-2+2 \sqrt{3})^2+(-2-2 \sqrt{3})^2} \text { unit }\) \(\begin{aligned}
& =\sqrt{(2 \sqrt{3}-2)^2+(2+2 \sqrt{3})^2 \text { unit }} \\
& =\sqrt{2\left\{(2)^2+(2 \sqrt{3})^2\right\}} \text { unit } \\
& =\sqrt{2(4+12)} \text { unit } \\
& =\sqrt{32} \text { unit }
\end{aligned}\)

∴ AB = BC = CA
∴ A, B, & C are the vertices of an equilateral triangle.

Question 12. Let us show that the points (-7, 2), (19, 8), (15, -6), and (-11, -12) form a parallelogram when they are joined orderly.

Solution: Let A = (-7,2); B = (19,8); C = (15,-6); D = (-11, -12).

 

 

\(The length of A B=\sqrt{(-7-19)^2+(2-8)^2} unit
\begin{aligned}
& =\sqrt{(-26)^2+(-6)^2} \text { unit } \\
& =\sqrt{676+36} \text { unit } \\
& =\sqrt{712} \text { unit }
\end{aligned}
\)

 

\(\begin{aligned}
\text { Length of } B C & =\sqrt{(19-15)^2+(8+6)^2} \text { unit } \\
& =\sqrt{(4)^2+(14)^2} \text { unit } \\
& =\sqrt{16+196} \text { unit } \\
& =\sqrt{212} \text { unit }
\end{aligned}\)

 

\(\begin{aligned}
\text { Length of } C D & =\sqrt{(15+11)^2+(-6+12)^2} \text { unit } \\
& =\sqrt{(26)^2+(6)^2} \text { unit } \\
& =\sqrt{26^2+6^2} \text { unit } \\
& =\sqrt{676+36} \text { unit }
& = \sqrt{712} \text { unit }
\end{aligned}\)

 

\(\begin{aligned}
\text { Length of } D A & =\sqrt{(-11+7)^2+(-12-2)^2} \text { unit } \\
& =\sqrt{(-4)^2+(14)^2} \text { unit } \\
& =\sqrt{4^2+14^2} \text { unit } \\
& =\sqrt{16+196} \text { unit } \\
& =\sqrt{212} \text { unit }
\end{aligned}\)

 

\(Length of Diagonal A C=\sqrt{(-7-15)^2+(2+6)^2} unit
\begin{aligned}
& =\sqrt{(-22)^2+8^2} \text { unit } \\
& =\sqrt{22^2+8^2} \text { unit } \\
& =\sqrt{484+64} \text { unit }
\end{aligned}\)

Class 9 Maths WB Board

\(\begin{aligned}
\text { Length of Diagonal } B D & =\sqrt{(19+11)^2+(8+12)^2}=\sqrt{(30)^2+(20)^2} \\
& =\sqrt{900+400} \text { unit } \\
& =\sqrt{1300} \text { unit }
\end{aligned}\)

 

In the quadrilateral ABCD, ABCD and BC = DA [opposite sides are equations]
Diagonal AC ≠ Diagonal BD
∴ ABCD is a parallelogram.

Question 13. Let us show that the points (2,-2), (8, 4), (5, 7), and (-1, 1) are the vertices of a rectangle.

Solution: Let P = (2,-2), Q = (8,4); R = (5,7) and S = (-1,1)
To prove the PQRS is a rectangle.

 

 

 

Here, Diagonal PR = Diagonal QS.
∴ In quadrilateral PQRS, opposite sides are equal, and diagonal PR diagonal
∴ PQRS is a rectangle.

Question 14. Let us show that the points (2,5), (5,9), (9,12), and (6,8) form a rhombus when they are joined orderly.

Solution: Let the points A (2, 5); B (5, 9), C (9, 12), and D (6, 8).

 

 

 

\(\text { Here, } \begin{aligned}
A B= & \sqrt{(2-5)^2+(5-9)^2} \text { unit } \\
& =\sqrt{9+16} \text { unit } \\
& =\sqrt{25} \text { unit } \\
& =5 \text { unit }
\end{aligned}\)

Class 9 Maths WB Board

\(\begin{aligned}
& B C=\sqrt{(5-9)^2+(9-12)^2} \text { unit } \\
&=\sqrt{16+9} \text { unit } \\
&=\sqrt{25} \text { unit } \\
&=5 \text { unit }
\end{aligned}\)

 

\(\begin{aligned}
& C D=\sqrt{(9-6)^2+(12-8)^2} \text { unit } \\
& =\sqrt{9+16} \text { unit } \\
& =\sqrt{25} \text { unit } \\
& =5 \text { unit } \\
&
\end{aligned}\)

 

\(\begin{aligned}
& D A=\sqrt{(6-2)^2+(8-5)^2} \\
&=\sqrt{16+9} \text { unit } \\
&=\sqrt{25} \text { unit } \\
&=5 \text { unit }
\end{aligned}\)

 

∴ AB BC CD = DA

\(Again, Diagonal A C=\sqrt{(2-9)^2+(5-12)^2} unit
\begin{aligned}
& =\sqrt{49+49} \text { unit } \\
& =\sqrt{98} \text { unit }
\end{aligned}
\)

 

\(\begin{aligned}
\text { Diagonal BD } & =\sqrt{(5-6)^2+(9-8)^2} \text { unit } \\
& =\sqrt{1+1} \text { unit } \\
& =\sqrt{2} \text { unit }
\end{aligned}\)

∴ Diagonals are not equal.
∴ In quadrilateral ABCD, all the sides are equal but the diagonals are not equal.
∴  ABCD is a rhombus.

Chapter 4 Co-ordinate Geometry Distance Formula  Multiple Choice Questions

1. The distance between the two points (a + b, c-d) and (a-b, c + d) is

(1)\(2 \sqrt{a^2+c^2}\)
(2)\(2 \sqrt{b^2+d^2}\)
(3)\(\sqrt{a^2+c^2}\)
(4)\(\sqrt{b^2+d^2}\)

Solution: The distance between (a + b, c-d) and (a-b, c + d)

\(\begin{aligned}
& =\sqrt{(a+b-a+b)^2+(c-d-c-d)^2} \text { unit } \\
& =\sqrt{(2 b)^2+(-2 d)^2} \text { unit } \\
& =\sqrt{4 b^2+4 d^2} \text { unit } \\
& =2 \sqrt{b^2+d^2} \text { unit }
\end{aligned}\)

∴(2)\(2 \sqrt{b^2+d^2}\)

2. If the distance between the two points (x, -7) and (3, -3) is 5 units, then the values of x are
(1)0 or 6
(2)2 or 3
(3)5 or 1
(4)-6 or 0

Solution: The distance between (x, -7) and (3,-3)

\(=\sqrt{(x-3)^2+(-7+3)^2} \text { unit }\) \(=\sqrt{(x-3)^2+16} \text { unit }\)

 

By the problem, \(\sqrt{(x-3)^2+16}=5\)

or, (x-3)²+16= (5)² (Squaring both sides)
or, (x-3)²=25-16
or, (x-3)² = 9
or, x-3 = ± 3
or, x = 3+3 or x = 3-3
or, x = 6 or x = 0

∴ (1)0 or 6

3. If the distance of the point (x, 4) from the origin is 5 units, then the values of x are
(1)±4
(2)±5
(3)±3
(4)None of these

Solution: The distance of the point (x, 4) from (0,0)

\(\begin{aligned}
& =\sqrt{\mathrm{x}^2+4^2} \text { unit } \\
& =\sqrt{\mathrm{x}^2+16} \text { unit }
\end{aligned}\)

 

By the problem, \(\sqrt{x^2+16}=5\)
or, x²+16= (5)² (Squaring both sides)
or, x²+16=25
or, x² = 25-16
or, x²= 9
or, x = +3

∴ (3)±3

4. The triangle formed by the points (3, 0), (-3, 0), and (0,3) is
(1)Equilateral
(2)Isosceles
(3)Scalene
(4)Isosceles right-angled.

Solution:  Let the co-ordinates of A, B & C be (3,0) (3,0); & (0,3) respectively.

\(\text { Here, } A B=\sqrt{(2-5)^2+(5-9)^2} \text { unit }\) \(=\sqrt{9+16} \text { unit }\) \(=\sqrt{25} \text { unit }\) \(=5 \text { unit }\)

 

\(\begin{aligned}
& =\sqrt{9+9} \text { unit } \\
& =\sqrt{18} \text { unit } \\
& =3 \sqrt{2} \text { unit }
\end{aligned}\)

 

\(\text { Length of } B C=\sqrt{(-3-0)^2+(0-3)^2} \text { unit }
& =\sqrt{9+9} \text { unit } \\
& =\sqrt{18} \text { unit } \\
& =3 \sqrt{2}\text { unit }
\end{aligned}\)

 

\(\begin{aligned}
A C & =\sqrt{(3-0)^2+(0-3)^2} \text { unit } \\
& =\sqrt{9+9} \text { unit } \\
& =\sqrt{18} \text { unit } \\
& =3 \sqrt{2} \text { unit }
\end{aligned}\)

 

∴ \(B C^2+A C^2=(\sqrt{18})^2+(\sqrt{18})^2\) = 18+ 18 = 36 = 6² = AB²

∴AC = BC and AB² = BC² + AC²
∴ The ΔABC formed by the points (3,0), (-3,0) & (0,3) is an isosceles right-angled triangle.

∴(4)Isosceles right-angled.

5. The co-ordinates of the centre of a circle are (0,0) and the co-ordinates of a point on the circumference are (3,4); the length of the radius of the circle is
(1)5 units
(2)4 units
(3)3 units
(4)None of these.

Solution: Distance between the points (0, 0) and (3, 4) is

\(\begin{aligned}
& =\sqrt{(0-3)^2+(0-4)^2} \text { units } \\
& =\sqrt{9+16} \text { units } \\
& =\sqrt{25} \text { units } \\
& =5 \text { units }
\end{aligned}\)

∴ (1)5 units

Question 16. Short answer type questions:

1. Let us write the value of y if the distance of the point (-4, y) from the origin is 5 units.

Solution:

\(Distance =\sqrt{(-4-0)^2+(0-y)^2} units
\sqrt{16+\mathrm{y}^2}=5 \text { (given) }
\)

 

 or, 16 + y² = 25
∴ y² = 9
∴ y =±3

2. Let us write the co-ordinates of a point on y-axis which is equidistant from the two points (2,3) and (-1,2).

Solution: Let the point on y-axis is (0,b).

\(\begin{aligned}
& =\sqrt{(2-0)^2+(3-b)^2} \\
& =\sqrt{(-1-0)^2+(2-b)^2} \\
& =4+(3-6)^2 \\
& =1+(2-b)^2
\end{aligned}\)

 

∴ b = 4
∴ The point is (0,4).

 

 

Class 9 Math Solution WBBSE Chapter 7 Polynomial Exercise 7

Question 1. If f(x) = x⁵ + 3x³ – 7x² + 6,h(x)=3x³-8x² + 7,g(x) = x + 1,p(x) = x⁴ – x² + 2 and q (y) = 7y³-y+ 10 then let us calculate and write what would be the following polynomials:

1. f(x)+g(x)

Solution: f(x)+g(x) =x⁵+3x³-7x²+6+x+1 =x⁵+3x³-7x²+x+7

2. f(x)-h(x)

Solution: f(x)-h(x) =x⁵+3x³-7x²+6-(3x³-8x²+7) =x⁵+3x³-7x²+6-3x³+8x²-7 = x⁵+x²-1

3. f(x) -p(x)

Solution: f(x)-p(x) =x⁵+3x³-7x²+6-(x⁴-x²+2) =x⁵+3x³-7x²+6-x⁴+x²-2 =x⁵-x⁴+3x³-6x²+4

4. f(x)+p(x)

Answer: f(x)+p(x) =x⁵+3x³-7x²+6+x⁴-x²+2 =x⁵+x⁴+3x³-8x²+8

5. p(x)+g(x)+f(x)

Solution: p(x)+g(x)+f(x) = x⁴-x²+2+x+1+x⁵+3x³-7x²+6 =x⁵+x⁴-3x³-8x²+x+7

Read and Learn More WBBSE Solutions For Class 9 Maths

6. p(x)-q(y)

Solution: p(x)-q(y) =x⁴-x²+2-(7y³-y+10) =x⁴-x²+2-7y³+y-10 =x⁴-7y³-x²+y-8

7. f(x).g(x)

Solution: f(x).g(x) = (x⁵+3x³-7x²+6)(x+1) =x⁶+3x⁴-7x³+6x+x⁵+3x³-7x²+6 =x+x⁵+3x⁴-4x³-7x²+6x+6

8. p(x).g(x)

Solution: p(x).g(x) = (x⁴-x²+2)(x+1) =x⁵-x³+2x+x⁴-x²+2 = x⁵+x⁴-x³-x²+2x+2

Ganit Prakash Class 9 Solutions Chapter 7 Polynomial Exercise 7.1

Question 1. Let us write which are the polynomials in the following algebraic expressions. Let us write the degree of each of the polynomials.

1. 2x⁶ – 4x⁵ +7x² +3

Solution: 2x⁶ – 4x⁵ +7x² +3 is a polynomial because the index of the variable is a whole number and the highest index of x is 6. So the degree of 2x⁶ – 4x⁵ +7x² +3 is 6.

2. x² + 2x^-1 +4

Solution: x² + 2x^-1 +4 is not a polynomial as the index of the variable is not a whole number.

3. \(y^3-\frac{3}{4} y+\sqrt{7}\)

Solution: \(y^3-\frac{3}{4} y+\sqrt{7}\) is a polynomial as the index of the variable is a whole number, and the highest index of y is 3, so the degree of \(y^3-\frac{3}{4} y+\sqrt{7}\) is 3.

4. \(\frac{1}{x}-x+2\)

Solution: \(\frac{1}{x}-x+2\)  is not polynomial as the index of the variable is not a whole number.

5. \(x^{51}-1\)

Solution: \(x^{51}-1\) is a polynomial as the index of the variable is a whole number, and the highest index of x is 51 so the degree of \(x^{51}-1\)  is 51.

6. \(3 \sqrt{t}+\frac{t}{27}\)

Solution: \(3 \sqrt{t}+\frac{t}{27}=t^{\frac{1}{3}}+\frac{t}{27}\) is not a polynomial as the index of the variable is not a whole number.

7. 15

Solution: 15 15.1 15.x° It is a monomial of degree 0.

8. 0

Solution: 0 Power of the polynomial is undefined.

9. \(z+\frac{3}{z}+2\)

Solution: \(z+\frac{3}{z}+2=z+3 z^{-1}+2\) is not a polynomial as the index of the variable is not a whole number.

10. y³+ 4

Solution: y3+4 is not a polynomial as the index of the variable is a whole number. So the degree of y³+ 4 is 3.

11. \(\frac{1}{\sqrt{2}} x^2-\sqrt{2} x+2\)

Solution: \(\frac{1}{\sqrt{2}} x^2-\sqrt{2} x+2\) is not a polynomial as the index of the variable is a whole number. So the degree of \(\frac{1}{\sqrt{2}} x^2-\sqrt{2} x+2\) is 2.

Question 2. In the following polynomials, let us write which are first-degree polynomials in one variable, which are second-degree polynomials in one variable, and which are third-degree polynomials in one variable.

1. 2x + 17

Solution: 2x + 17 One variable – Degree – 1

2. x³ + x² + x + 1

Solution: x³ + x² + x + 1 One variable – Degree – 3

3. – 3+2y²+5xy

Solution: – 3+2y²+5xy is Not one variable here; x & y are two variables.

4. 5-x-x³

Solution: 5-x-x³ One variable – Degree – 3

5. √2+t- t²

Solution: √2+t- t² One variable – Degree – 2

6. √5x

Solution: √5x One variable – Degree – 1

Class 9 Mathematics West Bengal Board

Question 3. Let us write the coefficients of the following polynomials according to the guidelines:

1. The co-efficient of x³ in 5x³ – 13x² + 2

Solution: In 5x³ – 13x² + 2, co-efficient of x3 is 5.

2. The coefficient of x in x²-x+2.

Solution: In x²-x+2, the co-efficient of x is 1.

3. The co-efficient of x² in 8x-19

Solution: In 8x-19= 0x² + 8x-19, co-efficient of x² is 0.

4. The co-efficient of x° in √11-3√11x+x²

Solution: In √11-3√11x+x² = √11x° -3√11x+x², co-efficient of x° is √11.

Question 4. I write the degree of each of the following polynomials :

1. x⁴ + 2x³ +x² + x

Solution: x⁴ + 2x³ +x² + x Degree – 4

2. 7x-5

Solution: 7×5 Degree – 1

3. 16

Solution: 16 = 16.1 16x° Degree – 0

4. 2-y-y³

Solution: 2-y-y³ Degree – 3

5. 7t

Solution: 7t Degree – 1

6. 5 – x² + x¹⁹

Solution: 5 – x² + x¹⁹ Degree – 19

Class 9 Mathematics West Bengal Board

Question 5. I write two separate binomials in one variable whose degrees are 17.

1. 5x¹⁷+1

Solution: 5x¹⁷+1 Binomial with two variables & degree 14.

2. 2y¹⁷-5

Solution: 2y¹⁷-5 Binomial with two variables & degree 17.

Question 6. I write two separate monomials in one variable whose degrees are 4.

1. 2x⁴

Solution: 2x⁴ Monomial with one variable & degree 4.

2. 3y⁴

Solution:: 3y⁴ Monomial with one variable & degree 4.

Question 7. I write two separate trinomials in one variable whose degrees are 3.

1.  2x³ + 3x² + 4x

Solution: 2x³ + 3x² + 4x Trinomial with one variable & degree 3.

2. y³+2y²+5

Solution: y³+2y²+5 Trinomial with one variable & degree 3.

Class 9 Math Solution WBBSE

Question 8. In the following algebraic expressions, which are polynomials in one variable, which are polynomials in two variables, and which are not polynomials – Let us write them.

1.  x² + 3x + 2

Solution: x² + 3x + 2, one variable

2. x² + y² + a²

Solution: x² + y² + a², one variable

3. y²– 4ax

Solution: y² – 4ax, one variable

4. x + y + 2

Solution: x + y +2, one variable

5. x⁸+y⁴+ x⁵y⁹

Solution: x⁸+y⁴+ x⁵y⁹, one variable

6. \(x+\frac{5}{x}\)

Solution: \(x+\frac{5}{x}=x+5 x^{-1}\) is not a polynomial as the degree of variable is not a whole number.

WBBSE Class 9 Maths Solutions Chapter 7 Polynomial Exercise 7.2

Question 1. If f(x) = x2+9x-6, then let us write by calculating the values of f(0), f(1) and f(3)

Solution: f(x) = x2+9x-6

\(\begin{aligned}
&f(0)=(0)^2+9.0-6=-6 \\
& f(1)=(1)^2+9.1-6=10-6=4 \\
& f(3)=(3)^2+9.3-6=9+27-6 \\
& =36-6=30 \\
&
\end{aligned}\)

Question 2. By calculating the following polynomials f(x), let us write the values of f(1) and f(-1).

1. f(x) = 2x³ + x² + x + 4

Solution: f(x) = 2x³ + x² + x + 4

\(\begin{aligned}
f(1) & =2(1)^3+(1)^2+1+4 \\
& =2+1+1+4=8 \\
f(-1) & =2(-1)^3+(-1)^2+(-1)+4 \\
& =-2+1-1+4 \\
& =5-3 \\
& =2
\end{aligned}\)

2. f(x) = 3x⁴– 5x³ + x² + 8

Solution: f(x) = 3x⁴– 5x³ + x² + 88

\(\begin{aligned}
f(1) & =3(1)^4-5(1)^3+(1)^2+8 \\
& =3-5+1+8 \\
& =7 \\
f(-1) & =3(-1)^4-5(-1)^3+(-1)^2+8 \\
& =3+5+1+8 \\
& =17
\end{aligned}\)

3. f(x) = 4 + 3x – x³ + 5x⁶

Solution: f(x) = 4 + 3x – x³ + 5x⁶

\(\begin{aligned}
f(1) & =4+3.1-(1)^3+5(1)^6 \\
& =4+3-1+5 \\
& =11 \\
f(-1) & =4+3(-1)-(-1)^3+5(-1)^6 \\
& =4-3+1+5 \\
& =7
\end{aligned}\)

4. f(x) = 6 + 10x – 7x²

Solution: f(x) = 6 + 10x – 7x²

\(\begin{aligned}
therefore f(1) & =6+10.1-7(1)^2 \\
& =6+10-7 \\
& =9 \\
f(-1) & =6+10(-1)-7(-1)^2 \\
& =6-10-7 \\
& =-11
\end{aligned}\)

Class 9 Maths WBBSE

Question 3. Let us check the following statements –

1. The zero of the polynomial P(x) = x – 1.

Solution: P(x)=x-1=0
∴ X = 1

2. The zero of the polynomial P(x) = 3-x is 3.

Solution: P(x)=3-x=0
∴ -x=-3 or, x = 3
∴ X=3,

3. The zero of the polynomial P(x) = 5x + 1 is

Solution: P(x)=5x+1=0
∴ 5x + 1 = 0 or, 5x = -1
∴ x = -1/5

4. The two zeroes of the polynomial P(x) = x² -9 are 3 and -3.

Solution: P(x) = x² -9=0
∴ x2-9=0 or, x2 = 9
or, x = ±3
∴ The two zeroes of the polynomial P(x) are 3 and -3.

5. The two zeroes of the polynomial P(x) = x²-5x are 0 and 5. Solve: P(x) = x²-5x=0

Solution: P(x) = x²-5x=0
∴ x² – 5x = 0 or, x(x-5)=0
or, x = 0 and, x-5=0; x = 5
∴ x = 0, 5
∴The two zeroes of the polynomial P(x) are 0 and (5).

6. The two zeroes of the polynomial P(x) = x²-2x-8 are 4 and (-2).

Solution: P(x) = x²-2x-8=0
∴ x²-2x-8=0
or, x²-4x+2x-8=0
or, x(x-4)+2(x-4)=0
or, (x-4) (x+2)=0
x = 0 and x + 2 = 0
∴ x = 4 and x = -2
∴ P(x) is a polynomial whose two zeroes are 4 and (-2).

Class 9 Maths WBBSE

Question 4. Let us determine the zeroes of the following polynomials –

1. f(x) = 2-x

Solution: f(x)=2-x … f(x) = 0
∴ 2-x=0
or, -x=-2
or, x = 2
∴ f(x) is a polynomial whose zero is 2.

2. f(x) = 7x + 2

Solution: f(x)=7x+2
∴ f(x) = 0
∴7x+2=0
or, 7x=-2
or, x = -2/7
∴ f(x) is a polynomial whose zero is -2/7

3. f(x) = x + 9

Solution: f(x)=x+9
∴f(x) = 0
∴x+9=0
or, x=-9
∴ f(x) is a polynomial whose zero is (-9).

4. f(x) = 62x

Solution: f(x) = 6-2x
∴ f(x) = 0
∴6-2x=0
or, -2x=-6
or, 2x = 6
or, x = 3
∴ f(x) is a polynomial whose zero is 3.

5. f(x) = 2x

Solution: f(x) = 2x
∴ f(x) = 0
∴2x = 0
or, x = 0
∴ f(x) is a polynomial whose zero is 0.

6. f(x) = ax + b, (a = 0)

Solution: f(x) = 0
∴ax + b = 0
∴ X = -a/b
∴ f(x) is a polynomial whose zero is (-b/a)

Class 9 Mathematics West Bengal Board Chapter 7 Polynomial Exercise 7.3

1. By applying Remainder Theorem, let us calculate and write the remainder that I shall get in each case, when x³-3x²+ 2x + 5 is divided by:

1. x-2

Solution: x-2
The zero of the linear polynomial x 2 = 0
∴ X = 2
From the Remainder Theorem, division of f(x) = x³-3x²+ 2x + 5 by (x-2) gives the remainder f(2).

∴ The required remainder = f(2) =(2)3-3 (2)² +2.2 +5 = 8-12+ 4+ 5 = 17-12 = 5

2. x+2

Solution: The zero of the linear polynomial x + 2 = 0
∴ X=-2
The required remainder = f(-2) =(-2)³-3(-2)² + 2 (-2)+5 =-8-12-4+5 =-19

3. 2x-1

Answer: The zero of the linear polynomial 2x-1=0 ∴x =1/2

\(\begin{aligned}
& =f\left(\frac{1}{2}\right) \\
& =\left(\frac{1}{2}\right)^3-3\left(\frac{1}{2}\right)^2+2 \cdot \frac{1}{2}+5 \\
& =\frac{1}{8}-\frac{3}{4}+1+5
\end{aligned}\)

 

\(\begin{aligned}
& =\frac{1}{8}-\frac{3}{4}+6 \\
& =\frac{1-6+48}{8} \\
& =\frac{43}{8} \\
& =5 \frac{3}{8}
\end{aligned}\)

 

4. 2x+1

Solution:

\(\begin{aligned}
& =f\left(-\frac{1}{2}\right) \\
& =\left(-\frac{1}{2}\right)^3-3\left(-\frac{1}{2}\right)^2+2\left(-\frac{1}{2}\right)+5 \\
& =-\frac{1}{8}-\frac{3}{4}-1+5 \\
& =\frac{-1-6-8+40}{8} \\
& =\frac{40-15}{8}=\frac{25}{8}=3 \frac{1}{8}
\end{aligned}\)

 

Question 2. By applying Remainder Theorem, let us calculate and write the remainders that I shall get when the following polynomials are divided by (x-1).

1. x³– 6x² + 13x + 60

Solution: Zero of the linear polynomial x -1 = 0
∴ X = 1
= f(1) = (1)³-6(1)² + 13.1 +60
= 1-6+13+60 =74-6=68

2. x³-3x² + 4x + 50

Solution: Let f(x) = x³-3x² + 4x + 50
= f(1) =(1)³-3(1)²+4.1 +50 = 1-3+4+ 50= 55 – 3 = 52

(3)4x³ + 4x²-x-1

Solution: Let f(x) = 4x³ + 4x²-x-1
= f(1)= 4(1)³-4(1)²-1-1= 4+4-1-1=8-2. = 6

(4)11x³– 12x² – x + 7

Solution: Let f(x) =11x³– 12x² – x + 7
= f(1) =11(1)³ – 12(1)²-1+7 11121+7 = 18 – 13= 5

Question 3. Applying Remainder Theorem, let us write the remainders, when –

1. The polynomial (x³-6x² + 9x-8) is divided by (x-3)

Solution: x-3=0
∴ X=3
(x) = x³-6x² + 9x-8

∴ Remainder = f(3) = (3)³-6(3)²+9(3)-8 =2754 +27-8 = 54-62=-8

2. The polynomial (x³ -ax² + 2x -a) is divided by (x-a).

Solution: x-a=0
∴ X = a
f(x) = x³ -ax² + 2x -a

∴ Remainder = f(a) = (a)³– a(a)² + 2.a – a = a³-a³+2a-a=a

Question 4. Applying Remainder Theorem, let us calculate whether the polynomial p(x) = 4x³+4x²-x -1 is a multiple of (2x + 1) or not.

Solution: 2x+1=0
∴ X=-1/2

∴ The zero of the linear polynomial (2x + 1) = -1/2
p(x) = 4x³+4x²-x -1
(2x+1) is a factor of P(x) if P(-1/2)=0

\(\begin{aligned}
& P\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^3+4\left(-\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)-1 \\
& =4\left(-\frac{1}{8}\right)+4\left(\frac{1}{4}\right)+\frac{1}{2}-1 \\
& =\frac{-1}{2}+1+\frac{1}{2}-1 \\
& =0
\end{aligned}\)

 

Question 5. For what value of a, the divisions of two polynomials (ax³+3x²-3) and (2x³-5x+a) by (x-4) give the same remainder – let us calculate and write it.

Solution: Let f(x) = ax³+3x²-3 and g(x) = 2x³– 5x + a.
If f(x) is divided by (x-4) the remainder is

\(\begin{aligned}
& f(4)=a(4)^3+3(4)^2-3 \\
& =64 a+48-3 \\
& =64 a+45
\end{aligned}\)

If g(x) is divided by'(x-4) the remainder is

\(\begin{aligned}
g(4) & =2(4)^3-5.4+a \\
& =128-20+a \\
& =108+a
\end{aligned}\)

∴ f(4) = g(4)
∴ 64a+45=108+a
or, 64a-a=108-45
or, 63a= 63
or, a = 1.
∴The value of a = 1

Question 6. The two polynomials x³ + 2x²-px-7 and x³ + px² – 12x + 6 are divided by (x + 1) and (x-2) respectively and if the remainder R₁ and R₂ are obtained such that 2R₁ + R₂=6, then let us calculate the value of p.

Solution: Let f(x) =x³ + 2x²-px-7
If f(x) is divided by (x + 1) the remainder is f(-1)=(-1)³ + 2(-1)²-P(-1)-7 =-1+2+P-7 =P-6
According to 1st condition, R₁ = P-6 …..(1)

Again, let g(x) = x³ + px² – 12x + 6
If g(x) is divided by (x-2) the remainder is g(2) = (2)³ + P(2)² – 12.2 +6 =8+4P-24 + 6 = 4P-10
According to 2nd condition, R₂ = 4P – 10 ………(2)

∴ 2R₁ + R₂ = 6
or, 2(P-6)+4P-10=6 or, 2P-12+ 4P-10=6
or, 6P 22=6
or, 6P=6+22
or, 6P=28

\(or, P=\frac{28}{6} or, P=\frac{14}{3}=4 \frac{2}{3}\)

∴The value of \(P=4 \frac{2}{3}\)

Question 7. The polynomial x⁴ – 2x³ + 3x²– ax + b is divided by (x-1) and (x + 1) and the remainders are 5 and 19 respectively. But if that polynomial is divided by x + 2, then what will be the remainder – Let us calculate.

Solution: Let f(x) = x⁴ – 2x³ + 3x²– ax + b
If f(x) is divided by (x-1) the remainder is f(1) =(1)⁴-2(1)³+3(1)-a.1+b = 1-2+3- a+b =2-a+b

According to 1st condition, 2-a+b=5
or,a+b=5-2
or, a – b = -3…(1)

Again, if f(x) is divided by (x + 1) the remainder is
f(-1)=(-1)⁴-2(-1)³+3(-1)²-a(-1)+b =1+2+3+a+b = a+b+6

According to 2nd condition, a+b+6=19
or, a+b=19-6
or, a+b=13 ……. (2)

from the eqation (2)/(1)
a+b=13/a – b = -3 = 2a =10

Adding, 2a = 10
or, a =10/2
or,a = 5

Putting value 4 in equation (2) we get, 5+b=13
b=13-5=8
∴ f(x) = x⁴ – 2x³ + 3x²– 5x + 8

If f(x) is divided by (x + 2) the remainder is
f(-2)=(-2)⁴ -2 (-2)³ + 3(-2)² -5(-2)+8 = =16+16 +12 + 10 + 8 = 62
The required remainder = 62.

Question 8. If \(f(x)=\frac{a(x-b)}{a-b}+\frac{b(x-a)}{b-a}\) then let us show that f(a) + f(b) = f(a + b).

Solution: \(f(x)=\frac{a(x-b)}{a-b}+\frac{b(x-a)}{b-a}\)

\(\begin{aligned}
& therefore f(a)=\frac{a(a-b)}{a-b}+\frac{b(a-a)}{b-a} \\
& =a+0 \\
& =a \\
& f(b)=\frac{a(b-b)}{a-b}+\frac{b(b-a)}{b-a} \\
& =0+b \\
& =b \\
&∴f(a)+f(b)=a+b \\
&
\end{aligned}\)

 

Again, \(\begin{aligned}
f(a+b) & =\frac{a(a+b-b)}{a-b}+\frac{b(a+b-a)}{b-a} \\
& =\frac{a^2}{a-b}+\frac{b^2}{-(a-b)} \\
& =\frac{a^2}{a-b}-\frac{b^2}{a-b} \\
& =\frac{a^2-b^2}{a-b} \\
& =\frac{(a+b)(a-b)}{(a-b)} \\
& =a+b
\end{aligned}\)

Question 9. If f(x) = ax + b and f(0) = 3, f(2) = 5, then let us determine the values of a and b.

Solution: f(x) = ax + b
∴f(0) = a.0+ b = 0 + b = b
∴b = 3
Again, f(2) = a.2 + b = 2a + b
∴2a + b = 5
or, 2a+35 [∴ b = 3]
or, 2a=5-3
or, 2a = 2
or, a = 1
∴ a = 1 and b = 3

Question 10. If f(x) = ax² + bx + c and f(0) = 2, f(1) = 1, and f(4) = 6, then let us calculate the values of a, b, and c.

Solution: f(x) = ax² + bx + c
∴ f(0) = a(0)²+ b.0+ c =0+0+c=c
∴ c = 2 ….(1)

Again, f(1)= a(1)²+b.1+ c= a+b+c
∴ a+b+c=1
or, a+b+2=1(∴c=2)
or, a+b=1-2
or, a+b=-1 ……(2)

Again, f(4) = a(4)2 + b.4 + c = 16a+ 4b + C
16a+ 4b+c=6
or, 16a+ 4b+2=6(c=2)
or, 16a+ 4b6-2
or, 16a+ 4b4
or, 4(4a + b) = 4

or, \(4 a+b=\frac{4}{4}\)

4a + b = 1 …..(3)

Divide equation (3)/(2)

\(\begin{aligned}
& 4 a+b=1 \\
& a+b=-1 \\
& (-) \quad(-) \quad(+) \\
& 3 a \quad=2
\end{aligned}\)

or, a =2/3

Putting the value of a in equation (2),

\(\frac{2}{3}+b=-1\)

or, \(b=-1-\frac{2}{3}\)

or,\(b=-\left(\frac{3+2}{3}\right)\)

or,\(b=\frac{-5}{3}\)

∴\(a=\frac{2}{3}, b=\frac{-5}{3}, c=2\)

Question 11. Multiple Choice Questions

1. Which of the following is a polynomial in one variable?

1. \(x+\frac{2}{x}+3\)

2. \(3 \sqrt{x}+\frac{2}{\sqrt{x}}+5\)

3. \(\sqrt{2} x^2-\sqrt{3} x+6\)

4. \(x^{10}+y^5+8\)

Solution: 3. \(\sqrt{2} x^2-\sqrt{3} x+6\)

2. Which of the following is a polynomial?

1. x-1

2. \(\frac{x-1}{x+1}\)

3. \(x^2-\frac{2}{x^2}+5\)

4. \(x^2+\frac{2 x^{\frac{3}{2}}}{\sqrt{x^2}}+6\)

Solution: 1. x-1

3. Which of the following is a linear polynomial?

1. x + x²
2. x + 1
3. 5x²-x+3
4. \(x+\frac{1}{x}\)

Solution:2. x + 1

4. Which of the following is a second-degree polynomial?

1. √x-4
2. x³ + x
3. x³+ 2x + 6
4. x²+ 5x + 6

Solution: 4. x²+ 5x + 6

5. The degree of the polynomial √3 is

1. 1/2
2.  2
3. 1
4. 0

Solution: 4. 0

Question 12. Short answer type questions :

1. Let us write the zero of the polynomial p(x) = 2x – 3.

Solution: 2x-3=0
∴ X =3/2
∴The zero of polynomial p(x) = 2x – 3 is 3/2.

2. If p(x) = x + 4, let us write the value of p(x) + P(-x).

Solution: p(x) = x+4
∴ p(x) + p(x) = x+4x+4=8.

3. Let us write the remainder, if the polynomial x³ + 4x² + 4x – 3 is divided by x.

Solution: Let f(x) =x³ + 4x² + 4x – 3
∴ The required remainder =f(0) = (0)³ + 4(0)²+4.0-3=-3.

4. If \((3 x-1)^7=a_7 x^7+a_6 x^6+a_5 x^5+\ldots \ldots \ldots+a_1 x+a_0\) then let us write the value of \(a_7+a_6+a_5+\ldots \ldots \ldots+a_0\) (where a₇, a₆……….a₀ are constants).

Solution: \((3 x-1)^7=a_7 x^7+a_6 x^6+a_5 x^5+\ldots \ldots \ldots+a_1 x+a_0\)

or, (3.1-1)⁷ = a₇(1)7+ a₆(1)⁶ + a₅(1)⁵+. . ..+a₁x+a₀ (Putting x = 1 on both sides)
∴ (2)⁷ = a₇+ a₆+ a₅… a₁+ a₀
∴ a₇+ a₆ + a₅ + ……. a₁+ a₀ = 128.

 

Chapter 7 Polynomial Exercise 7.4

 

Question 1. Let us calculate and write, which of the following polynomials will have a factor (x+1).

Solution: x + 10 ∴x=-1
∴ The zero of polynomial (x + 1) is – 1.

1. 2x³ + 3x² – 1

Solution: Let f(x) =2x³ + 3x² – 1
∴f(-1) =2(-1)³+3(-1)²-1 =-2+3-1=3-3 = 0
∴ The factor of 2x³ + 3x² – 1 is (x + 1).

2. x⁴ + x³– x² + 4x + 5

Solution: Let f(x) = x⁴ + x³– x² + 4x + 5
∴ f(-1)=(-1)⁴+(-1)³-(-1)² + 4(-1)+5 1-1-1-4+5 =6-6 = 0
∴ One factor of x⁴ + x³– x² + 4x + 5 is x + 1.

3. 7x³ + x² + 7x + 1

Solution: Let f(x) = 7x³ + x² + 7x + 1 =7(-1)³+(-1)²+7(-1)+1 =-7+1-7+1 = 2 – 14 = 12
∴ (x + 1) is a factor of (7x³ + x² + 7x + 1).

4. 3+3x-5x³– 5x⁴

Solution: Let f(x)=3+3x-5x³– 5x⁴
f(-1)= 3+3(-1)-5(-1)³-5(-1)⁴ = 3-3+5-5 = 8-8 = 0
∴ One factor of 3+3x-5x³– 5x⁴ is (x + 1).

5. x⁴ + x² + x +1

Solution: Let f(x) =x⁴ + x² + x +1
f(-1) = (-1)⁴+(-1)²+(-1)+1 =1+1 1+1 = 3-1 = 2
∴ (x + 1) is a factor of (x⁴ + x² + x +1).

6. x³ + X² + X +1

Solution: f(-1) = (-1)³+(-1)²+(-1)+1=-1+1+1+1=2-2 = 0
∴ One factor of x³ + X² + X +1 is (x + 1).

Question 2. By using Factor Theorem, let us write whether g(x) is a factor of the following polynomials f(x):

1. f(x) = x⁴-x²-12 and g(x) = x + 2

Solution: g(x) = x + 2 is a factor of f(x) if f(-2) = 0
∴ f(x)= x⁴-x²-12 = (-2)⁴-(-2)²-12=16- 4- 12 = 16- 16 = 0
∴ g(x) is a factor of f(x).

2. f(x) = 2x³ +9x²-11x-30 and g(x) = x + 5

Solution: x+5=0
∴ X=-5
∴ f(x) = 2x³ +9x²-11x-30
f(-5) = 2(-5)³ +9(-5)²-11x-30=250+225 +55 – 30 = 280 280= 0
∴ g(x) is a factor of f(x).

3. f(x) = 2x³ + 7x²-24x-45 and g(x) = x – 3

Solution: x-3=0
∴ X=3
∴ f(x) = 2x³ + 7x²-24x-45
∴ f(3) = 2(3)³+7(3)²-24.3-45=54+63 72-45= 117 117 = 0
∴ g(x) is a factor of f(x).

4. f(x) = 3x³ + x² – 20x + 12 and g(x) = 3x – 2

Solution:3x-2=0
∴ f(x) = 3x³ + x² – 20x + 12

\(\begin{aligned}
f\left(\frac{2}{3}\right) & =3\left(\frac{2}{3}\right)^3+\left(\frac{2}{3}\right)^2-20 \cdot \frac{2}{3}+12 \\
& =3 \cdot \frac{8}{27}+\frac{4}{9}-\frac{40}{3}+12 \\
& =\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12 \\
& =\frac{8+4-120+108}{9} \\
& =\frac{120-120}{9}=\frac{0}{9}=0
\end{aligned}\)

∴ g(x) is a factor of f(x).

Question 3. Let us calculate and write the value of k for which the polynomial 2x + 3×3+2kx2 + 3x + 6 is divided by x + 2.
Solution: Let f(x) = 2x + 3x³+2kx² + 3x + 6
The zero of (x + 2) is – 2.
∴ (x+2) is a factor of f(x).
∴ f(-2)=0
∴ 2(-2)⁴+3(-2)³+2.k.(-2)²+3(-2)+6=0
or, 32-24+8k-6+6=0
or, 8+ 8k = 0
or, 8k=-8
or, k=-1

Question 4. Let us calculate the value of k for which g(x) will be a factor of the following polynomials f(x):

1.  f(x) = 2x³ +9x²+ x + k and g(x) = x-1

Solution: g(x) = (x-1) is a factor of f(x) if x = 1 (x-1=0. x = 1)
∴ g(x) is a factor of f(x)
∴ f(1) = 0
∴ 2(1)³+9(1)²+1+k=0
or, 2+9+1+k=0
or, k=-12
∴ If k = 12, g(x) is a factor of f(x).

2. f(x)=kx²-3x+k and g(x)=x-1

Solution: g(x) = (x-1) is a factor of f(x) if x = 1 (x-1=0. x = 1)
∴g(x) is a factor of f(x)
∴f(1) = 0
∴ k(1)²-3.1+k=0
or, 2k-3=0
or, k = 3/2
∴ If k = 3/2,g(x) is a factor of f(x).

3. f(x) =2x⁴ + x³ – kx²-x+6 and g(x) = 2x-3

Solution: g(x) = 2x-3.
∴ The zero of g(x) is 3/2(2x-3=0 ∴X =3/2)
∴ g(x) is a factor of f(x)

\(\begin{aligned}
& f\left(\frac{3}{2}\right)=0 \\
&2\left(\frac{3}{2}\right)^4+\left(\frac{3}{2}\right)^3-k\left(\frac{3}{2}\right)^2-\frac{3}{2}+6=0 \\
& \text { or, } 2 \cdot \frac{81}{16}+\frac{27}{8}-\frac{9 k}{4}-\frac{3}{2}+6=0 \\
& \text { or, } \frac{81}{8}+\frac{27}{8}-\frac{9 k}{4}-\frac{3}{2}+\frac{6}{1}=0 \\
& \text { or, } \frac{81+27-18 k-12+48}{8}=0 \\
& \text { or, } 144-18 k=0 \\
& \text { or, }-18 k=-144 \\
& \text { or, } k=\frac{-144}{-18} \\
& k=8
\end{aligned}\)

∴If k= 8 then g(x) will be a footer of W

4. f(x) =2x³ + kx² + 11x+ k + 3 and g(x) = 2x – 1

Solution: The zero of polynomial g(x) = (2x-1) is 1/2(2x-1=0 ∴x = 1/2)
∴ g(x) is the factor of f(x)

\(\begin{aligned}
& f\left(\frac{1}{2}\right)=0 \\
& 2\left(\frac{1}{2}\right)^3+K\left(\frac{1}{2}\right)^2+11 \cdot \frac{1}{2}+K+3=0
\end{aligned}\)

 

\(\text { 2. } \frac{1}{8}+k \cdot \frac{1}{4}+\frac{11}{2}+k+3=0\)

 

\(\frac{1}{4}+\frac{k}{4}+\frac{11}{2}+k+3=0\)

 

\(\frac{1+k+22+4 k+12}{4}=0\)

 

or, 5k + 35 = 0
or, 5k = -35
or, k = -35/5 = -7
∴ If k=-7, g(x) is a factor of f(x).

Question 5. Let us calculate and write the values of a and b if x²-4 is a factor of the polynomial ax⁴ + 2x³-3x²+ bx-4.

Solution: x²-4=0
or, (x+2) (x-2)=0
x+2=0 or, x-2=0
Let f(x) = ax⁴ + 2x³-3x²+ bx-4

∴ (x²-4) is a factor of f(x)
∴ f(2) = 0 and f(-2) = 0
f(2) = 0
∴ a(2)⁴+2(2)³-3(2)²+ b.2-4=0
or, 16a+ 16-12+2b-4=0
or, 16a+2b=0
or, 8a+ b = 0 …..(1)

f(-2)=0
∴a(-2)⁴+2(-2)³-3(-2)²+ b.(-2)-4=0
or, 16a-16-12-2b-4=0
or, 16a-2b= 32
or,8a-b=16 ….(2)

8a+ b = 0

Divide by the equation (2)/(1)

\(\begin{aligned}
& 8 a-b=16 \\
& 8 a+b=0
\end{aligned}\)

 

or, a =16/16
or, a =1

Putting the value of a in equation (1),
8.1+b=0
or, b=-8
∴ a = 1, b=-8

Question 6. If (x + 1) and (x+2) are two factors of the polynomial x³ + 3x²+2ax + b, then let us calculate and write the values of a and b.

Solution: Let f(x) = x³ + 3x²+2ax + b
The zero of (x + 1) is -1 (x+1=0  ∴x=-1)
∵ (x + 1) is a factor of f(x)
∴ f(-1)-0
∴ (-1)³+3(-1)²+2a(-1)+b=0
or, -1+3-2a+b=0
or, 2-2a+b=0
or, 2a+b=-2
or, 2a-b=2 ….(1)

Again, the zero of (x + 2) is -2
∵ (x+2) is a factor of f(x)
∴ f(-2)=0
∴ (-2)³+3(-2)²+2a(-2)+b=0
or, -8+12-4a+b=0
or, 4-4a+b=0
or, 4a+b=-4
or, 4a-b = 4 …(2)

(2)-(1)

\(\begin{aligned}
& 4 a-b=4 \\
& 2 a-b=2 \\
& (-)(+)(-) \\
& 2 a=2
\end{aligned}\)

Subtracting, 2a=2
or, a = 1

Putting the value of in equation (1), 2.1-b=2
or, -b = 2-2
or, – b = 0
or, b = 0
∴ a = 1, b = 0.

Question 7. If the polynomial ax³ + bx²+x-6 is divided by x-2 and the remainder is 4, then et us calculate the values of a and b when x + 2 is a factor of this polynomial.

Solution: Let f(x) = ax³ + bx²+x-6
∴ If f(x) is divided by (x-2), the remainder is 4
∴f(2) = 4[ The zero of (x-2) is 2]
or, 8a+ 4b-4 = 4
or, 4(2a + b) = 4+4
or, (2a + b) = 8/4
or, 2a + b = 2

Again, (x+2) is a factor of f(x)
∴ f(-2) = 0 (∵x+2=0 ∴x=-2)
∴ a(-2)³+(-2)2+(-2)-6-0
or, -8a+ 4b-2-6=0
or, -8a+ 4b = 8
or, -4(2a – b) = 8

or, \(2 a-b=\frac{8}{-4}\)

or, 24-b =-2 ….(2)

Adding, (2)+(1)

\(\begin{aligned}
& 2 a-b=-2 \\
& 2 a+b=2 \\
& 4 a=0
\end{aligned}\)

or, a = 0
∴ 2.0 + b = 2
or, 0+b=2
or, b = 2
∴ a = 0, b = 2

Question 8. Let us show that if n is any positive integer (even or odd), x-y is a factor of the polynomial xⁿ – yⁿ

Solution: x – y = 0
∴ x = y
Let f(x) = xⁿ – yⁿ
∴f(y) = yⁿ – yⁿ= 0
(x-y) is a factor of xⁿ – yⁿ.

Question 9. Let us show that if n is any positive odd integer, then x + y is a factor of xⁿ + yⁿ.

Solution: x + y = 0
∴ x = – Y
Let f(x) = xⁿ + yⁿ
∴ f(-y)= (-y)ⁿ+ yⁿ =-yⁿ+ yⁿ= 0
∴ (x + y) is a factor of xⁿ + yⁿ.

Question 10. Let us show that if n be any positive integer (even or odd), then x – y will never be a factor of the polynomial xⁿ + yⁿ

Solution: x-y=0
∴ x = y
Let f(x) = xⁿ + yⁿ
f(y) = (y)ⁿ+ yⁿ
f(y) = 2yⁿ ≠0
∴ (x-y) can never be a factor of f(x) = (xⁿ + yⁿ).

Question 11. Multiple Choice Questions

1. x³ + 6x² + 4x + k (x + 2)

(1)-6
(2)-7
(3)-8
(4) 10

Solution: Let f(x) = x³ + 6x² + 4x + k
The zero of (x+2) is -2
∴ (x+2) is a factor of f(x)
∴ f(-2) = 0
∴ (-2)³+6(-2)²+4(-2)+k=0
or, 8+24-8+k=0
or, 8+ k = 0.
or, k=-8

∴(3) – 8

2. In the polynomial f(x) if \(f\left(-\frac{1}{2}\right)=0\)  then a factor of f(x) will be

(1) 2x-1
(2) 2x + 1
(3) x-1
(4) x + 1

Solution: \(f\left(-\frac{1}{2}\right)=0\)

∴X =-1/2
or, 2x = -1
or, 2x+1=0
∴(2x+1) is a factor of f(x).

∴ (2)2x + 1

3. (x-1) is a factor of the polynomial f(x) but it is not a factor of g(x). So (x-1) will be a factor of

(1)f(x) g(x)
(2)- f(x) + g(x)
(3)f(x) = g(x)
(4){f(x) + g(x)}g(x)

Solution: (x-1) is a factor of f(x) g(x)
∴ (1)f(x) g(x)

4. (x + 1) is a factor of the polynomial xⁿ + 1 when

(1) n is a positive odd integer
(2) n is a positive even integer
(3) n is a negative integer
(4) n is a positive integer

Solution: We know if n is an odd positive integer, (x+y) is a factor of (xⁿ + yⁿ )
∴ (x + 1) is a factor of xⁿ+ 1, i.e., xⁿ+1ⁿ when n is an odd positive integer.

∴(1) n is a positive odd integer

5. If n²-1 is a factor of the polynomial an⁴ + bn³ + cn² + dn + e

(1) a +c+e=b+d
(2) a+b+e=c+d
(3) a+b+c=d+e
(4) b+c+d=a+e

Solution: Let f(n) = an⁴ + bn³ + cn² + dn + e
Factor of 1(n) = n²-1-(n + 1) (n-1)
∴ f(-1)=0 and f(1) = 0 When f(-1) = 0.

or, a(-1)⁴+b(-1)³+c(-1)² +d(-1)+e=0
or, a-b+c-d+e=0
∴ a+c+e=b+d

Again, f(1)=0
∴ a(1)⁴+b(1)³+c(1)²+d.1+e=0
or, a+b+c+d+e=0
∴ a+c+e-b-d

∴(1) a +c+e=b+d

Question 12. Short answer type questions:

1. Let us calculate and write the value of a for which x + a will be a factor of the polynomial x³+ax²-2x+a-12.

Solution: Let f(x) = x³+ax²-2x+a-12
Factor of f(x) is (x + a)
∴f(-a) = 0
∴(-a)³ + a(-a)²-2(-a) + a-12=0
or,(-a)³+ a³+2a + a-12=0
or, 3a= 12
or, a = 12/3
or, a = 4

2. Let us calculate and write the value of k for which x-3 will be a factor of the polynomial k²x³-kx²+3kx-k.

Solution: Let f(x) = k²x³-kx²+3kx-k
Factor of f(x) is (x-3)
∴f(3) = 0
∴k²(3)³-k(3)²+3k.3-k=0
or, 27k²-9k+9k – k=0
or, 27k²-k=0
or, k(27k-1)=0 = k = 0 and 27k10.
∴ k = 1/27
∴ k = 0, 1/27

3. Let us write the value of f(x) + f(-x) when f(x) = 2x + 5.

Solution: f(x) = 2x + 5
∴ f(x) + f(x) = 2x+5+2(-x)+5 =2x+5-2x+5=10

4. Both (x-2) and \(\left(x-\frac{1}{2}\right)\) are factors of the polynomial px² + 5x + r, let us calculate and write the relation between p and r.

Solution: Let f(x) = px² + 5x + r
Factor of f(x) is (x-2)
∴ f(2) = 0 (x-2=0; x=2) .
∴ p(2)²+ 5.2 + r=0
or, 4p+r+10=0
or, 4p+r=-10  ….(1)

Again, factor of f(x) is (x) is \(\left(x-\frac{1}{2}\right)\)

\(\begin{aligned}
&  f\left(\frac{1}{2}\right)=0\left(because x-\frac{1}{2}=0 therefore x=\frac{1}{2}\right) \\
& p\left(\frac{1}{2}\right)^2+5 \cdot \frac{1}{2}+r=0 \\
& \text { or, } \frac{p}{4}+\frac{5}{2}+r=0 \\
& \text { or, } \frac{p+10+4 r}{4}=0
\end{aligned}\)

or, p +4r + 10 = 0

or, p + 4r = -10 .,…..(2)

(1) and (2)

4p + r = p + 4r
or, 4p – p = 4r -r
or, 3p = 3r
or, p = r

5. Let us write the roots of the linear polynomial f(x) = 2x + 3. Solve: 2x+3=0

Solution: 2x=-3
or, 2x = -3
or, x = -3/2
∴ The roots of f(x) = 2x + 3 = -3/2

 

Class IX Maths Solutions WBBSE Chapter 8 Factorisation: Formulae of Factorisation

1. a² – b² = (a + b) (a – b)
2. a³+b³ = (a + b) (a² – ab + b²)
3. a³-b³ (a – b) (a² + ab + b²)
4. a³ + b + c³-3abc = (a+b+c) (a²+ b²+c²-ab-bc-ca)

Read and Learn More WBBSE Solutions For Class 9 Maths

Ganit Prakash Class 9 Solutions Chapter 8 Factorisation Exercise 8.1

 

Let us factorise the following polynomials

Question 1. x³-3x+2

Solution: x³-3x+2

\(\begin{aligned}
& =x^3-1-3 x+3 \\
& =(x)^3-(1)^3-3(x-1) \\
& =(x-1)\left(x^2+x+1\right)-3(x-1) \\
& =(x-1)\left(x^2+x+1-3\right) \\
& =(x-1)\left(x^2+x-2\right) \\
& =(x-1)\left(x^2+2 x-x-2\right) \\
& =(x-1)\{x(x+2)-1(x+2)\} \\
& =(x-1)(x+2)(x-1) \\
& =(x-1)^2(x+2)
\end{aligned}\)

 

Question 2. x³ + 2x + 3

Solution: x³ + 2x + 3

\(\begin{aligned}
& =x^3+1+2 x+2 \\
& =(x)^3+(1)^3+2(x+1) \\
& =(x+1)\left(x^2-x+1\right)+2(x+1) \\
& =(x+1)\left(x^2-x+1+2\right) \\
& =(x+1)\left(x^2-x+3\right)
\end{aligned}\)

Ganit Prakash Class 9 Solutions

Question 3.a³-12a-16

Solution: a³-12a-16

\(\begin{aligned}
& =a^3+8-12 a-24 \\
& =(a)^3+(2)^3-12(a+2) \\
& =(a+2)\left(a^2-2 a+4\right)-12(a+2) \\
& =(a+2)\left(a^2-2 a+4-12\right) \\
& =(a+2)\left(a^2-2 a-8\right) \\
& =(a+2)\left(a^2-4 a+2 a-8\right) \\
& =(a+2)\{a(a-4)+2(a-4)\} \\
& =(a+2)(a+2)(a-4) \\
& =(a+2)^2(a-4)
\end{aligned}\)

 

Question 4. x³– 6x + 4

Solution: x³– 6x + 4

\(\begin{aligned}
& =x^3-8-6 x+12 \\
& =(x)^3-(2)^3-6(x-2) \\
& =(x-2)\left(x^2+2 x+4\right)-6(x-2) \\
& =(x-2)\left(x^2+2 x+4-6\right) \\
& =(x-2)\left(x^2+2 x-2\right)
\end{aligned}\)

Ganit Prakash Class 9 Solutions

Question 5. x³-19x-30

Solution: x³-19x-30

\(\begin{aligned}
& =x^3+8-19 x-38 \\
& =(x)^3+(2)^3-19(x+2) \\
& =(x+2)\left(x^2-2 x+4\right)-19(x+2) \\
& =(x+2)\left(x^2-2 x+4-19\right) \\
& =(x+2)\left(x^2-2 x-15\right) \\
& =(x+2)\left(x^2-5 x+3 x-15\right) \\
& =(x+2)\{x(x-5)+3(x-5)\} \\
& =(x+2)(x+3)(x-5)
\end{aligned}\)

 

Question 6. 4a³-9a²+ 3a +2

Solution: 4a³-9a²+ 3a +2
If a =1, 4a³-9a²+ 3a +2=0
∴ (a-1) is a factor of 4a³-9a²+ 3a +2

\(\begin{aligned}
& =4 a^3-9 a^2+3 a+2 \\
& =4 a^3-4 a^2-5 a^2+5 a-2 a+2 \\
& =4 a^2(a-1)-5 a(a-1)-2(a-1) \\
& =(a-1)\left(4 a^2-5 a-2\right)
\end{aligned}\)

Ganit Prakash Class 9 Solutions

Question 7. x³-9x²+23x-15

Solution: If x = 1, x³-9x²+23x-15=0
∴ (x-1) is a factor of  x³-9x²+23x-15

\(\begin{aligned}
& =x^3-x^2-8 x^2+8 x+15 x-15 \\
& =x^2(x-1)-8 x(x-1)+15(x-1) \\
& =(x-1)\left(x^2-8 x+15\right) \\
& =(x-1)\left(x^2-5 x-3 x+15\right) \\
& =(x-1)\{x(x-5)-3(x-5)\} \\
& =(x-1)(x-3)(x-5)
\end{aligned}\)

 

Question 8. 5a³ +11a²+ 4a-2

Solution: If a=-1,5a³ +11a²+ 4a-2=0
∴ (a + 1) is a factor of 5a³ +11a²+ 4a-2

\(\begin{aligned}
& =5 a^3+5 a^2+6 a^2+6 a-2 a-2 \\
& =5 a^2(a+1)+6 a(a+1)-2(a+1) \\
& =(a+1)\left(5 a^2+6 a-2\right)
\end{aligned}\)

Class 9 Math Solution WBBSE In English

Question 9. 2x³– x² + 9x + 5

Solution: If x = -1/2, 2x³– x² + 9x + 5=0
∴(2x + 1) is a factor of 2x³– x² + 9x + 5

\(\begin{aligned}
& =2 x^3+x^2-2 x^2-x+10 x+5 \\
& =x^2(2 x+1)-x(2 x+1)+5(2 x+1) \\
& =(2 x+1)\left(x^2-x+5\right)
\end{aligned}\)

 

Question 10. 2y³-5y²-19y+ 42

Solution: Putting, y = 2 then 2y³-5y²-19y+ 42= 0
∴ (y-2) is a factor of 2y³-5y²-19y+ 42

\(\begin{aligned}
& =2 y^3-4 y^2-y^2+2 y-21 y+42 \\
& =2 y^2(y-2)-y(y-2)-21(y-2) \\
& =(y-2)\left(2 y^2-y-21\right) \\
& =(y-2)\left(2 y^2-7 y+6 y-21\right) \\
& =(y-2)\{y(2 y-7)+3(2 y-7)\} \\
& =(y-2)(y+3)(2 y-7)
\end{aligned}\)

Class 9 Math Solution WBBSE In English Chapter 8 Factorisation Exercise 8.2

Question 1. \(\frac{x^4}{16}-\frac{y^4}{81}\)

Solution: \(\frac{x^4}{16}-\frac{y^4}{81}\)

\(\begin{aligned}
& =\left(\frac{x^2}{4}\right)^2-\left(\frac{y^2}{9}\right)^2 \\
& =\left(\frac{x^2}{4}+\frac{y^2}{9}\right)\left(\frac{x^2}{4}-\frac{y^2}{9}\right) \\
& =\left(\frac{x^2}{4}+\frac{y^2}{9}\right)\left\{\left(\frac{x}{2}\right)^2-\left(\frac{y}{3}\right)^2\right\} \\
& =\left(\frac{x^2}{4}+\frac{y^2}{9}\right)\left(\frac{x}{2}+\frac{y}{3}\right)\left(\frac{x}{2}-\frac{y}{3}\right)
\end{aligned}\)

 

Question 2. \(m^2+\frac{1}{m^2}+2-2 m-\frac{2}{m}\)

Solution: \(\begin{aligned}
& \left(m^2+\frac{1}{m^2}+2-2 m-\frac{2}{m}\right. \\
& (m)^2+\left(\frac{1}{m}\right)^2+2-2\left(m+\frac{1}{m}\right)
\end{aligned}\)

\(\begin{aligned}
& =\left(m+\frac{1}{m}\right)^2-2 m \cdot \frac{1}{m}+2-2\left(m+\frac{1}{m}\right) \\
& =\left(m+\frac{1}{m}\right)^2-2+2-2\left(m+\frac{1}{m}\right) \\
& =\left(m+\frac{1}{m}\right)^2-2\left(m+\frac{1}{m}\right) \\
& =\left(m+\frac{1}{m}\right)\left(m+\frac{1}{m}-2\right)
\end{aligned}\)

 

Question 3. \(9 p^2-24 p q+16 q^2+3 a p-4 a q\)

Solution: \(9 p^2-24 p q+16 q^2+3 a p-4 a q\)

\(\begin{aligned}
& =(3 p)^2-2.3 p .4 q+(4 q)^2+a(3 p-4 q) \\
& =(3 p-4 q)^2+a(3 p-4 q) \\
& =(3 p-4 q)(3 p-4 q+a)
\end{aligned}\)

Class 9 Maths WBBSE

Question 4. 4x⁴+81

Solution: 4x⁴+81

\(\begin{aligned}
& =\left(2 x^2\right)^2+(9)^2 \\
& =\left(2 x^2+9\right)^2-2.2 x^2 .9 \\
& =\left(2 x^2+9\right)^2-36 x^2 \\
& =\left(2 x^2+9\right)^2-(6 x)^2 \\
& =\left(2 x^2+9+6 x\right)\left(2 x^2+9-6 x\right) \\
& =\left(2 x^2+6 x+9\right)\left(2 x^2-6 x+9\right)
\end{aligned}\)

 

Question 5. x⁴ – 7x² + 1

Solution: x⁴ – 7x² + 1

\(\begin{aligned}
& =\left(x^2\right)^2+(1)^2-7 x^2 \\
& =\left(x^2+1\right)^2-2 \cdot x^2 \cdot 1-7 x^2 \\
& =\left(x^2+1\right)^2-9 x^2 \\
& =\left(x^2+1\right)^2-(3 x)^2 \\
& =\left(x^2+1+3 x\right)\left(x^2+1-3 x\right) \\
& =\left(x^2+3 x+1\right)\left(x^2-3 x+1\right)
\end{aligned}\)

 

Question 6. p⁴-11p²q²+q⁴

Solution: p⁴-11p²q²+q⁴

\(\begin{aligned}
& =\left(p^2\right)^2+\left(q^2\right)^2-11 p^2 q^2 \\
& =\left(p^2-q^2\right)^2+2 \cdot p^2 \cdot q^2-11 p^2 q^2 \\
& =\left(p^2-q^2\right)^2-9 p^2 q^2 \\
& =\left(p^2-q^2\right)^2-(3 p q)^2 \\
& =\left(p^2-q^2+3 p q\right)\left(p^2-q^2-3 p q\right) \\
& =\left(p^2+3 p q-q^2\right)\left(p^2-3 p q-q^2\right)
\end{aligned}\)

Class 9 Maths WBBSE

Question 7. a² + b² -c² -2ab

Solution: a² + b² -c² -2ab

\(\begin{aligned}
& =a^2-2 a b+b^2-c^2 \\
& =(a-b)^2-(c)^2 \\
& =(a-b+c)(a-b-c)
\end{aligned}\)

 

Question 8. 3a(3a+2c) – 4b(b + c)

Solution: 3a(3a+2c) – 4b(b + c) = 9a2+6ac4b2-4bc

\(\begin{aligned}
& =9 a^2+6 a c-4 b^2-4 b c \\
& =9 a^2-4 b^2+6 a c-4 b c \\
& =(3 a)^2-(2 b)^2+2 c(3 a-2 b) \\
& =(3 a+2 b)(3 a-2 b)+2 c(3 a-3 b) \\
& =(3 a-2 b)(3 a+2 b+2 c)
\end{aligned}\)

 

Question 9. a²-6ab + 12bc-4c²

Solution: a²-6ab + 12bc-4c²

\(\begin{aligned}
& =(a)^2-(2 c)^2-6 a b+12 b c \\
& =(a+2 c)(a-2 c)-6 b(a-2 c) \\
& =(a-2 c)(a+2 c-6 b) \\
& =(a-2 c)(a-6 b+2 c)
\end{aligned}\)

Class 9 Maths WBBSE

Question 10. 3a²+4ab+b²-2ac- c²

Solution: 3a²+4ab+b²-2ac- c²

\(\begin{aligned}
& =4 a^2+4 a b+b^2-a^2-2 a c-c^2 \\
& =(2 a)^2+2 \cdot 2 a \cdot b+(b)^2-\left(a^2+2 a c+c^2\right) \\
& =(2 a+b)^2-(a+c)^2 \\
& =(2 a+b+a+c)(2 a+b-a-c) \\
& =(3 a+b+c)(a+b-c)
\end{aligned}\)

Class 9 Maths WBBSE

Question 11. x² -y²-6ax + 2ay + 8a²

Solution: x² -y²-6ax + 2ay + 8a²

\(\begin{aligned}
& =x^2-6 a x+9 a^2-a^2+2 a y-y^2 \\
& =(x)^2-2 \cdot x \cdot 3 a+(3 a)^2-\left(a^2-2 a y+y^2\right) \\
& =(x-3 a)^2-(a-y)^2 \\
& =(x-3 a+a-y)(x-3 a-a+y) \\
& =(x-2 a-y)(x-4 a+y)
\end{aligned}\)

 

Question 12. a²-9b²+4c²-25d²-4ac+30bd

Solution: a²-9b²+4c²-25d²-4ac+30bd

\(\begin{aligned}
& =a^2-4 a c+4 c^2-9 b^2+30 b d-25 d^2 \\
& =(a)^2-2 \cdot a \cdot 2 c+(2 c)^2-\left(9 b^2-30 b d+25 d^2\right) \\
& =(a-2 c)^2-\left\{(3 b)^2-2 \cdot 3 b \cdot 5 d+(5 d)^2\right\} \\
& =(a-2 c)^2-(3 b-5 d)^2 \\
& =(a-2 c+3 b-5 d)(a-2 c-3 b+5 d) \\
& =(a+3 b-2 c-5 d)(a-3 b-2 c+5 d)
\end{aligned}\)

WBBSE Class 9 Maths Solutions

Question 13. 3a²b²-c²+2ab-2bc + 2ca

Solution: 3a²b²-c²+2ab-2bc + 2ca

\(\begin{aligned}
& =3 a^2-a b-a c+3 a b-b^2-b c+3 a c-b c-c^2 \\
& =a(3 a-b-c)+b(3 a-b-c)+c(3 a-b-c) \\
& =(3 a-b-c)(a+b+c)
\end{aligned}\)

 

Question 14. x² -2x-22499

Solution: x²-2x-22499

\(\begin{aligned}
& =x^2-(151-149) x-22499 \\
& =x^2-151 x+149 x-22499 \\
& =x(x-151)+149(x-151) \\
& =(x-151)(x+149)
\end{aligned}\)

WBBSE Class 9 Maths Solutions

Question 15. (x² – y²) (a² – b²) + 4abxy

Solution: (x² – y²) (a² – b²) + 4abxy

\(\begin{aligned}
& =a^2 x^2-b^2 x^2-a^2 y^2+b^2 y^2+4 a b x y \\
& =a^2 x^2+2 a b x y+b^2 y^2-b^2 x^2+2 a b x y-a^2 y^2 \\
& =(a x)^2+2 \cdot a x \cdot b y+(b y)^2-\left(b^2 x^2-2 a b x y+a^2 y^2\right) \\
& =(a x+b y)^2-(b x-a y)^2 \\
& =(a x+b y+b x-a y)(a x+b y-b x+a y)
\end{aligned}\)

WBBSE Class 9 Maths Solutions Chapter 8 Factorisation Exercise 8.3

Question 1. t⁹-512

Solution: t⁹-512

\(\begin{aligned}
& =\left(t^3\right)^3-(8)^3 \\
& =\left(t^3-8\right)\left\{\left(t^3\right)^2+t^3 .8+(8)^2\right\} \\
& =\left\{(t)^3-(2)^3\right\}\left(t^6+8 t^3+64\right) \\
& =(t-2)\left(t^2+2 t+4\right)\left(t^6+8 t^3+64\right)
\end{aligned}\)

Question 2. 729p⁶-q⁶

Solution: 729p⁶-q⁶

\(\begin{aligned}
& =\left(27 p^3\right)^2-\left(q^3\right)^2 \\
& =\left(27 p^3+q^3\right)\left(27 p^3-q^3\right) \\
& =\left\{(3 p)^3+(q)^3\right\}\left\{(3 p)^3-(q)^3\right\} \\
& =(3 p+q)\left\{(3 p)^2-3 p \cdot q+(q)^2\right\}(3 p-q)\left\{(3 p)^2+3 p . q+(q)^2\right\} \\
& =(3 p+q)\left(9 p^2-3 p q+q^2\right)(3 p-q)\left(9 p^2+3 p q+q^2\right) \\
& =(3 p+q)(3 p-q)\left(9 p^2-3 p q+q^2\right)\left(9 p^2+3 p q+q^2\right)
\end{aligned}\)

Question 3. 8(p-3)³+343

Solution: 8(p-3)³+343

\(\begin{aligned}
& =\left\{2(p-3\}^3+(7)^3\right. \\
& =(2 p-6)^3+(7)^3 \\
& =(2 p-6+7)\left\{(2 p-6)^2-(2 p-6) 7+(7)^2\right\} \\
& =(2 p+1)\left\{(2 p)^2-2.2 p \cdot 6+(6)^2-14 p+42+49\right\} \\
& =(2 p+1)\left(4 p^2-24 p+36-14 p+42+49\right) \\
& =(2 p+1)\left(4 p^2-38 p+127\right)
\end{aligned}\)

Question 4. \(\frac{1}{8 a^3}+\frac{8}{b^3}\)

Solution: \(\frac{1}{8 a^3}+\frac{8}{b^3}\)

\(\begin{aligned}
& =\left(\frac{1}{2 a}\right)^3+\left(\frac{2}{b}\right)^3 \\
& =\left(\frac{1}{2 a}+\frac{2}{b}\right)\left\{\left(\frac{1}{2 a}\right)^2-\frac{1}{2 a} \cdot \frac{2}{b}+\left(\frac{2}{b}\right)^2\right\} \\
& =\left(\frac{1}{2 a}+\frac{2}{b}\right)\left(\frac{1}{4 a^2}-\frac{1}{a b}+\frac{4}{b^2}\right)
\end{aligned}\)

Question 5.(2a³-b³)³-b⁹

Solution: (2a³-b³)³-b⁹

\(\begin{aligned}
& =\left(2 a^3-b^3\right)^3-\left(b^3\right)^3 \\
& =\left(2 a^3-b^3-b^3\right)\left\{\left(2 a^3-b^3\right)^2+\left(2 a^3-b^3\right) b^3+\left(b^3\right)^2\right\} \\
& =\left(2 a^3-2 b^3\right)\left\{\left(2 a^3\right)^2-2 \cdot 2 a^3 \cdot b^3+\left(b^3\right)^2+2 a^3 b^3-b^6+b^6\right\} \\
& =2\left(a^3-b^3\right)\left(4 a^6-4 a^3 b^3+b^6+2 a^3 b^3\right) \\
& =2(a-b)\left(a^2+a b+b^2\right)\left(4 a^6-2 a^3 b^3+b^6\right)
\end{aligned}\)

Question 6. AR³-Ar³+ AR²h – Ar²h

Solution: AR³-Ar³+ AR²h – Ar²h

\(\begin{aligned}
& =A\left(R^3-r^3\right)+A h\left(R^2-r^2\right) \\
& =A(R-r)\left(R^2+R r+r^2\right)+A h(R+r)(R-r) \\
& =A(R-r)\left\{R^2+R r+r^2+h(R+r)\right\} \\
& =A(R-r)\left(R^2+R r+r^2+h R+h r\right)
\end{aligned}\)

Question 7. \(a^3+3 a^2 b+3 a b^2+b^3-8\)

Solution:

\(\begin{aligned}
&a^3+3 a^2 b+3 a b^2+b^3-8 \\
& =(a+b)^3-(2)^3 \\
& =(a+b-2)\left\{(a+b)^2+(a+b) \cdot 2+(2)^2\right\} \\
& =(a+b-2)\left(a^2+2 a b+b^2+2 a+2 b+4\right)
\end{aligned}\)

Question 8. \(32 x^4-500 x\)

Solution:

\(\begin{aligned}
& 32 x^4-500 x \\
& =4 x\left(8 x^3-125\right) \\
& =4 x\left\{(2 x)^3-(5)^3\right\} \\
& =4 x(2 x-5)\left\{(2 x)^2+2 x .5+(5)^2\right\} \\
& =4 x(2 x-5)\left(4 x^2+10 x+25\right)
\end{aligned}\)

Class 9 Mathematics West Bengal Board

Question 9. \(8 a^3-b^3-4 a x+2 b x\)

Solution:

\(\begin{aligned}
&8 a^3-b^3-4 a x+2 b x \\
& =(2 a)^3-(b)^3-2 x(2 a-b) \\
& =(2 a-b)\left\{(2 a)^2+2 a \cdot b+(b)^2\right\}-2 x(2 a-b) \\
& =(2 a-b)\left(4 a^2+2 a b+b^2-2 x\right)
\end{aligned}\)

 

Question 10. \(x^3-6 x^2+12 x-35\)

Solution:

\(\begin{aligned}
&x^3-6 x^2+12 x-35 \\
& =(x)^3-3 \cdot x^2 \cdot 2+3 \cdot x \cdot(2)^2-(2)^3-27 \\
& =(x-2)^3-(3)^3 \\
& =(x-2-3)\left\{(x-2)^2+(x-2) 3+(3)^2\right\} \\
& =(x-5)\left(x^2-4 x+4+3 x-6+9\right) \\
& =(x-5)\left(x^2-x+7\right)
\end{aligned}\)

Class 9 Mathematics West Bengal Board Chapter 8 Factorisation Exercise 8.4

 

Question 1. \(8 x^3-y^3+1+6 x y\)

Solution:

\(\begin{aligned}
& =(2 x)^3+(-y)^3+(1)^3-3.2 x(-y) \cdot 1 \\
& =(2 x-y+1)\left\{(2 x)^2+(-y)^2+(1)^2-2 x(-y)-(-y) \cdot 1-1.2 x\right\} \\
& =(2 x-y+1)\left(4 x^2+y^2+1+2 x y+y-2 x\right) \\
& =(2 x-y+1)\left(4 x^2+y^2+1-2 x+y+2 x y\right)
\end{aligned}\)

 

Question 2. \(8 a^3-27 b^3-1-18 a b\)

Solution:

\(\begin{aligned}
& 8 a^3-27 b^5-1-18 a d \\
& =(2 a)^3+(-3 b)^3+(-1)^3-3 \cdot 2 a \cdot(-3 b)(-1) \\
& =(2 a-3 b-1)\left\{(2 a)^2+(-3 b)^2+(-1)^2-2 a(-3 b)-(-3 b)(-1)-(-1) \cdot 2 a\right\} \\
& =(2 a-3 b-1)\left(4 a^2+9 b^2+1+6 a b-3 b+2 a\right) \\
& =(2 a-3 b-1)\left(4 a^2+9 b^2+1+2 a-3 b+6 a b\right) \\
&
\end{aligned}\)

Class 9 Mathematics West Bengal Board

Question 3. \(1+8 x^3+18 x y-27 y^3\)

Solution:

\(\begin{aligned}
& =8 x^3-27 y^3+1+18 x y \\
& =(2 x)^3+(-3 y)^3+(1)^3-3 \cdot 2 x(-3 y) .1 \\
& =(2 x-3 y+1)\left\{(2 x)^2+(-3 y)^2+(1)^2-2 x \cdot(-3 y)-(-3 y) .1-1.2 x\right\} \\
& =(2 x-3 y+1)\left(4 x^2+9 y^2+1+6 x y+3 y-2 x\right) \\
& =(2 x-3 y+1)\left(4 x^2+9 y^2+1-2 x+3 y+6 x y\right)
\end{aligned}\)

 

Question 4. \(x^3+y^3-12 x y+64\)

Solution:

\(\begin{aligned}
& =x^3+y^3+64-12 x y \\
& =(x)^3+(y)^3+(4)^3-3 \cdot x \cdot y \cdot 4 \\
& =(x+y+4)\left\{(x)^2+(y)^2+(4)^2-x \cdot y-y \cdot 4-4 \cdot x\right\} \\
& =(x+y+4)\left(x^2+y^2+16-x y-4 y-4 x\right) \\
& =(x+y+4)\left(x^2+y^2+16-4 x-4 y-x y\right)
\end{aligned}\)

Class 9 Mathematics West Bengal Board

Question 5. \((3 a-2 b)^3+(2 b-5 c)^3+(5 c-3 a)^3\)

Solution:

 

 

Question 6. \((2 x-y)^3-(x+y)^3+(2 y-x)^3\)

Solution:

 

Let a = 2x-y, b= -(x + y), c = 2y –x

a + b + c = 2x – y – x – y + 2y – x =0

a³ + b³ = c³ = 3abc

= 3(2x –y) {-(x + y)} (2y –x)

= 3(2x –y) (x + y) (x–2y)

 

Class 9 Mathematics West Bengal Board

Question 7. \(a^6+32 a^3-64\)

Solution:

\(\begin{aligned}
& a^6+32 a^3-64 \\
& =\left(a^2\right)^3+(2 a)^3+(-4)^3-3 a^2 .2 a(-4) \\
& =\left(a^2+2 a-4\right)\left\{\left(a^2\right)^2+(2 a)^2+(-4)^2-a^2 .2 a-2 a(-4)-(-4) a^2\right\} \\
& =\left(a^2+2 a-4\right)\left(a^4+4 a^2+16-2 a^3+8 a+4 a^2\right) \\
& =\left(a^2+2 a-4\right)\left(a^4-2 a^3+8 a^2+8 a+16\right)
\end{aligned}\)

 

Question 8. \(a^6-18 a^3+125\)

Solution:

\(\begin{aligned}
&a^6-18 a^3+125 \\
& =\left(a^2\right)^3+(3 a)^3+(5)^3-3 \cdot a^2 \cdot 3 a \cdot 5 \\
& =\left(a^2+3 a+5\right)\left\{\left(a^2\right)^2+(3 a)^2+(5)^2-a^2 \cdot 3 a-3 a \cdot 5-5 \cdot a^2\right\} \\
& =\left(a^2+3 a+5\right)\left(a^4+9 a^2+25-3 a^3-15 a-5 a^2\right) \\
& =\left(a^2+3 a+5\right)\left(a^4-3 a^3+4 a^2-15 a+25\right) \\
&
\end{aligned}\)

 

Question 9. \(p^3(q-r)^3+q^3(r-p)^3+r^3(p-q)^3\)

Solution:

\begin{aligned}
& p^3(q-r)^3+q^3(r-p)^3+r^3(p-q)^3 \\
& =\{p(q-r)\}^3+\{q(r-p)\}^3+\{r(p-q)\}^3 \\
& \text { Let } a=p(q-r), b=q(r-p), c=r(p-q) \\
& ∴ a+b+c=p(q-r)+q(r-p)+r(p-q) \\
& =p q-p r+q r-p q+p r-q r \\
& =0 \\
& ∴a^3+b^3+c^3-3 a b c=0 \\
& ∴ a^3+\dot{b}^3+c^3=3 a b c \\
&∴ \{p(q-r)\}^3+\{q(r-p)\}^3+\{r(p-q)\}^3 \text {. } \\
& =3 p(q-r) \cdot q(r-p) r(p-q) \\
& =3 p q r(p-q)(q-r)(r-p) \\
&
\end{aligned}

Class 9 Mathematics West Bengal Board

Question 10. \(p^3+\frac{1}{p^3}+\frac{26}{27}\)

Solution:

\(\begin{aligned}
& p^3+\frac{1}{p^3}+\frac{26}{27} \\
& =p^3+\frac{1}{p^3}+\frac{27-1}{27} \\
& =p^3+\frac{1}{p^3}+\frac{27}{27}-\frac{1}{27} \\
& =p^3+\frac{1}{p^3}+1-\frac{1}{27}
\end{aligned}\)

Class 9 Maths WBBSE

\(\begin{aligned}
& =(p)^3+\left(\frac{1}{p}\right)^0+\left(-\frac{1}{3}\right)^0-3 \cdot p \cdot \frac{1}{p}\left(-\frac{1}{3}\right) \\
& =\left(p+\frac{1}{p}-\frac{1}{3}\right)\left\{(p)^2+\left(\frac{1}{p}\right)^2+\left(-\frac{1}{3}\right)^2-p \cdot \frac{1}{p}-\frac{1}{p} \cdot\left(-\frac{1}{3}\right)-\left(-\frac{1}{3}\right) \cdot p\right\}
\end{aligned}\)

 

\(\begin{aligned}
& =\left(p+\frac{1}{p}-\frac{1}{3}\right)\left(p^2+\frac{1}{p^2}+\frac{1}{9}-1+\frac{1}{3 p}+\frac{p}{3}\right) \\
& =\left(p+\frac{1}{p}-\frac{1}{3}\right)\left(p^2+\frac{1}{p^2}+\frac{p}{3}+\frac{1}{3 p}-\frac{8}{9}\right)
\end{aligned}\)

Class 9 Maths WBBSE Chapter 8 Factorisation Exercise 8.5

Question 1. \((a+b)^2-5 a-5 b+6\)

Solution:

\(
\begin{aligned}
& (a+b)^2-5 a-5 b+6 \\
& =(a+b)^2-5(a+b)+6
\end{aligned}
Let a+b=x\)

 

\(\begin{aligned}
& ∴ x^2-5 x+6 \\
& =x^2-(3+2) x+6 \\
& =x^2-3 x-2 x+6 \\
& =x(x-3)-2(x-3) \\
& =(x-3)(x-2)
\end{aligned}\)

pitting the value of x the given expression becomes (a+b-3)(a+b-2)

Question 2. (x+1)(x+2)(3x-1)(3x-4) +12

Solution: (x+1)(x+2)(3x-1)(3x-4) +12
=(x+1)(3x-1)(x+2)(3x-4)+12

\(\begin{aligned}
& =\left(3 x^2-x+3 x-1\right)\left(3 x^2-4 x+6 x-8\right)+12 \\
& =\left(3 x^2+2 x-1\right)\left(3 x^2+2 x-8\right)+12
\end{aligned}\)

 

\(\begin{aligned}
& \text { Let } 3 x^2+2 x=a \\
& \qquad \begin{array}{l}
therefore(a-1)(a-8)+12 \\
\quad=a^2-8 a-a+8+12 \\
\quad=a^2-9 a+20 \\
\quad=a^2-5 a-4 a+20 \\
\quad=a(a-5)-4(a-5) \\
\quad=(a-5)(a-4)
\end{array}
\end{aligned}\)

putting the value of a,

\(\begin{aligned}
& \left(3 x^2+2 x-5\right)\left(3 x^2+2 x-4\right) \\
& =\left(3 x^2+5 x-3 x-5\right)\left(3 x^2+2 x-4\right) \\
& =\{x(3 x+5)-1(3 x+5)\}\left(3 x^2+2 x-4\right) \\
& =(3 x+5)(x-1)\left(3 x^2+2 x-4\right)
\end{aligned}\)

Class 9 Maths WBBSE

Question 3. \(x\left(x^2-1\right)(x+2)-8\)

Solution:

\(\begin{aligned}
& x\left(x^2-1\right)(x+2)-8 \\
& =x(x+1)(x-1)(x+2)-8 \\
& =\left(x^2+x\right)\left(x^2+2 x-x-2\right)-8 \\
& =\left(x^2+x\right)\left(x^2+x-2\right)-8
\end{aligned}\)

 

Let,\(\begin{aligned}
& x^2+x=a \\
& therefore \mathrm{a}(\mathrm{a}-2)-8 \\
& =\mathrm{a}^2-2 \mathrm{a}-8 \\
& =a^2-(4-2) a-8 \\
& =a^2-4 a+2 a-8 \\
& =a(a-4)+2(a-4) \\
& =(a-4)(a+2) \\
&
\end{aligned}\)

 

Pitting the value of a, the given expression becomes \(\left(x^2+x-4\right)\left(x^2+x+2\right)\)

Question 4. \(7\left(a^2+b^2\right)^2-15\left(a^4-b^4\right)+8\left(a^2-b^2\right)^2\)

Solution:

 

\(\begin{aligned}
& =7 x^2-7 x y-8 x y+8 y^2 \\
& =7 x(x-y)-8 y(x-y) \\
& =(x-y)(7 x-8 y)
\end{aligned}\)

Putting the value of x and y, the given expression becomes

\(\begin{aligned}
& \left(a^2+b^2-a^2+b^2\right)\left(7 a^2+7 b^2-8 a^2+8 b^2\right) \\
& =2 b^2\left(15 b^2-a^2\right)
\end{aligned}\)

Class 9 Maths WBBSE

Question 5. \(\left(x^2-1\right)^2+8 x\left(x^2+1\right)+19 x^2\)

Solution:

\(\begin{aligned}
&\left(x^2-1\right)^2+8 x\left(x^2+1\right)+19 x^2 \\
& =\left(x^2+1\right)^2-4 \cdot x^2 \cdot 1+8 x\left(x^2+1\right)+19 x^2 \\
& =\left(x^2+1\right)^2+8 x\left(x^2+1\right)+15 x^2
\end{aligned}\)

 

Let,\(\begin{aligned}
& \text { Let } x^2+1=a \\
& \quad therefore a^2+8 a x+15 x^2 \\
& =a^2+(5+3) a x+15 x^2 \\
& =a^2+5 a x+3 a x+15 x^2 \\
& =a(a+5 x)+3 x(a+5 x) \\
& =(a+5 x)(a+3 x)
\end{aligned}\)

 

Putting the value of a the given expression becomes

\(\begin{aligned}
& \left(x^2+1+5 x\right)\left(x^2+1+3 x\right) \\
& =\left(x^2+5 x+1\right) \cdot\left(x^2+3 x+1\right)
\end{aligned}\)

 

Question 6. \((a-1) x^2-x-(a-2)\)

Solution:

\(\begin{aligned}
& (a-1) x^2-x-(a-2) \\
&=(a-1) x^2-\{(a-1)-(a-2)\} x-(a-2) \\
&=(a-1) x^2-(a-1) x+(a-2) x-(a-2) \\
&=(a-1) x(x-1)+(a-2)(x-1) \\
&=(x-1)\{(a-1) x+(a-2)\} \\
&=(x-1)(a x-x+a-2)
\end{aligned}\)

Class 9 Maths WBBSE

Question 7. \((a-1) x^2+a^2 x y+(a+1) y^2\)

Solution:

\(\begin{aligned}
&(a-1) x^2+a^2 x y+(a+1) y^2 \\
& =(a-1) x^2+\left\{\left(a^2-1\right)+1\right\} x y+(a+1) y^2 \\
& =(a-1) x^2+\left(a^2-1\right) x y+x y+(a+1) y^2 \\
& =(a-1) x^2+(a+1)(a-1) x y+x y+(a+1) y^2 \\
& =(a-1) x\{x+(a+1) y\}+y\{x+(a+1) y\} \\
& =\{x+(a+1) y\}\{(a-1) x+y\} \\
& =(x+a y+y)(a x-x+y)
\end{aligned}\)

 

Question 8. \(x^2-q x-p^2+5 p q-6 q^2\)

Solution:

\(\begin{aligned}
& x^2-q x-p^2+5 p q-6 q^2 \\
& =x^2-q x-\left(p^2-5 p q+6 q^2\right) \\
& =x^2-q x-\left(p^2-3 p q-2 p q+6 q^2\right) \\
& =x^2-q x-\{p(p-3 q)-2 q(p-3 q)\} \\
& =x^2-q x-(p-3 q)(p-2 q) \\
& =x^2-\{(p-2 q)-(p-3 q)\} x-(p-3 q)(p-2 q) \\
& =x^2-(p-2 q) x+(p-3 q) x-(p-3 q)(p-2 q) \\
& =x(x-p+2 q)+(p-3 q)(x-p+2 q) \\
& =(x-p+2 q)(x+p-3 q)
\end{aligned}\)

 

Question 9. \(2\left(a^2+\frac{1}{a^2}\right)-\left(a-\frac{1}{a}\right)-7\)

Solution:

\(2\left(a^2+\frac{1}{a^2}\right)-\left(a-\frac{1}{a}\right)-7
\begin{aligned}
& =2\left\{(a)^2+\left(\frac{1}{a}\right)^2\right\}-\left(a-\frac{1}{a}\right)-7 \\
& =2\left\{\left(a-\frac{1}{a}\right)^2+2 \cdot a \cdot \frac{1}{a}\right\}-\left(a-\frac{1}{a}\right)-7
\end{aligned}
\)

 

\(\begin{aligned}
& \text { Let } a-\frac{1}{a}=x \\
& \qquad \begin{aligned}
therefore & 2\left(x^2+2\right)-x-7 \\
= & 2 x^2+4-x-7 \\
= & 2 x^2-x-3 \\
= & 2 x^2-3 x+2 x-3 \\
= & x(2 x-3)+1(2 x-3) \\
= & (2 x-3)(x+1)
\end{aligned}
\end{aligned}\)

 

Putting the value of a, the given expression becomes

\(\begin{aligned}
& =\left\{2\left(a-\frac{1}{a}\right)-3\right\}\left(a-\frac{1}{a}+1\right) \\
& =\left(2 a-\frac{2}{a}-3\right)\left(a-\frac{1}{a}+1\right)
\end{aligned}\)

 

Question 10. \(\left(x^2-x\right) y^2+y-\left(x^2+x\right)\)

Solution:

\(\begin{aligned}
& \left(x^2-x\right) y^2+y-\left(x^2+x\right) \\
& =x(x-1) y^2+y-x(x+1) \\
& =x(x-1) y^2+\left\{x^2-\left(x^2-1\right)\right\} y-x(x+1) \\
& =x(x-1) y^2+x^2 y-\left(x^2-1\right) y-x(x+1) \\
& =x y\{(x-1) y+x\}-(x+1)\{(x-1) y+x\} \\
& =\{(x-1) y+x\}\{x y-(x+1)\} \\
& =(x y-y+x)(x y-x-1)
\end{aligned}\)

 

Question 11. Multiple Choice Questions

1)If a² – b²= 11 x 9; a and b are positive integers (a> b) then

(1) a = 11, b = 9
(2) a = 33, b = 3
(3) a = 10, b = 1
(4) a = 100, b = 1

Solution: a² – b²= 11 x 9
or, (a + b) (a – b) (10+ 1) (10-1)
∴ a = 10, b = 1

∴ (3) a = 10, b = 1

2) If \(\frac{a}{b}+\frac{b}{a}=1\) then the value of a³ + b³ is

(1)1
(2)a
(3)b
(4)0

Solution: \(\frac{a}{b}+\frac{b}{a}=1\)

or, \(\frac{a^2+b^2}{a b}=1\)

 

or,\(\begin{aligned}
& a^2+b^2=a b \\
& a^2-a b+b^2=0 \\
& therefore a^3+b^3 \\
& =(a+b)\left(a^2-a b+b^2\right) \\
& =(a+b) \times 0 \\
& =0
\end{aligned}\)

∴ (4) 0

3)The value of \(25^3-75^3+50^3+3 \times 25 \times 75 \times 50\) is

(1) 150
(2) 0
(3) 25
(4) 50

Solution: Let a = 25, b=-75 and c =
∴ a+b+c=25-75+ 50 = 0

\(\begin{aligned}
& therefore a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right) \\
& =0 \times\left(a^2+b^2+c^2-a b-b c-c a\right) \\
& =0 \\
& therefore 25^3-75^3+50^3+3 \times 25 \times 75 \times 50 \\
& =(25)^3+(-75)^3+(50)^3-3.25 .(-75) .50 \\
& =0
\end{aligned}\)

4) If a + b + c = 0, then the value of \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\) is

(1) 0
(2) 1
(3) 1
(4) 3

Solution: a+b+c=0
∴ a³+b³+ c³-3abc = 0
or, a³+b³+ c³= 3abc

or, \(\frac{a^3+b^3+c^3}{a b c}=\frac{3 a b c}{a b c}\)

or, \(\frac{a^3}{a b c}+\frac{b^3}{a b c}+\frac{c^3}{a b c}=3\)

or, \(\frac{a^2}{b c}+\frac{b^2}{a c}+\frac{a^2}{a b}=3\)

∴ (4)3

5) If p²– px + 12 = (x-3) (x – a) is an identity, then the values of a and p are respectively
(1) a = 4, p = 7
(2) a = 7, p = 4
(3) a = 4, p = -7
(4) a = 4, p = 7

Soltion:  p²-px+12= (x-3) (x – a)
or,  p²-px+12=x²-ax-3x + 3a
or, p²-px+12= x²– (a + 3)x + 3a
∴ 3a= 12

or, a =12/3 =4
Again, a + 3 = p
or, 4+ 3 = p
or, p=7

(1) a = 4, p = 7

Question 12. Short answer type questions:

(1) Let us write the simplest value of

\(\frac{\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3+\left(a^2-b^2\right)^3}{(b-c)^3+(c-a)^3+(a-b)^3}\)

Solution:

 

 

(2) Let us write the relation between a, b, and c if a³ + b³ + c³-3abc = 0 and a+b+c ≠ 0

Solution:

 

 

(3 ) If a²-b² = 224 and a and b are negative integers (a<b), then let us write the values of a and b.

Solution :

 

 

(4) Let us write the value of \((x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)\) if 3x= a+b+c.

Solution: 3x=a+b+c
or, 3x-a-b-c=0
or, x-a+x-b+x-c=0
Let x-a = p, x-b=q and x-c=r
∴ p+q+r=0

\(\begin{aligned}
& therefore p^3+q^3+r^3-3 p q r=0 \\
& therefore(x-a)^3+(x-b)^3+(x-c)^3-3(x-a)(x-b)(x-c)=0
\end{aligned}\)

 

(5) Let us write the values of a and p if 2x²+ px + 6 = (2x – a) (x-2) is an identity.

Solution:

\(2 x^2+p x+6=(2 x-a)(x-2)\)

 

\(\text { or, } 2 x^2+p x+6=2 x^2-4 x-a x+2 a\)

 

\(\text { or, } 2 x^2+p x+6=2 x^2-(4+a) x+2 a\)

 

\(\begin{aligned}
& therefore 2 a=6 \\
& \text { or, } a=\frac{6}{2}=3
\end{aligned}\)

 

Again,\(p=-(4+a)=-(4+3) or, p=-7
therefore a=3, p=-7
\)

 

 

Class IX Maths Solutions WBBSE Chapter 9 Transversal And Mid Point Theorem Exercise 9

Question 1. In the triangle ABC, D is the midpoint of the side BC. From point D, the parallel straight lines CA and BA intersect the sides BA and CA at points E and F respectively, Let us prove that, \(E F=\frac{1}{2} B C\)

Solution: In ΔABC the midpoint of side D is BC. From point D, CA and BA parallel to CA and BA are drawn which intersect BA at point E and CA at point F.

Prove that \(E F=\frac{1}{2} B C\)

 

 

Proof: DF II BA (Given) and D are the midpoint of BC
∴ Point F is the midpoint of side AC.
Similarly, DE II CA and point D is the midpoint of BC.
∴ point E is the midpoint of AB

Read and Learn More WBBSE Solutions For Class 9 Maths

Now E and F points respectively are mid points of AB and AC sides ΔABC.

∴ \(E F=\frac{1}{2} B C\)

Question 2. D and E lie on AB and AC respectively of the triangle ABC such that \(A D=\frac{1}{4} A B\) and \(A E=\frac{1}{4} A C\). Let us prove that DE II BC and \(D E=\frac{1}{4} B C\)

Solution: In ΔABCD and F lie on sides AB and AC respectively such that \(A D=\frac{1}{4} A B\) and \(A E=\frac{1}{4} A C\).

Prove that DE II BC and \(D E=\frac{1}{4} B C\)

 

 

Construction: Midpoints of sides F AB and AC, G and are joined.

Proof: In ΔABC the midpoints of sides AB and AC are F and G.

∴ FG II BC and \(F G=\frac{1}{2} B C\)

∴ F is the and midpoint of AB \(A D=\frac{1}{4} A B\)

Class IX Maths Solutions WBBSE

∴ D is the midpoint of AF.
Similarly, E is the midpoint of AG.
So, in ΔAFG the midpoints of sides AF and AG are respectively D and E

∴ DE II FG and FG II BC.
∴ DE II BC.

Again, \(D E=\frac{1}{2} F G=\frac{1}{2} \times \frac{1}{2} B C=\frac{1}{4} B C\)

Question 3. In the triangle PQR, the midpoints of the sides QR and QP are X and Z respectively. The side QP is extended upto the point S so that PS = ZP. SX intersects the side PR at the point Y. Let us prove that PY = \(P Y=\frac{1}{4} P R\)

Solution: In ΔPQR the midpoints of sides QR and PQ are X and Z. Side QP is extended up to points S so that PS = ZP. SX, at Y, intersects the side PR.

Prove that PY = \(P Y=\frac{1}{4} P R\)

 

 

Construction: X, and Y are joined.

Proof: Of QP and QR the midpoints are Z and X.

∴ ZX II PR and \(Z X=\frac{1}{2} P R\)

Now, in ΔSZX the midpoint SZ is and P and PY II XZ.
∴ Y is the midpoint of SX.

∴ \(P Y=\frac{1}{2} Z X\)       \(\left[because \mathrm{ZX}=\frac{1}{2} \mathrm{PR}\right]\)

or, \(P Y=\frac{1}{2} \times \frac{1}{2} P R\)

or, \(P Y=\frac{1}{4} P R\)

Class IX Maths Solutions WBBSE

Question 4. Let us prove that the quadrilateral formed by joining mid-points of the consecutive sides of a parallelogram is a parallelogram.

Solution: Let in the quadrilateral of ABCD the midpoints sides AB, BC, CD, and DA are P, Q, R, and S respectively joining. PQ, QR, RS and SP we get PQRS quadrilateral.

Prove that PQRS is a parallelogram.

 

 

 

Construction: diagonal BD is drawn.

Proof: In ΔABD the midpoints of sides AB and AD are P and S respectively.

∴ PS II BD and \(P S=\frac{1}{2} B D\)

Similarly, in ΔCBD the midpoints of sides CB and CD are Q and R respectively.

∴ QR II BD and \(\mathrm{QR}=\frac{1}{2} \mathrm{BD}\)

∴ PS = QR
Now, in quadrilateral PQRS, PS II QR and PS = QR. So, PQRS is a parallelogram.

Class IX Maths Solutions WBBSE

Question 5. Let us prove that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangular figure is not a square figure but a rhombus.

Solution: Let ABCD is a rectangle whose midpoints of sides AB, BC, CD, and DA are P, Q, R, and S respectively P, Q; Q, R; R, S; and S, P are joined.

Prove that PQRS is a rhombus, not a square.

 

 

Construction: P, R, and S, Q are joined.

Proof: In ΔAPS and ΔPBQ,
AP = BP, AS BQ [By construction]
∠PAS ≅ ∠PBQ [Both are right angles]

∴ ΔPAS ≅ ΔPBQ (S-A-S congruency)
∴ PS = PQ

Similarly, in ΔPBQ and ΔRCQ, PQ = QR
In ΔQRC and ΔRCD, QR = RS
In ΔRSD and ΔAPS, RS = PS
∴ PQ = QR RS = SP

Now, in quadrilateral PQRS, PQ = QR = RS = SP
But PR > SQ.
∴ The four sides of quadrilateral PQRS are equal but the diagonals are not. So is a rhombus but not a square.

Class IX Maths Solutions WBBSE

Question 6. Let us prove that the quadrilateral formed by joining the mid-points of consecutive sides of a square is a square.

Solution: Let the midpoints of sides AB, BC, CD, and DA of the square ABCD be P, Q, R & S respectively. P, Q; Q, R; R, S, and S, P are joined.

Prove that PQRS is a square.

 

 

Construction: Diagonals AC and BD are joined which intersect each other at O. In □PQRS, PQ cuts diagonal BD at point F and PS cuts diagonal AC at point E.

Proof: In ΔADC the midpoint of AD is S and the midpoint DC is R.

∴ SR II AC and \(S R=\frac{1}{2} A C\)  ….(1)

Similarly, in ΔABC, PQ II AC and \(P Q=\frac{1}{2} A C\) …(2)

∴ PQ II SR and PQ = SR.
∴ PQRS is a parallelogram.

Class 9 Mathematics West Bengal Board

∴ In ΔABD the midpoints of sides AB and AD are P and S respectively.

∴ \(P S=\frac{1}{2} B D\). But ACBD.

∴ \(P Q=\frac{1}{2} A C=\frac{1}{2} B D=P S\)

∴ In Parallelogram PQRS the length of adjacent sides are equal.
∴PQRS is a rhombus.
∴ PE II FO and PF II EO.
∴ ∠EOF= 90°
∴ ∠EPF = 90°
∴ PQRS is a square.

Question 7. Let us prove that the quadrilateral formed by joining the mid-points of a rhombus is a rectangle.

Solution: Let in rhombus ABCD the midpoints of sides AB, BC, CD, and DA are E, F, G, and H respectively. E, F; F, G; G, H, and H, E are joined.

Prove that EFGH is a rectangle.

 

 

Construction: B, D; A, C; E, G and H, F are joined.

Proof: In ΔABD the midpoints of sides AB and AD are E and H respectively.

∴ EH II BD and \(\mathrm{EH}=\frac{1}{2} \mathrm{BD}\)

Class 9 Mathematics West Bengal Board

Similarly, in ΔBCD the midpoints of sides BC and CD are F and G respectively.∴ FG II BD and FG = 1BD

∴ EH II FG and \(\)

∴ EFGH is a parallelogram.

Now, in □ABFH, AH II BF and AH = BF
∴ ABFH is a parallelogram.
∴ AB = HF

Similarly, ADGE is a parallelogram.
∴ AD = EG
But AB = AD ( ABCD is a rhombus)
∴ FH = EG

∴ Diagonals of the parallelogram EFGH are equal in length.
∴ EFGH is a rectangle.

Question 8. In the triangle ABC, the mid-points of AB and AC are D and E respectively; the mid-points of CD and BD are P and Q respectively. Let us prove that BE and PQ bisect each other.

Solution: In ΔABC the midpoints of sides AB and AC are D and E respectively and P & Q are the midpoints of CD and BD respectively.

Prove that BE and PQ bisect each other.

 

 

Construction: B, P; P, E; E, Q and P, Q bisect each other.

Proof: In ΔACD the midpoints of sides AC and CD are E and P respectively.

∴ PE II AD and \(\mathrm{PE}=\frac{1}{2} \mathrm{AD}\)

Again, AD = BD and \(B Q=\frac{1}{2} B D\)

(∵ D is the midpoint of AB and Q the midpoint of BD)

∴ \(P E=\frac{1}{2} A D=\frac{1}{2} B D=B Q\)

In □BPEQ, PE II BQ and PE = BQ
∴ BPEQ is a parallelogram whose diagonals are BE and PQ.
∴ BE and PQ bisect each other.

Class 9 Mathematics West Bengal Board

Question 9. In the triangle ABC, AD is perpendicular on the bisector of ZABC. From point D, a straight line DE parallel to side BC is drawn which intersects side AC at point E. Let us prove that AE = EC.

Solution: In triangle ABC on the bisector of ZABC AD is perpendicular. From point D parallel to BC of parallel straight line DE is drawn which intersects AC side at point E.

Prove that AE = EC.

 

 

Construction: AD is extended forwards which at intersects BC point F.

Proof: In ΔABD and ΔBDF,
∠BDA = ∠BDF (∵ AD ⊥ BD)
∠ABD = ∠FBD (∵ ∠ZABC is the bisector of BD) and BD is a common side.

∴ ΔABD ≅ ΔBDF (A-A-S Congrency)
∴ AD DF, that is D is the midpoint of AF

In ΔAFC of AF D is the midpoint and DE II FC.
∴ E is the midpoint of AC
∴ AE = EC

Class 9 Maths WB Board

Question 10. In the triangle ABC, AD is a median. From points B and C, two straight lines BR and CT, parallel to AD are drawn, which meet extended BA and CA at points T and R respectively. Let us prove that \(\frac{1}{A D}=\frac{1}{R B}+\frac{1}{T C}\)

Solution: In a triangle, ABC AD is the median. from points BC and parallel to AD straight lines BR and CT are drawn which intersect extended BA and CA at points T and R.

Prove that = \(\frac{1}{A D}=\frac{1}{R B}+\frac{1}{T C}\)

 

 

Proof: In AD is the median
∴ D, is the midpoint of BC

In ΔBCT the midpoint of BC is D and DA II CT
∴ A is the midpoint of BT

∴ AD II CT and \(\mathrm{AD}=\frac{1}{2} \mathrm{TC}\)
∴ TC= 2AD

Now, in ΔBCR the mid point of BC is D and DA II BR
∴ A is the midpoint of RC

Class 9 Maths WB Board

∴ AD II RB and \(A D=\frac{1}{2} R B\)

∴ RB = 2AD

\(\frac{1}{\mathrm{RB}}+\frac{1}{T C}=\frac{1}{2 A D}+\frac{1}{2 A D}=\frac{2}{2 A D}=\frac{1}{A D}\) \(\frac{1}{A D}=\frac{1}{R B}+\frac{1}{T C}\)

 

Question 11. In the trapezium ABCD, AB II DC and AB > DC; the mid-points of two diagonals AC and BD are E and F respectively. Let us prove that \(E F=\frac{1}{2}(A B-D C)\)

Solution: In trapezium ABCD AB II DC and AB > DC; E and F are the midpoints of AC and BD diagonals respectively.

Prove that \(E F=\frac{1}{2}(A B-D C)\)

 

 

Construction: E, and F are joined. CF is extended forwards which intersect AB at point G point.

Proof: In ΔCFD and ΔBFG,
∠CDF = alternate ∠FBG (∴ DC II AB and BD is transversal)
∠CFD = VOA ∠BFG and DF = BF (∵ F is the midpoint of BD).
∴ ΔCFD ≅ ΔBFG (A -A – S congruency)
∴ DC = GB and CF = FG

Now, in ΔACG the midpoints of sides AC and CG are E and F respectively

∴ EF II AG and \(E F=\frac{1}{2} A G\)

∴ \(E F=\frac{1}{2} A G=\frac{1}{2}(A B-G B)\)

∴ \(=\frac{1}{2}(A B-D C)(because G B=D C)\)

Class 9 Maths WB Board

12. C is the mid-point of the line segment AB and PQ is any straight line. The minimum distances of the line PQ from points A, B, and C are AR, BS, and CT respectively; let us prove that AR + BS = 2CT.

Solution: The midpoint of st. line AB is C and PQ is any st. line. from points A, B, and C the minimum distance of straight line PQ are respectively AR, BS, and CT.

Prove that AR + BS = 2CT

 

 

Proof:  ∵AR, BS, and CT are perpendicular on PQ
∴AR II BS II CT

In ΔABR of side AB, the midpoint is C and CM II AR
∴ M, BR is the midpoint

Again, in ΔBRS of BR is the midpoint M and MT II BS
∴ T, RS is the midpoint of
∴ RT=TS

In ΔABR we get, \(C M=\frac{1}{2} A R\)
∴ AR = 2CM

Again, in ΔBRS we get \(M T=\frac{1}{2} B S\)
Or, BS = 2MT
∴ AR+ BS 2CM + 2MT = 2(CM + MT) = 2CT Proved

Class 9 Math Solution WBBSE

Question 13. In a triangle ABC, D is the mid-point of the side BC; through point A, PQ is any straight line. The perpendiculars from the points B, C, and D on PQ are BL, CM, and DN respectively; let us prove that, DL = DM.

Solution:

 

 

Construction: D, Land D, and M are joined.

Proof:  On a straight line PQ BL, CM, and DN are perpendicular.
∴ BL II DN II CM
∵ BD = DC ( ∵D, is the midpoint of BC)
∴ LN = NM

In ΔDLN and ΔDMN, LN = NM Proved
∠DNL = ∠DNM  and DN is a common side. (∵ DN ⊥ PQ)
∴ ΔDLN ≅ ΔDMN (S-A-S Congruency)
∴ DL = DM

Ganit Prakash Class 9 Solutions

Question 14. ABCD is a square figure. The two diagonals AC and BD intersect each other at point O. The bisector of ∠BAC intersects BO at the point P and BC at point Q. Let us prove that \(O P=\frac{1}{2} C Q\)

Solution:

 

 

Construction from point C parallel to OP an st. line CR is drawn which cuts extended AQ at point R.

Proof: In ΔACR midpoint AC the is O and OP II CR.

\(O P=\frac{1}{2} C R\)

∴ ∠AOP = 90° ( ∵The diagonals of a square bisect each other at right angles.) and ∠AOP = ∠ACR = 90° (corresponding angles).

In ΔACR and ΔABQ,
∠ACR = ∠ABQ ( ∵ Each angle is a right angle)
∠CAR = ∠BAQ ( ∵ ∠BAC is the bisector of AQ)

Remaining ∠ARC= remaining ∠AQB
But ∠AQB VOA ∠CQR
∠CRQ = ∠CQR ( Each is equal to ∠AQB)
∴ CR = CQ

\(O P=\frac{1}{2} C R=\frac{1}{2} C Q\)

Ganit Prakash Class 9 Solutions

Question 15. Multiple Choice Question

1. In the triangle PQR, ∠PQR = 90° and PR = 10 cm. If S is the midpoint of PR, then the length of QS is

 

 

 

(1)4 cm
(2)5 cm
(3)6 cm
(4)3 cm

Solution: PR= 10 cm
We know that in a right-angled triangle the straight line from the vertex of the right angle to the midpoint of the hypotenuse is equal to half of the hypotenuse.

∴ \(\begin{aligned}
& Q S=\frac{1}{2} P R \\
& =\frac{1}{2} \times 10 \mathrm{~cm}
\end{aligned}\) = 5

∴ (2)5 cm.

2. In the trapezium ABCD, AB II DC and AB = 7 cm and DC = 5 cm. The mid-points of AD and BC are E and F respectively, the length of EF is

 

 

(1)5 cm
(2)6 cm
(3)7 cm
(4)12 cm

Solution: EF =1/2(AB+DC)

\(\begin{aligned}
& =\frac{1}{2}(7+5) \mathrm{cm} \\
& =\frac{1}{2} \times 12 \mathrm{~cm}=6 \mathrm{~cm}
\end{aligned}\)

Ganit Prakash Class 9 Solutions

∴ 2. 6cm

3. In the triangle ABC, E is the mid-point of the median AD; the extended BE intersects AC at the point F. If AC = 10.5 cm then the length of AF is

 

 

(1)3 cm
(2)3.5 cm
(3)2.5 cm
(4)5 cm

Solution: In triangle ABC the midpoint of median AD is E. The extended part of BE AC cuts at F. From point D parallel to BF a straight line is drawn which cuts AC at G point.

\(\begin{aligned}
& A F=\frac{1}{3} A C=\frac{1}{3} \times 10.5 \mathrm{~cm} \\
& =3.5 \mathrm{~cm}
\end{aligned}\)

∴(2) 3.5 cm

4. In the triangle ABC, the mid-points of BC, CA, and AB are D, E, and F respectively; 3E and DF intersect at the point X and CF and DE intersect at the point Y, the length of XY is equal to

 

 

(1)1/2 BC
(2)1/4 BC
(3)1/3 BC
(4)1/8 BC

Solution: E and F are joined.

∴ \(\mathrm{FE}=\frac{1}{2} \mathrm{BC}\) and FE II BC, i.e., FE II BD

∴ ED II AB, i.e., ED II FB
∴ FEDB is a parallelologram.
∴ Diagonals of a paralelogram bisect each other.
∴ BE and DF bisect each other at point X.
∴  X is the midpoint of FD
∴ Y, is the midpoint of the ED

\(X Y=\frac{1}{2} F E=\frac{1}{2} \cdot \frac{1}{2} B C=\frac{1}{4} B C\)

∴ (2)1/4 BC

Ganit Prakash Class 9 Solutions

5. In the parallelogram ABCD, E is the mid-point of the side BC; DE and extended AB meet at point F. The length of AF is equal to

 

 

 

(1) 3/2 AB
(2)2AB
(3)3AB
(4)5/4 AB

Solution: ΔDEC and ΔBEF
CE = BE ( ∵E, is the midpoint of BC)
∠CED = VOA ∠BEF and ∠DCE = alternate ∠EBF
∴ΔDEC ≅ ΔBEF (A-A-S congruency)
∴ DC = BF
∴ AF = AB+ BF =AB+ DC ( ∵BF = DC)
= AB+ AB (∵DC = AB)
= 2AB

∴ (2)2AB

Question 16. Short answer type questions

1. In the triangle ABC, AD and BE are two medians, and DF parallel to BE, meets AC at point F. If the length of the side AC is 8 cm, then let us write the length of the side CF.

 

 

Solution: In ΔABC, AD, and BE are medians.
∴ D and E are the midpoints of BC and AC respectively.

∴ \(C E=\frac{1}{2} A C\)

In ΔBCE the midpoint of BC is D and DF II BE.
∴ F, is the midpoint of CE

\(\begin{aligned}
& therefore C F=\frac{1}{2} C E=\frac{1}{2} \times \frac{1}{2} A C \\
& =\frac{1}{4} \times 8 \mathrm{~cm}=2 \mathrm{~cm}
\end{aligned}\)

 

2. In the triangle ABC, the mid-points of BC, CA, and AB are P, Q, and R respectively; if AC = 21 cm, BC= 29 cm, and AB = 30 cm, then let us write the perimeter of the quadrilateral ARPQ.

 

 

 

Solution: Q, is the midpoint of AC

∴ \(A Q=\frac{1}{2} A C=\frac{1}{2} \times 21 \mathrm{~cm}\) = 10.5 cm

∵ R is the midpoint of AB

∴ \(A R=\frac{1}{2} A B=\frac{1}{2} \cdot 30 \mathrm{~cm}\)= 15 cm

In ΔABC the midpoints of sides BC and AB are P and R respectively.

\(P R=\frac{1}{2} A C=\frac{1}{2} \times 21 \mathrm{~cm}=10.5 \mathrm{~cm}\)

Similarly,\(P Q=\frac{1}{2} A B=\frac{1}{2} \times 30 \mathrm{~cm}=15 \mathrm{~cm}\)

Perimeter of quadrilateral ARPQ = AQ + PQ+ PR + AR=(10.5+15+ 10.5+15) cm = 51 cm

3. In the triangle ABC, D is any point on the side AC. The mid-points of AB, BC, AD, and DC are P, Q, X, Y respectively. If PX = 5 cm, then let us write the length of the side QY.

 

 

Solution: B, and D are joined.
In ΔABD the midpoints of AB are ads are P and X respectively.

\(P X=\frac{1}{2} B D\)

∴ BD = 2PX = 2 x 5 cm = 10 cm

In ΔBCD the midpoints of BC and CD are Q and Y respectively.

\(Q Y=\frac{1}{2} B D=\frac{1}{2} \times 10 \mathrm{~cm}=5 \mathrm{~cm}\)

 

4. In the triangle ABC, the medians BE and CF intersect at point G. The mid-points of BG and CG are P and Q respectively. If PQ is 3 cm, then let us write the length of BC.

 

 

Solution: In ΔGBC the midpoints of BG and CG are P and Q respectively.

\(P Q=\frac{1}{2} B C\)

∴ BC= 2PQ = 2 x 3 cm = 6 cm

5. In the triangle ABC, the mid-points of BC, CA, and AB are D, E, and F respectively; FE intersects AD at the point O. If AD = 6 cm, let us write the length of AO.

 

 

Solution: D, E, and DF are joined.
In □AEDF, AF II ED, and FD II AE.
∴ AEDF is a parallelogram.

∵ Diagonals of a parallelogram bisect each other.
∴ AO = OD
∴O, is the midpoint  of AD

∴ \(A O=\frac{1}{2} A D=\frac{1}{2} \times 6 \mathrm{~cm}=3 \mathrm{~cm}\)

 

 

Class IX Maths Solutions WBBSE Chapter 10 Profit And Loss

Definitions:

Cost price: The price at which a product is bought is called cost price.

Selling price: The price at which a product is sold is called the selling price.

Profit: If the selling price is greater the cost price there is profit.

Loss: In the selling price is less than the cost price there is losses.

Production cost: The cost of making a product is called production cost.

Marked price: The price which is written on a product or on its packet is called the marked price.

Discount: The difference between the market price and the selling price is called a discount.

Successive Discount: When a product is sold at a discount n, on the marked price, n, on remaining, n, on remaining; then ni, ng, ng…… are called successive discounts.

Read and Learn More WBBSE Solutions For Class 9 Maths

Note:

1. Profit or loss is calculated on the cost price.
2. Discount is always given on marked price.

Important Formulae :

1. (Profit) = (SP) – (CP)
2. (Loss) = (CP) – (SP)
3. (CP) = (SP) – (Profit)
4. (CP) = (SP) + (Loss)
5. (SP) = (CP) + (Profit)
6. (SP) = (CP) – (Loss)
7. \(\text { Profit } \%=\frac{\text { Profit } \times 100}{\text { C.P. }}\)
8. \(\text { Loss } \%=\frac{operatorname{Loss} \times 100}{C P}\)∴ \(\frac{C P}{S P}=\frac{100}{100+\text { Profit } \%}\)
9. \(\mathrm{CP}=\frac{100}{100+\text { Profit } \%} \times \mathrm{SP}\)
10. \(\mathrm{SP}=\frac{100+operatorname{Pr} \text { ofit } \%}{100} \times \mathrm{CP}\)
11. \(S P=\frac{100}{100+105 S} \% \times P\)

 

Class 9 Mathematics West Bengal Board Chapter 10 Profit And Loss Exercise 10.1

 

Question 1. Let us fill up the following table:

Cost price	Selling price	Profit Moss	Percentage profit\ loss
Rs. 500			Profit 25
Rs. 300			Loss 7
Rs. 1250			Loss 8
	Rs. 23000		Profit 15

 

Solution:

Cost price	Selling price	Prof it Moss	Percentage profit \ loss
Rs. 500	Rs. 625	Rs. 125	Profit 25
Rs. 300	Rs. 279	Rs. 21	Loss 7
Rs. 1250	Rs. 1150	Rs. 100	Loss 8
Rs. 20,000	Rs. 23000	Rs. 3000	Profit 15

 

Question 2. From the graph, let us find out the answers of the following questions:

1. Let us write the relation between cost price and selling price by observing the graph.
2. Let us write the selling price if the production cost of the jute bag is Rs. 60.
3. Let us write the production cost if the selling price of the jute bag is Rs. 125 by observing the graph.
4. Let us calculate and write the percentage of profit/loss from the graph.
5. Let us write the percentage of profit/loss on the selling price from the graph.

 

 

Solution: We measure the production cost of the jute bag on the x-axis and cost price of the jute bag on the y-axis; and the length of a side of the smallest square on the x-axis = Rs. 4; and the length of a side of the smallest square on y-axis = Rs. 5.

1. Here cost price & selling price are proportional.
2. Here the production cost is Rs. 60 & from the graph the selling price is Rs. 75.
3. Here selling price is Rs. 125 & from the graph cost price is Rs. 100.
4. Here profit = Rs. (125-100) Rs. 25.
   ∴ Profit = Rs. (125-100) = Rs. 25
   ∴ Profit = 25%.
5. Selling price = Rs 100 & production cost = Rs. 80.
   ∴ Profit = Rs. (100-80) = Rs. 20
   ∴ Profit = 20%.

Question 3. Subir’s uncle has sold a clock at the price of Rs 176. If the loss of Subir’s uncle is 12% by selling the clock, let us calculate and observe with how much money he has bought the clock.

Solution: The selling price of the clock = Rs. 176
Loss = 12%

∴ Cost price of the clock = is Rs. \(\frac{100}{100-12} \times 176\)

= \(\text { Rs. }=\frac{100}{88} \times 176\)

= Rs. 200

Question 4. Anoarabibi has sold each dozen lemons at Rs. 42 by buying 10 lemons at Rs. 30. Let us calculate and observe the percentage profit or loss Anoarabibi has made. [Hints: The cost price of 1 lemon = Rs. 1; the selling price of 1 lemon = Rs. \(\) paisa.]

Solution: Cost price of 10 lemons = Rs. 30

∴ Cost price of 1 lemon = Rs. \(\frac{42}{12}\) = Rs. 3

∵ Selling price of 12 lemons = Rs. 42

∴ The selling price of 1 lemon = Rs. \(\frac{42}{12}\) = Rs. \(\frac{7}{2}\)

Profit = Rs. \(\left(\frac{7}{2}-3\right)\) = Rs. \(\frac{1}{2}\)

% profit = \(\frac{1 \times 100}{2 \times 3}\)

Profit = \(\frac{50}{3}=16 \frac{2}{3} \%\)

∴ Profit = \(16 \frac{2}{3} \%\)

Question 5. Amalbabu sold a picture at 20% loss. But he made a profit of 5% if he sold it with Rs. 200 more. Let us calculate and observe the cost price of the picture he has bought.

Solution: Let the cost price = Rs. x, loss 20%

∴ S.P. = Rs. \(\frac{80}{100} \times x\)

= Rs. \(\frac{4 x}{5}\)

Next time S.P. = Rs. \(\left(\frac{4 x}{5}+200\right)\)

S.P. at 5% profit= \(\frac{105}{100} x x=\text { Rs. } \frac{21 x}{20}\)

\(\frac{21 x}{20}=\frac{4 x}{5}+200\)

or, \(\frac{21 x}{20}-\frac{4 x}{5}=200\)

or, \(\frac{21 x-16 x}{20}=200\)

or, \(\frac{5 x}{20}=200\)

or, \(\frac{x}{4}=200\)

or, x =800
∴ Cost price = Rs. 800

Question 6. Supriya has bought a clock. If she sells the clock at Rs. 370, her profit will be equal to the loss for selling it at Rs. 210. Let us calculate and write the cost price of the clock.

Solution: Let C.P. of the clock = Rs. x.
Profit = Rs. (370-x)
Loss = Rs.s (x-210)
X-210 = 370-x
or, x + x = 370 + 210
or, 2x = 580

or, =\(\frac{580}{2}\)

or, x = 290
=Rs. 290

Class 9 Maths WB Board

Question 7. My elder sister bought an umbrella from Arun’s uncle’s shop for Rs. 255. If Arun’s uncle gave a 15% discount on the market price, then let us write after calculating the marked price of the umbrella.

Solution: Let the marked price of the umbrella be Rs. x.

∴ Discount 15%, of Rs. \(x=\frac{15}{100} x\)

∴According to the question,

= \(\begin{aligned}
& =x-\frac{10}{100}=255 \\
& =\frac{100 x-15 x}{100}=255 \\
& =\frac{85 x}{100}=255
\end{aligned}\)

 

∴ \(x=\frac{255 \times 100}{85}=300\)

∴ The marked price of the umbrella = is Rs. 300.

Question 8. My friend bought a storybook at a 25% discount on the written price. If he sells the book at the written price then let us write the profit percentage after calculation.

Solution: Let the written price of the book = be Rs. 100.

∴ Discount of Rs. 100 = Rs. \(\frac{100 \times 25}{100}\) = Rs.25

Cost price Rs. (100-25)= Rs. 75

Profit by selling at written price = Rs. (100-75) = Rs. 25.

\(\begin{aligned}
& \text { Profit } \%=\frac{\text { Profit } \times 100}{\text { Cost price }} \\
& \begin{aligned}
\text { Profit } & =\frac{25 \times 100}{75} \% \\
& =\frac{100}{3} \% \\
& =33 \frac{1}{3} \%
\end{aligned}
\end{aligned}\)

∴ \(\text { Profit }=33 \frac{1}{3} \%\)

Class 9 Maths WB Board

Question 9. Niyamatchacha has bought 150 eggs at the rate of Rs 5 each. But after bringing to the shop, he saw that 8 eggs are broken and 7 eggs are rotten. If he sells each egg at Rs. 6, then what will be the profit/loss percentage of Niyamatchacha – let us calculate and write.

Solution: C.P. of 150 eggs at Rs. 5 each = Rs 150 x 5 = Rs. 750.
Now 8 eggs are broken & 7 eggs are rotten.
Remaining no. of eggs (150-15) = 135
S.P. of 135 eggs at Rs. 6 each = Rs. 135 x 6 Rs. 810.
Profit Rs. (810-750) = Rs. 60

Profit % = \(\frac{60 \times 100}{750}\)

Profit = 8%

Question 10. Asifchacha sold a toy at 5% profit. If the cost price of the toy is less by 20% and the selling price is less by Rs 34, then Asifchacha would make a 10% profit. Let us calculate the cost price of the toy.

Solution: Profit = 5%

∴ The selling price of the toy = Rs. \(\frac{105}{100} \times x=\text { Rs. } \frac{21 x}{20}\)

Less in cost price = Rs. x 20%

= \(\text { Rs. } \frac{x \times 20}{100}=\text { Rs. } \frac{x}{5}\)

∴ New cost price = \(\left(x-\frac{x}{5}\right)=\operatorname{Rs} \cdot \frac{4 x}{5}\)

2nd time selling price = \([\text { Rs. }\left(\frac{21 x}{20}-34\right)\)

New selling price at 10% profit on cost profit = \(\text { Rs. } \frac{110}{100} \times \frac{4 x}{5}=\text { Rs. } \frac{22 x}{25}\)

According to the problem, \(\frac{21 x}{20}-34=\frac{22 x}{5}\)

or, \(\frac{21 x}{20}-\frac{22 x}{25}=34\)

or, \(\frac{105 x-88 x}{100}=34\)

or, \(\frac{17 x}{100}=34\)

or, \(x=\frac{34 \times 100}{17}\)

or, x = 200

=Rs.200

Class 9 Maths WB Board

Question 11. There is a loss of 4% by selling 12 commodities at Rs. 1. To make 44% profit how many commodities have to be sold at Rs. 1?

Solution: S.P. of 12 commodities = Rs.1.
Loss = 4%

∴ C.P. of 12 commodities = \(\text { Rs. } \frac{100}{96} \times 1 \mathrm{Rs} .=\frac{25}{24}\)
Profit = 44%

=\(\frac{1.44}{100} \times \frac{100}{96}=\text { Rs. } \frac{3}{2}\)

∴ Rs. 3/2 is the S.P. of 12 commodities

∴ 1 is the S. P. of \(\frac{12 \times 2}{3}\) = 8 commodities

Question 12. By producing two saris, Rama aunty sold a sari at 15% profit and another at 20% profit. She has made a total profit of Rs. 262.50. If the ratio of production costs of two saris is 1 : 3, what will be the production cost of each of the two saris?

Solution: Let the production price of two series be Rs. x & Rs. 3x respectively. Total profit 15% on Rs. x + 20% on Rs. 3x

\(\begin{aligned}
& =\text { Rs. }\left(\frac{x \times 15}{100}+\frac{3 x \times 20}{100}\right) \\
& =\text { Rs. } \frac{15 x+60 x}{100} \\
& =\text { Rs. } \frac{75 x}{100} \\
& =\text { Rs. } \frac{3 x}{4}
\end{aligned}\)

Class 9 Math Solution WBBSE

According to the problem,
\(\begin{aligned}
& \frac{3 x}{4}=262.50 \\
& \text { or, } x=\frac{26250 \times 4}{100 \times 3}
\end{aligned}\)

or, x = 350
∴ C.P. of 1st Sari = Rs. 350
& C.P. of 2nd sari = Rs. 3 x 350 = Rs. 1050

Question 13. One man bought some toffees at the rate of Rs. 2 for 15 pieces. He sold them at the rate of half of the money for 5 pieces and at the rate of the remaining half of the money for 10 pieces. What will be his profit/loss percentage?

Solution: Let the man buy x toffees at Rs. 2 for 15 toffees.

∴ Cost price of x toffees = \(\text { Rs. } \frac{2 x}{15}\)

∵ 1st the time he sold \(\frac{x}{2}\) toffees at the rate of Re. 1 for 5 toffees.

∴ Selling price \(\frac{x}{2}\) toffees = \(\text { Rs. } \frac{1}{5} \times \frac{x}{2}=\text { Rs. } \frac{x}{10}\)

∵ 2nd time he sold \(\frac{x}{2}\) toffees at the rate of Re. 1 for 10 toffees.

∴ Selling price of \(\frac{x}{2}\) toffees = \(\text { Rs. } \frac{x}{2} \times \frac{1}{10}=\text { Rs. } \frac{x}{20}\)

Total selling price = \(\text { Rs. }\left(\frac{x}{10}+\frac{x}{20}\right)\)

\(=\text { Rs. } \frac{2 x+x}{20}=\text { Rs. } \frac{3 x}{20}\)

Profit Selling price – cost price

\(\begin{aligned}
& =\text { Rs. }\left(\frac{3 x}{20}-\frac{2 x}{15}\right) \\
& =\text { Rs. } \frac{9 x-8 x}{60}=\text { Rs } \frac{x}{60}
\end{aligned}\)

Class 9 Math Solution WBBSE

∴ Profit % = \(\frac{x}{60} \frac{1}{2 x / 15} \times 100\)

∴ Profit = \(\frac{25}{2} \%=12 \frac{1}{2} \%\)

= \(12 \frac{1}{2} \%\)

Question14. Afsarchacha made two wooden chairs at same price and he put the marked price for each chair as Rs. 1250. He made a profit of 15% by selling one chair at 8% discount. If he sold the second chair at Rs. 1120, then let us calculate his overall percentage of profit.

Solution: Let the marked price of each chair = Rs. 1250

S.P. of one chair at 8% discount \(=\frac{1250 \times 8}{100}\) = Rs. 100

S.P of 1st chair = Rs. (1250-100) = Rs. 1150
Profit = 15%

The production cost of 1st chairs = \(\text { Rs. } \frac{100}{115} \times 1150\) = Rs. 1000

∴ Total cost of two chair = Rs. (1000+ 1000) = Rs. 2000

Total selling price = Rs. (1150+1120) = Rs. 2270
Profit Rs. (2270-2000) = Rs. 270

Profit % = \(\text { Rs. } \frac{270 \times 100}{2000}\)

Profit =\(\frac{27}{2} \%\)= 13.5%

Overall profit = 13.5%.

Ganit Prakash Class 9 Solutions

Question 15. The market price of a special type of pen is Rs. 36.50. By selling the pen to Shuvam with a discount of Rs. 2.90, Rafikchacha makes a profit of 12%. If he sold a pen of that type to Mita at Rs. 34.50, then let us find out his percentage profit in the second pen.

Solution: Marked price of a pen = Rs. 36.50
Discount Rs. 2.90

∴ S.P. Rs. (36.50 -2.90) = Rs. 33.60
Profit = 12%

∴ C.P. of the pen = \(\text { Rs. } \frac{100}{112} \times\) = Rs. 30

S.P. of the 2nd pen = Rs. 34.50
Profit Rs. (34.5030) Rs. 4.50

Profit \(\begin{aligned}
& =\frac{4.50 \times 100}{30} \% \\
& =\frac{450 \times 100}{30 \times 100}=15 \%
\end{aligned}\) =15%

Question 16. A publisher spent Rs. 3,875 for buying papers, Rs. 3,315 for printing, and Rs. 810 for binding of 2000 copies of books. He sold to booksellers and made a profit 20% after giving discount of 20%. Let us determine the marked price of each book.

Solution: Total production price of 2000 books = Rs. (3,875 + 3,315+ 810) = Rs. 8000.
Profit 20% of Rs. 8,000

= \(\frac{8000 \times 20}{100}=1600\)

Selling price of 2000 books = Rs. (8000+1600) = Rs. 9600.
Let the marked price of each book = Rs. 100.
∴ Discount of Rs. 100

= \(\frac{100 \times 20}{100}\) = Rs. 20

∴ Selling price = Rs. (100-20) = Rs. 80.
∵ When Rs. 80 is S.P., the marked price = Rs. 100

∴ When Re. 1 is the S.P., the Marked price = Rs. \(\frac{100}{80}\)

∴ When Rs. 9600 is the S.P., the Marked price = Rs. \(\frac{100}{80} \times 9600\) = 12000

∴ Marked price of 200 books = Rs 12000

∴ Marked price of 1 book = Rs. \(\frac{12000}{2000}\) = 6

∴ Marked price of each book = Rs. 6.

Ganit Prakash Class 9 Solutions

Question 17. Hasimabibi sold each of two handloom factories at Rs. 1248. She makes a profit of 4% for the first, but makes a loss of 4% for the 2nd. What is her overall profit or loss?

Solution: S.P. of each handloom = Rs. 1248

Is 1st case, profit = 4%

∴ C.P = Rs. \(\frac{100}{104} \times 1248\) Rs. 1200

In 2nd case, loss = 4%

∴ C.P. = Rs. \(\frac{100}{96} \times 1248\) = Rs. 1300

Total C.P. = Rs. (1200+ 1300) = Rs. 2500
Total S.P. = Rs. 2 x 1248 = Rs. 2496
Loss = Rs.(2500-2496) = Rs. 4

Question 18. Karim makes a loss of 19% by selling a mobile phone to Mohan at Rs. 4860. If Mohan sells to Rahim at the same price in which Karim sells to Mohan, then Karim makes a profit of 17%. What is the percentage profit of Mohan?

Solution:
∴ Karim’s SP = Rs. 4860
Loss = 19%

∵ C.P. = \(\frac{100}{81} \times 4860\) = Rs. 6000

Now if Karim sells it at a profit of 17%

His S.P. will be Rs. \(\frac{117}{100} \times 6000\) = Rs. 7020

Mohan’s profit = Rs. (7020-4860) = Rs. 2160

∴ Mohan’s profit% = \(\frac{2160}{4860} \times 100 \%\)

Profit = \(44 \frac{4}{9} \%\)

Question 19. Firojchacha got total Rs. 719.50 by selling a pant at 20% profit and a shirt at 15% profit. If he would sell the pant at 25% profit and the shirt at 20% profit, then he would get Rs. 30.50 more. Let us calculate the cost prices of the pant and the shirt.

Solution: Let the cost of one pant = Rs. x & the cost price of the shirt = Rs. y.

Total S.P. of the pant at 20% and the shirt at 15% profit

\(\begin{aligned}
& =\text { Rs. }\left(\frac{120}{100} \times x+\frac{115}{100} \times y\right) \\
& =\text { Rs. }\left(\frac{120 x+115 y}{100}\right)
\end{aligned}\)

Ganit Prakash Class 9 Solutions

Total S.P. of the pant at 25% profit and the shirt at 20% profit

\(\begin{aligned}
& =\text { Rs }\left(\frac{125}{100} \times x+\frac{120}{100} \times y\right) \\
& =\text { Rs. }\left(\frac{125 x+120 y}{100}\right)
\end{aligned}\)

 

According to 1st condition, \(\frac{120 x+115 y}{100}=719.50\)

or, \(\frac{120 x+115 y}{100}=\frac{71950}{100}\)

or, 120x+115y = 71950 ….(1)

According to 2nd condition, \(\frac{125 x+120 y}{100}=719.50+30.50=750\)

or, 125x+120y = 75000 …..(2)

\(\begin{aligned}
& 125 x+120 y=75000 \\
& 120 x+115 y=71950 \\
& (-) \quad(-)
\end{aligned}\)

 

Subtracting, 5x+5y=3050
or, 5(x+y)= 3050

or, x + y = \(\frac{3050}{5}\)

or, x + y = 610
or, y = 610-x

Putting the value of y in equation (2) we get,
125x+120 (610 – x) = 75000
or, 125x+73200-120x=75000
or, 5x= 75000-73200
or, 5x = 1800

or, \(x=\frac{1800}{5}\)

or, x = 360

Putting the value of x in equation (3) we get,
∴ y = 610-360
or, y = 250

∴ Cost price of one pants = Rs. 360
Cost price of one pant = Rs. 250

WBBSE Class 9 Maths Solutions

Question 20. Rabi uncle bought rice at Rs. 3000. He sold \(\frac{1}{3}\) rd part of the rice at a 20% loss and \(\frac{2}{5}\) th part of the rice at 25% profit. At what percentage profit the remaining part of rice is to be sold to get an overall 10% profit?

Solution: Total cost price = 3000

Profit = 10% = \(\frac{10}{100} \times 3000\)=3000

∴ Total selling price = Rs. (3000+300) = 3300

1st time he sold \(\frac{1}{3}\) part of rice = \(\frac{1}{3}\) x 3000 = Rs. 1000 at a loss 20%

∴ Selling price = Rs. \(1000 \times \frac{80}{100}\) = 800

2nd time he sold \(\frac{2}{5}\) the part of rice at a profit = \(\frac{2}{5}\) Χ 300 = 1200 = 25%

∴ Selling price = Rs. \(1200 \times \frac{125}{100} \times 1500\)

Price of remaining rice = Rs. 3000 (1000+ 1200) Rs. 800
Remaining selling price = 3300 – (1000 + 1500) = Rs (33002300) = Rs 1000

Profit = Rs (1000 – 800) Rs. 200

Profit % = \(\frac{200}{800} \times 100=25 \%\)

Profit = 25%

Question 21. A trader by selling one kind of tea at Rs. 80/kg makes a loss of 20% and makes a profit of 25% by selling another kind of tea at Rs. 200/kg. At what ratio these two types of tea should be mixed so that by selling it at Rs. 150/ kg the profit would be 25%?

Solution : Selling price of 1st type of tea = Rs. 80 kg
Loss = 20%

∴ C.P. of 1st type of tea = Rs. \(\frac{100}{80} \times 80=100\)= 100 per kg

Again, S.P. of 2nd type of tea = Rs. 200 per kg
Profit = 25%

∴ The cost price of 2nd type of tea = is Rs. \(\frac{100}{25} \times 200\) = \(\frac{4}{5} \times 200=\text { Rs. } 160\)

Let the two types of tea be mixed in the ratio x: y.
∴ C.P. of x kg 1st type of tea = Rs. 100x
& CP of y kg 2nd type of tea = Rs. 160y

Total CP of (x + y). kg tea = Rs.(100x + 160y) (100x +160y)

∴ C.P. of 1 kg mixed tea = Rs. \(\frac{(100 x+160 y)}{x+y}\)

∵ S.P. of 1 kg mixed tea at 25% profit = Rs. \(\frac{100 x+160 y}{x+y}=\frac{125}{100}\)

According to the problem,

or, \(\frac{100 x+160 y}{x+y}=\frac{125}{100}=150=30\)

or, 120x + 120y= 100x + 160y
or, 120x100x = 160y-120y
or, 20x = 40y

or, \(\frac{x}{y}=\frac{40}{20}\)

or, \(\frac{x}{y}=\frac{2}{1}\)

or, x: y = 2:1
∴ Required ratio = 2:1

WBBSE Class 9 Maths Solutions Chapter 10 Profit And Loss Exercise 10.2

Question 1. Subalbabu of Antpur, by producing rice, sells it to a wholesaler Sahanabibi at 20% profit. Sahanabibi sells that rice to the shopkeeper Utpalbabu at 10% profit. But if Utpalbabu sells this rice at 12% profit, then let us find out the answers of the following questions by drawing pictures on a straight line 

(1)Subalbabu has spent Rs. 7500 to produce some amount of rice. Let us calculate and write with how much money Sahanabibi hás bought that. amount 6 : rice.

(2)To produce the same amount of rice Subalbabu has spent Rs. 2500, with how much money Utpalbabu will sell that amount of rice let us calculate and write it.

(3) The price at which Utpalbabu sells rice to us, if Subalbabu sells directly at that price then what will be the profit percentage of Subalbabu – let us calculate and write it.

Solution:

Subalbabu sells rice at a 20% profit		Sahanabibi sells rice at a 10% profit		Utpalbabu sells rice at 12% profit.
C. P.	S. P.	C. P.	S. P.	C. P.	S. P.
Rs. 100	Rs. 120	Rs. 120	Rs. 132	Rs. 132	\(\text { Rs. } 147 \frac{21}{25}\)

 

First we draw a straight line.

(1)Cost price of Subalbabu = Rs. 7500
Profit = 20%

S.P. = Rs. \(\frac{120}{100} \times 7500\) = Rs. 9000

Sahanabibi’s cost price = Rs. 9000

(2)Cost price of Subalbabu= Rs. 2500
Profit = 20%

∴ S.P. of Subalbabu = Rs. \(\frac{120}{100} \times 2500\) = Rs. 3000

C.P. of Sahanabibi = Rs. 3000,
Profit = 10%

S.P. of Sahanabibi = Rs. \(\frac{110}{100} \times 3000\) = Rs. 330

C.P. of Utpalbabu = Rs. 3300

∴ Profit = 12%

S.P. of Utpalbabu = \(\frac{112}{100} \times 3300\) = Rs. 3696

∴ Utpalbabu sells that amount of rice at Rs. 3696.

(3)C.P. of Subalbabu = Rs. 2500
S.PRs. 3696
Profit = Rs. (3696-2500) = Rs. 1196

Profit = \(\frac{1196 \times 100}{2500} \%=47 \frac{21}{25} \%\)

Profit of Subalbabu = \(47 \frac{21}{25} \%\)

Question 2. In a bazaar, at the time of selling jute bags, the producer, wholesaler, and retailer make profits of 15%, 20%, and 25% respectively. Now if a bag reaches the buyer through the producer, wholesaler, and retailer, then let us find out the answers of the following questions 

1. Let us calculate and write the production cost of a bag that is bought by a buyer at Rs. 138.
2. Let us calculate and write the price of the bag at which the buyer will buy when its production cost is Rs 140.
3. The bag which a retailer has bought at Rs. 98, let us calculate and write how much money will be given by a buyer for that bag.
4. The bag which the wholesaler has bought at Rs. 175, let us calculate and write how much money a buyer will give to buy that bag.
5. The bag which a buyer has bought at Rs. 276 if that bag would have been bought directly from the wholesaler then how much money would be saved-let us calculate and write it.

Solution: Let the production cost of a jute bag = Rs. x.
Profit of the producer = 15%

∴ S.P. of the producer = \(\frac{115}{100} \times x\) = Rs. \(\frac{23 x}{20}\)

∴ C.P. of the wholesaler = Rs. \(\frac{23 x}{20}\)

Profit = 20%

S.P. of the wholesaler Rs. = \(\frac{120}{100} \times \frac{23 x}{20}\) = Rs. \(\frac{138 x}{100}\)

∴ C.P. of the retailer = Rs. \(\frac{138 x}{100}\)

∴ Profit = 25%

S. P. of the retailer = Rs.\(\frac{125}{100} \times \frac{138 x}{100}\) = Rs. \(\frac{69 x}{40}\)

According to the problem, = Rs. \(\frac{69 x}{40}=138\)

∴ \(x=\frac{138 \times 40}{69}=80\)

∴ The production price of the bag = Rs. 80.

2. 2nd time production cost = Rs. 140

∴ C.P. of the buyer = Rs. \(\frac{69 x}{40}\)

= Rs. \(\frac{69}{40} \times 140\) = Rs. 241.50

3. C.P. of retailer = Rs. 98
Profit = 25%

∴ S.P. = Rs. \(\frac{125}{100} \times 98\) = Rs. 122.50.

Buyer has to pay Rs. 122.50

4. C.P. of the wholesaler = Rs. 175.
Profit = 20%

∴ S.P. of the wholesaler = Rs. \(\frac{120}{100} \times 175\) = Rs. 210.

C.P. of retailer = Rs.
Profit = 25%

S.P. of the retailer = Rs. \(\frac{125}{100} \times 210\) = Rs. 262.50

The bag which has been bought by the wholesaler at Rs. 175, has to be bought by consmer at a price of Rs. 262.50.

5. C.P. of the buyer = Rs. 276

∴ \(\frac{69 x}{40}=276\)

or, \(x=\frac{216 \times 40}{69}\)

or, x = 160

∴ Production price = Rs. 160

∴ S.P. of wholesaler = Rs. \(\frac{138 x}{100}\)

= Rs. \(\frac{138 \times 160}{100}\). = Rs. 220.80

∵ C.P. of the buyer = Rs. 276
∴ If the bag is bought from a wholesaler the buyer will save = Rs. (276-220.80) = Rs. 55.20

Question 3. The production cost and the cost prices of a cycle at different levels are-

Production Cost (Rs.)	C.P.of wholesaler (Rs.)	C.P. of Retailer	C.P.of Buyer (Rs.)
1050	1260	1449	1666.35

 

1. Let us calculate, by selling cycle, how much profit percentage the retailer has made.
2. Let us calculate and observe that by selling cycle, what the profit percentage the wholesaler has made.
3. Let us calculate and write the profit percentage that the producer has made by the selling cycle.
4. Let us calculate and write that how much profit percentage has to be given more by a buyer than the production cost to buy a cycle.
5. If buyer buys a cycle directly from the producer and the producer he a profit of 30%, then how much money the buyer will save- let us calculate and write it.

Solution:

1. Profit of the retailer = Rs. (1666.351449) = Rs. 217.35

Profit = \(\frac{217.35 \times 100}{1449}=15 \%\)

Profit of the retailer = 15%

2. Profit of the wholesaler = Rs. (14491260) = Rs. 189

Profit % = \(\frac{189 \times 100}{1260}=15 \%\)

Profit of the producer = 20%

3. Profit of the producer Rs. (1260-1050) = Rs. 210

Profit %= \(\frac{210 \times 100}{1050}=20 \%\)

Profit of the producer = 20%

4. Buyer pays more than the production cost = Rs. (1666.35-1050) = Rs. 616.35

Profit % given more than the production cost = \(\frac{616.35 \times 100}{1050}=58.7 \%\)

Buyer gives 58.7% more than the production cost.

5. S.P. at 30% profit of the production cost = Rs. \(\frac{130}{100} \times 1050=\text { Rs. } 1365\)

∴ Buyer saves = Rs. (1666.35 – 1365) = Rs. 301.35
If the buyer buys from the producer then he saves Rs. 301.38.

Question 4. Multiple choice questions

1. The ratio of cost price and selling price is 10: 11, the profit percentage is

1.  9
2. 11
3. 101
4. 10

Solution: Let the C.P. & S.P. be Rs. 10x & Rs. 11x respectively.
Profit Rs. (11x-10x) = Rs. x

Profit% \(\frac{x \times 100}{10 x}=10\)

∴ 4. 10

2. By buying a book at Rs. 40 and selling it at Rs. 60, the profit percentage will be

1. 50
2. 33
3. 20
4. 30

Solution: Profit = Rs. (60-40)= Rs. 20

Profit % = \(\frac{20 \times 100}{40}=50\)

∴ 1. 50

3. A shirt is sold at Rs. 360 and there is a loss of 10%. The cost price of the shirt is

1. Rs. 380
2. Rs. 400
3. Rs. 420
4. Rs. 450

Solution: S.P. of shirt = Rs. 360
Loss = 10%

C.P. of shirt = Rs. \(\frac{100}{90} \times 360\) = Rs. 400

∴ 2. Rs. 400

4. After 20% discount the selling price of a geometry box becomes Rs. 48. The marked price of the geometry box is

1. Rs. 60
2. Rs. 75
3. Rs. 80
4. Rs. 50

Solution: Let the marked price of the geometry box is Rs. x.

Discount of 20% on Rs. x = Rs. \(\frac{x \times 20}{100}=\text { Rs. } \frac{x}{5}\)

∴ Selling price = Rs. \(\text { Rs. }\left(x-\frac{x}{5}\right)=\text { Rs. } \frac{4 x}{5}\)

According to the problem, \(\frac{4 x}{5}=48\)

or, \(x=\frac{48 \times 5}{4}\)

∴ Marked price of the geometry box = Rs. 60

∴ 1. Rs. 60

5. A retailer buys medicine at 20% discount on the marked price and sells to buyer at a marked price. The retailer makes a profit percentage

1.  20
2.  25
3. 10
4. 30

Solution: Let marked price = Rs. 100

Discount 20% of Rs. 100 = Rs. \(\frac{100 \times 20}{100}=\text { Rs. } 20\)

∴ C.P. of the medicine = Rs. (100-20) = Rs. 80
S.P. of the medicine = Marked price.
Profit = Rs. 100
Profit Rs. (100-80) = Rs. 20

Profit% = \(\frac{20 \times 100}{80}=25\)

∴ 2. 25

5. Short answer type questions :

1. If 20% of profit is on cost price, what is the profit percentage on selling price?

Soltion: Let cost price = Rs. 100

Profit 20% of Rs. 100 = \(\frac{100 \times 20}{100}=\text { Rs. } 20\)

Selling price Rs. (100 +20) = Rs. 120

Profit on S.P. = \(\frac{20 \times 100}{120}\)

Profit = \(16 \frac{2}{3} \%\)

∴ Profit on selling price = \(16 \frac{2}{3} \%\)

2. If 20% profit is on the selling price, what is the profit percentage on cost price?

Solution: Let selling price = Rs. 100

Profit = Rs. of Rs. 100 = \(\frac{100 \times 12}{100}=\text { Rs. } 20\) = Rs. 20

Cost price = Rs. (100-20) = Rs. 80
∴ Profit percentage on cost price = 25%

3. By selling 110 mangoes, if the cost price of 120 mangoes has been got, what will be the profit percentage?

Soltion: Let C.P. of 1 mango = Re. 1
∴ C.P. of 110 mangoes Rs. 110 x 1 = Rs. 110
∴ C.P. of 120 mangoes Rs. 120
∵ S.P. of 110 mangoes = C.P. of 120 mangoes
∴ Profit Rs. (120-110) = Rs. 10

Profit % = \(\frac{10 \times 100}{110}=\frac{100}{11}=9 \frac{1}{11}\)

∴ Profit % = \(9 \frac{1}{11}\)

4. To submit the electricity bill in due time, 15% discount can be obtained. Sumonbabu has got Rs. 54 as a discount for the submission of the electricity bill in due time. How much was his electricity bill?

Solution: Let electricity bill = Rs. x

∴ Discount = Rs. x of 15% = Rs. \(\frac{x \times 15}{110}=\text { Rs. } \frac{3 x}{20}\)

\(\frac{3 x}{20}=54\)

or, \(x=\frac{54 \times 20}{3}\)

or, x = 360
His electricity bill = 360.

5. A commodity is sold at 480 with a loss of 20% on the selling price, what is the cost price of the commodity?

Solution: Let S.P. = Rs. 100

Loss 20% of Rs. 100 \(=\frac{100 \times 20}{100}=\text { Rs. } 20\)

∴ C.P. = Rs. (100+ 20) = Rs. 120
∴ When S.P. = Rs. 100, C.P. = Rs. 120

∴ When S.P. Re 1, C.P. = Rs. \(\frac{120}{100}\)

∴ When S.P. Rs. 480, C.P. = Rs. \(\frac{120}{100} \times 480\)

∴ Cost price of the commodity = Rs. 576.

6. If a commodity is sold with successive discounts of 20% and 10%, what will be the equivalent discount?

Solution: Let the price = Rs. 100
1st discount = 20% of Rs. 100

= Rs. \(\frac{100 \times 20}{100}=\text { Rs. } 20\)

∴ S.P. = Rs. (100 – 20) = Rs. 80
2nd discount = 10% of Rs. 80

\(\frac{80 \times 10}{100}=\text { Rs. } 8\)

 

Total discount = Rs. (20+8)= Rs. 28
∵ As total discount (Rs. 20+ Rs. 8) = Rs. 28 on Rs. 100
∴ Equivalent discount = 28%.

 

Class IX Maths Solutions WBBSE Chapter 6 Properties Of Parallelogram Exercise 6.1

Question 1. By calculating let us write the angles of the parallelogram ABCD, when ZB=60°.

Solution:

 

 

In parallelogram ABCD, ∠B = 60°
∴ ∠D = ZB = 60°
∴∠A+ZB=180°
∴∠A +60° 180°
∴∠A=180°- 60° 120°.
∴∠CZA =120°

Read and Learn More WBBSE Solutions For Class 9 Maths

Question 2. In the picture of the parallelogram aside, let us calculate and write the value of ∠PRQ of PQRS.

Solution:

 

In parallelogram PQRS
∠PQR + ∠QRS =180°
or, ∠PQR +∠PRQ+∠PRS =180°
or, 55° + PRQ+70°=180°
or, ∠PRQ=180°-125°
∴ ∠PRQ = 55°

Class IX Maths Solutions WBBSE

Question 3. In the picture aside, if AP and DP are the bisectors of ∠BAD and ∠ADC respectively of the parallelogram ABCD, then by calculating value of ∠APD.

Solution:

 

In parallelogm ABCD in AP, ∠BAD is the bisector.
∴ ∠PAB = ∠PAD

Similarly, ∠PDC=∠PDA
ABCD is a parallelogram.
∴ ∠DAB + ∠ADC = 180°
∴ ∠PAD + ∠PAB +∠PDC + ∠PDA=180°
∴ ∠PAD + ∠PAD +∠PDA + ∠PDA =180°
or, 2 ∠PAD +2 ∠PDA =180°
or, ∠PAD +∠PDA =90°

Now in ΔAPD
∠APD + ∠PAD + ∠PDA = 180°
or, ∠APD+90°-180°
or, ∠APD = 180°-90°-90°

Question 4. By calculating, I write the values of the angles X and Y in the following rectangle PQRS.

Solution:

 

(1)∵∠SQR=25°
∴ ∠PSQ = ∠SQR=25°
∵ ∠PQRS

∴ ∠PSQ +∠QSR = 90°
or, 25°+∠QSR = 90°
or, ∠QSR = 90°- 25° = 65°

We know the Δ QPS & PSR are similar.
∴ ∠ PSQ = ∠SPR = 25°
∴ y=180° – (25°+25°)=130°
∴ x° = ∠QSR = 65°
∠x=65° and y =130°

Class IX Maths Solutions WBBSE

(2) In PQRS rectangle ∠QOR=100°

 

 

∴ ∠POS=100°
∴ 2y°=80°
or, y°=40°
x°-90°-40°= 50°
∴ x=50° and y=40°.

Question 5. In the figure aside, ABCD and ABEF are two parallelograms. I prove with the help of reason that CDFE is also a parallelogram.

Solution:

 

 

 

∵ ABCD is a parallelogram.
∴ AB||CD and AB=CD

Again, ABEF is a parallelogram.
∴ AB||EF and AB=EF
∵ AB-CD and AB=EF
∴ CD=EF (both are equal to AB)
∵ ABCD and AB||EF
∴ CD||EF

□CDEF in CD=EF and CD||EF
∴ CDEF is a parallelogram.

Class IX Maths Solutions WBBSE

Question 6. If in the parallelogram ABCD, AB > AD, then I prove with the help of reason that ∠BAC < ∠DAC.

Solution:

 

 

In parallelogram, AB>AD
∴ ∠BAC < ∠DAC

As AD||BC & AC is the transversal.
∴ ∠DAC = alternate ∠ACB

We know that the angle opposite to the greater side is greater.
∴ In ΔABC in ∠ACB>∠BAC
∴ ∠DAC>∠BAC
i.e., ∠BAC<∠DAC Proved.

 

Class 9 Mathematics West Bengal Board Chapter 6 Properties Of Parallelogram Exercise 6.2

 

Question 1. Firoz has drawn a quadrilateral PQRS whose PQ = SR and PQ II SR. I prove with reason that PQRS is a parallelogram.

Solution:

 

 

In quadrilaleral PQRS, PQ = SR &PQ||SR.
To prove PQRS is a parallelogram.
Construction diagonal PR is drawn.

Proof: In ΔPQR and ΔPSR

(1)PQ = SR (given)
∠QPR alternate ∠PRS (∴ PQ||SR & PR is a transversal) PR is the common side.
∴ ΔPQRA = ΔPSR
∴ ∠PRQ = ∠SPR

They are alternate angles.
∴ QR||PS
∴ In quadrilateral PQRS, PQ||SR & QRIPS.
∴ PQRS is a parallelogram.

Question 2. Sabba has drawn two straight lines such that, AD II BC and AD = BC. I prove with reason that AB = DC and AB II DC.

Solution:

 

 

Two straight lines AB & CD are such that AD||BS and AD=BC. To prove (1) AB = DC; (2) AB II DC.

Proof: Join A, B and C, D and B, D
To prove AB=DC and AB||DC.

Construction: Join B & D.

Proof:  ∵ AD||BC & BD is a transversal.
∴ ∠ADB alternate ∠CBD

In ΔABD and ΔCBD,
∠ADB = ∠CBD (Proved)

AD = BC (given) and BD is the common side
∴ ΔABDA ≅ ΔCBD
∴ AB = DC……(1)

Again ∠ABD = ∠BDC
As they are alternate angles
∴ AB||DC…….. (2) Proved

Class 9 Mathematics West Bengal Board

Chapter 6 Properties Of Parallelogram Exercise 6.3

Question 1. Let us prove that, if the lengths of two diagonals of a parallelogram are equal, then the parallelogram will be a rectangle.

Solution:

 

 

In parallelogram ABCD, diagonals AC & BD are equal, i.e., AC BC. To prove, ABCD is a rectangle.

Proof: In ΔADC and  ΔBCD,
AD =  BC (opposite side of parallelogram ABCD)
AC = BD (given)
CD is the common side.

∴ Δ ADC ≅ΔBCD (S-S-S)
∴ ∠ADC = ∠BCD

Again, ∠ADC = ∠ABC (opposite angles of the parallelogram)
∴  In parallelogram ABCD, ∴ ∠ADC=∠BCD =∠ABC = ∠BAD
∴ ∠A+∠B+∠C+∠D = 360°
∴ 4∠A = 4 right angle
∴ ∠A = 1 right angle

∴ In parallelogram ABCD ∠A = ∠B = ∠C= ∠D=1 right angle
∴ ABCD is a rectangle. Proved

Question 2. Let us prove that, if in a parallelogram, the diagonals are equal in lengths and intersect at right angles, the parallelogram will be a square.

Solution:

 

 

In parallelogram ABCD, diagonals AC and BD are equal & they intersect each other at right angles.
To prove that ABCD is a square.

Proof: In parallelogram ABCD, diagonals AC & BD bisect each other at O at right angles.
∴ ∠AOB = ∠AOD = 90°
ΔAOB and ΔAOD, ∠AOB = ∠AOD

OB = OD & OA is a common side.
∴ Δ ΑΟΒ ≅ Δ AOD
∴ AB = AD  AB=BC & BC=CD

∴ In parallelogram ABCD, AB = BC = CD = DA
∵ OA=OB
∴ ∠OAB = ∠OBA

Similarly, ∠OAD = ∠ODA

∵ ∠DAB+ ∠ABC = 180°
∴ 2∠DAB = 180°
∴ ∠DAB = 90°
∴ Parallelogram □ABCD is a square.

Question 3. Let us prove that, a parallelogram whose diagonals intersect at right angles, is a rhombus.

Solution:

 

 

In parallelogram ABCD diagonals AC & BD cut each other perpendicularly.
To prove ABCD is a rhombus.

Proof: In ΔAOB and ΔBOC.
OA=OC ( ∵ Diagonals of a parallelogram bisect each other.)
∠AOB = ∠BOC (∵ Each angle is a right angle) and OB is a common side.

∴ ΔOBA ≅  ΔBOC (S-A-S Congruency)
∴AB = BC, similarly BC = CD, and CD = AD.
∴AB = BC = CD = DA
∴ ABCD is a rhombus.

Class 9 Maths WB Board

Question 4. The two diagonals of a parallelogram intersect each other at point O. A straight line passing through O intersects the sides AB and DC at points P and Q respectively. Let us prove that OP=OQ.

Solution:

 

 

Two diagonals AC & BD of the parallelogram ABCD intersect each other O.
Any straight line passing through & O cut at P & Q OP = OQ.

Proof: In ΔAOP and ΔCOQ,
OA=OC (Diagonals of a parallelogram bisect each other)
∴ ∠OAP= alternate ∠OCQ ( AB||DC and AC is a traversal) and ∠AOP vertically opposite ∠COQ (A-A-S)
∴ ΔOPA ≅ ΔCOQ
∴ OP = OQ. Proved

Question 5. Let us prove that in an isosceles trapezium, the two angles adjacent to the parallel sides are equal.

Solution:

 

 

ABCD is an isosceles trapezium whose sides sides AD = BC.
To prove ∠DAB = ∠CBA.

Construction: Draw straight line from parallel to DA, which cuts AB at Q.

Proof: ∵AB||DC
∴  AQ||DC
∵ ADIIQC (acc. to construction)
∴ AQCD is a parallelogram.

∴ AD = QC
∴ AD = BC (given)
∵ QC = BC (both are equal to AD)

In ΔBCQ,∠CQB = ∠CBQ (∵ BC = QC)
∠DAQ = ∠CQB (∵ DA||CQ and AB is the tranversal)
∴ ∠DAQ = ∠CBQ
∴ ∠DAB = ∠CBA. Proved

Class 9 Maths WB Board

Question 6. In a square ABCD, P is any point on the side BC. The perpendicular drawn on AP from point B intersects the side DC at point Q. Let us prove that AP = BQ.

Solution:

 

 

In a square ABCD, P is any point on the side BC.
The perpendicular drawn on AP from the point B intersects the side DC at the point Q. AP and BQ cut each other at O.

To prove: AP = BQ

Proof: ∵ BO ⊥ AP
∴ ∠BOP =1 right angle
∴ ∠OBP+∠OPB = 1 right angle
or, ∠QBC+∠APB = 1 right angle

Again, ∠QBC+∠BQC = 1 right angle
∴ ∠QBC+∠APB = ∠QBC + ∠BQC
∴ ∠APB = ∠BQC

Now, in ΔAPB and ΔBQC,
∠APB = ∠BQC (Proved)
∠ABP =∠BCQ (90°) and AB = BC (∵ ABCD is a square)

∴ ΔAPB ≅ ΔBQC
∴ AP = BQ. Proved

Question 7. Let us prove that if two opposite angles and two opposite sides of a quadrilateral are equal, then the quadrilateral will be a parallelogram.

Solution:

 

 

In quadrilateral ABCD, ZABC = ∠ADC & AB||DC
To prove ABCD is a parallelogram. Join A and C.

Construction: AC diagonal is drawn.
Proof: ∵ AB||DC & AC is the transversal .
∴ ∠BAC = alternate ∠ACD

In ΔABC and ΔADC,
∠ABC = ∠ADC (given)
∠BAC = ∠ACD And AC is the common side.

∴ Δ ABC ≅ ADC (A-A-S Congruency)
∴ ∠ACB = ∠CAD
They are alternate angles.

∴ BC||AD
In □ABCD, AB|DC & BC||AD
∴ ABCD is a parallelogram.

Class 9 Maths WB Board

Question 8. In AABC, the two medians BP and CQ are so extended upto the points R and S respectively that BP = PR and CQ = QS. Let us prove that S,A,R are collinear.

Solution:

 

 

In ΔABC, BP, and CQ medians are produced to R & S points such that BP= PR and CQ = QS.
To prove S, A, and R are collinear.

Construction: S, A; A, R; R, C and S, B are joined. S

Proof: ∵BP is median
∴ AP = PC and BP = PR (given)

In □ABCR diagonals BR and AC bisect each other at P.
∴ ABCR is a parallelogram.
∴ BC||AR

Similarly, ACBS is a parallelogram.
∴ BC||SA

Two straight lines passing through A are parallel to BC.
∴ Two line segments are one straight line.
∴  S, A, and R are collinear. Proved

Question 9. The diagonal SQ of the parallelogram PQRS is divided into three equal parts at points K and L. PK intersects SQ at point M and RL intersects PQ at point N. Let us prove that PMRN is a parallelogram.

Solution:

 

 

To prove PMRN is a parallelogram.
Proof: In ΔPKS and ΔQLR, ∠PSK alternate ∠LQR (∵PS||QR & SQ is the transversal)

SK QL (given)and PS QR (∵ PORS is a parallelogram)
∴ ΔPKS ≅ ΔQLR
∴ ∠PKS = ∠QLR
∴ ∠SKM = ∠QLN

Now, in ΔNQL and ΔSKM,
∠NQL = alternate ∠KSM (∵ PQ||SR & SQ is the transversal)
∠QLN = ∠SKM (proved)

and QL = SK (given)
∴ ΔNQL ≅ Δ SKM
∴ QN = SM
∴ PQ = SR (∵ PQRS is a parallelogram)
∴ PQ – QN = SR – SM
∴ PN = RM

□PMRN PN||RM & PN = RM
∴ PMRN is a parallelogram.

Question 10. In two parallelograms ABCD and AECF, AC is a diagonal. If B, E,D,F are not collinear, then let us prove that, BEDF is a parallelogram.

Solution:

 

 

In ABCD and AECF parallelograms AC is diagonal. If B, E, D, F are not collinear, prove that BEDF is a parallelogram.

Construction: In parallelogram ABCD the diagonal BD and in parallelogram AECF the diagonal EF are drawn. Both diagonals intersect each other at O.

Proof: In parallelogram ABCD diagonals AC and BD intersect each other at O.
∴ OA = OC and OB OD….(1)

Similarly, in parallelogram AECF diagonals AC and EF intersect each other at O.
∴ OA = OC and OE = OF…(2)

BEDF is a parallelogram and BD & EF are two diagonals.
OB = OD [from (1)]
OE = OF [from (2)]

∴ □BEDF diagonals BD and EF bisect each other.
∴ BEDF is a parallelogram. Proved

Class 9 Math Chapter 6 WBBSE

Question 11. ABCD is a quadrilateral. The two parallelograms ABCE and BADF are drawn. Let us prove that, CD and EF bisect each other.

Solution:

 

 

ABCD is quadrilateral. Two parallelograms ABCF & BADF are drawn.
To prove CD & EF bisect each other.

Construction : D, E, and C, F are joined.

Proof: AB = CE (∵ ABCE is a parallelogram \([/square\)) and AB DF (∵BADF is a parallelogram \([/square\))
∴ DF = CF (∵ AB)

Again, AB||DF (∵ ABDF is a parallelogram □)
AB||CE (∵ ABDF is a parallelogram □)
∴ DF||CE (∵ ABCE is a parallelogram □)

Now, opposite sides of □DFCE are parallel.
∴ DFCF is a parallelogram.
∴ CD & EF are the diagonals of the parallelogram DFCE.
∴ CD & EF bisect each other.

Class 9 Math Chapter 6 WBBSE

Question 12. In parallelogram ABCD, AB = 2AD. Let us prove that the bisectors of ∠BAD and ∠ABC meet at the mid-point of the side DC in right angle.

Solution:

 

 

ABCD is a parallelogram and the bisectors of ∠BAD & ∠ABC meet at P on DC such that AB = 2AD.

To prove: (1)P is the mid point of DC;
(2)∠APB = 1 right angle.

Proof: AP is the bisector of ∠BAD
∴ ∠DAP = ∠PAB = ∠DPA (∵ DC||AB)

Again, DP = DA = \(\frac{1}{2} A B=\frac{1}{2} D C\)

∴ PC = DP
∴ P is the mid point of DC …..(1)

Now, ∠PAB + ∠ABP = \(\frac{1}{2}\)∠DAB + \(\frac{1}{2}\)∠ABC
=\(\frac{1}{2}\)(∠DAB+∠ABC)
=\(\frac{1}{2}\)X180° = 90°

∴ ∠APB =180°- (∠PAB+ ∠ABP) =180°-90° = 90° = 1 right angle…..(2) Proved

Question 13. The two squares ABPQ and ADRS drawn on two sides AB and AD of the parallelogram ABCD respectively are outside of ABCD. Let us prove that PRC is an isosceles triangle.

Solution:

 

The two squares ABPQ and ADRS drawn on two sides AB and AD of the parallelogram ABCD respectively are outside of ABCD.

Proof: ∵ABPQ is a square
∴ AB = PB and AB = CD (∵ABCD is a parallelogram □)
∴ PB = CD = AB

Again, ADRS is a square
∴ AD = RD & AD = BC (∵ ABCD is a parallelogram □)
∴ RD BC (∵ As both are equal to AD)

ΔPRC PC = PB + BC = CD + RD (∵ PB = CD and BC = RD) = CR
∴ PRC is an isosceles triangle.

Question 14. In the parallelogram ABCD, ∠BAD is an obtuse angle; the two equilateral triangles ABP and ADQ are drawn on the two sides AB & AD outside of it. Let us prove that CPQ is an equilateral triangle.

Solution:

 

 

In parallelogram ABCD ∠BAD is on an obtuse angle. On two sides AB and AD two equilateral triangles ABP and ADQ are drawn which are outside the parallelogram. Prove that CPQ is an equilateral triangle.

Proof: In ΔAPQ and ΔDCQ,
AQ = DQ (∵ ADQ is an equilateral triangle)
AP CD (∵ AP = AB and AB = CD ∴AP = CD) and PQ = QC [∵ ∠PAQ and ∠CDQ are obtuse angles]

Similarly, it can be proved that, PQ = PC
∴ PQ = CQ = PC
∴ CPQ is an equilateral triangle.

Class 9 Math Chapter 6 WBBSE

Question 15. OP, OQ, and QR are three straight lines. The three parallelograms OPAQ, OQBR, and ORCP are drawn. Let us prove that AR, BP, and CQ bisect each other.

Solution:

 

 

OP, OQ, and OR are 3 straight lines and 3 parallelograms. OPAQ, OQBR, and ORCP is drawn.
Prove that AR, BP, and CQ bisect each other.

Construction: Q, R; A, C; Q, P; and B, C are joined.

Proof: In quadrilateral AQRC
QA||RC (∵ In OPAQ parallelogram □QA||OP and in ORCP parallelogram □OP||RC) and QA RC ( ∵ QA OP, OP = RC)
∴ AQRC is a parallelogram.

∵ In parallelogram AQRC AR and CQ are two diagonals.
∴ AR and CQ bisect each other.

In BCPQ, □BQ||CP (∵ ORCP is a parallelogram CP||OR and OQBR parallelogram OR||BQ) and BQ = CP ( CP = OR, OR = BQ)
∴ BCPQ is a parallelogram.

∴ In BCPQ two diagonals are BP & CQ.
∴ BP & CQ bisect each other.
∴ AR, BP & CQ bisect each other.

Question 16. Multiple Choice Question

1.In parallelogram ABCD, ∠BAD=75° and CBD=60°; the value of ∠BDC is

(1)60°
(2)75°
(3)45°
(4)50°

 

 

Solution: ∵ ABCD is a parallelogram
∴ ∠ABC + ∠BAD = 180°
∵ ∠ABD+ ∠DBC+75° = 180°

or, ∠ABD+60° +75° = 180°
∴ ∠ABD 180°-60°-75° = 45°
∵  AB||DC & BD is the transversal
∴ ∠BDC alternate ∠ABD = 45°
∴ ∠BDC = 45°

∴ (3)45°

2. Let us write which of the following geometric figures has diagonals equal in length:

(1)Parallelogram
(2)Rhombus
(3)Trapezium
(4)Rectangle

Solution: (4) Rectangle

Class 9 Math Chapter 6 WBBSE

3. In the parallelogram ABCD, ∠BAD = ∠ABC, the parallelogram ABCD is a

(1)Rhombus
(2) Rectangle
(3)Trapezium
(4)None of them

Solution:(3)Trapezium

4. In the parallelogram ABCD, M is the mid-point of the diagonal BD; if BM bisects ∠ABC, then the value of ∠AMB is
(1) 45°
(2) 60°
(3) 90°
(4) 75°

 

 

Solution:  ∵ ABCD is a parallelogram.
∴ ∠DAB+ ∠ABC = 180°

or, \(\frac{1}{2}\) ∠DAB+ ∠ABC = 90°

or, \(\frac{1}{2}\) ∠MAB+ ∠MBA = 90°

∴ ∠AMB=90°

∴ (3) 90°

Class 9 Maths Exercise 6.1

5. In the rhombus ABCD, ∠ACB=40°, the value of ∠ADB is

(1) 50°
(2) 110°
(3) 90°
(4) 120°

 

 

 

Solution: ∠ACB = 40°
∴ ∠CBD = 50°
∵ AD||BC & BD is the transversal.
∴ ∠ADB = alternate ∠CBD = 50°

∴ (1) 50°

Question 17. Short answer type questions:

1. In the parallelogram ABCD,∠A: ∠B 3:2. Let us write the measures of the angles of the parallelogram.

Solution:

 

 

 

∵ ABCD is a parallelogram.
∴ ∠A = ∠B= 180°
∴ ∠A: ∠B 3:2

Sum of the ratios = 3 + 2 = 5

∴ ∠A = \(\frac{3}{5}\) x180° = 1080
∴ ∠C = \(\frac{2}{5}\)x180° = 72°

∵  The opposite angles of a parallelogram are equal.
∴ ∠C = ∠A 108° and ∠D = ∠B = 72°
∴ In parallelogram □ABCD ∠A = 108°, ∠B = 72°, ∠C = 108°, ∠D = 72°.

Class 9 Maths Exercise 6.1

2. In the parallelogram ABCD, the bisectors of ∠A and ∠B meet CD at the point E. The length of the side BC is 2 cm. Let us write the length of the side AB.

Solution:

 

 ∵ ABCD is a parallelogram
∴ ∠DAB+ ∠ABC = 180°

or, \(\frac{1}{2}\) ∠DAB+ \(\frac{1}{2}\) ∠ABC = 90°

∴ ∠AEB 90°
∵  In parallelogram ABCD bisectors of ∠A & ∠B meet perpendicularly at E on CD.
∴ AB = 2AD = 2 x 2 cm = 4 cm.

3. The equilateral triangle AOB lies within the square ABCD. Let us write the value of COD.

Solution:

 

 

∵ AOB is an equilateral triangle.
∴ ∠AOD and ∠BOC are isosceles triangles ( ∵AO = AD & BO = BC)
∵ AOB is an equilateral triangle.

∴ ∠OAB = ∠OBA = ∠AOB = 60°
∴ ∠OAD = ∠OBC = 90°-60°-30° [∵ ∠DAB = ∠ABC = 90°]

∴ ∠BCO = ∠ADO = \(\frac{180^{\circ}-30}{2}\) = 75°

In ΔCOD, ∠OCD = 90°-75° 15° and ∠ODC = 90°-75° 15°

∴ ∠COD = 180°- (150+ 15°) = 180°-30°-150°

4. In the square ABCD, M is a point on AD so that ∠CMD=30°. The diagonal BD intersects CM at the point P. Let us write the value of ∠DPC.

Solution:

 

In a square diagonal bisects the angles.
∴ In ΔPMD, ∠PDM = 45°

In ΔDMP, external ∠DPC = ∠PDM + ∠DMP = 45° + 30°=75°.

Class 9 Maths Exercise 6.1

5. In the rhombus ABCD, the length of the side AB is 4 cm and ∠BCD=60°. Let us write the length of the diagonal BD.

Solution:

 

 

In ΔBCD, ∠BCD = 60°
∴ ∠CBD + ∠BDC = 180°-60° = 120°
∵ BC=CD

∴ ∠ BDC= ∠CBD = \(\frac{120^{\circ}}{2}\) = 60°

∴ ΔBCD is an equitateral triangle.
∴ BC = CD = BD
∵ AB = 4 cm
∴BC= 4 cm
∴ BD 4 cm

 

 

Ganit Prabha Class 9 Solutions Chapter 2 Laws Of Indices

Important Formula:
1. x^m.xⁿ = x^m+n

2. \(x^m \div x^n=\frac{x^m}{x^n}=x^{m-n}\)

3. (x^m)ⁿ = x^mn

4. (xy)^m = x^my^m

5. \(\left(\frac{x}{y}\right)^m=\frac{x^m}{y^m}\)

6. \(\frac{1}{x^{-m}}=x^m\)

7. x⁰=1

8. \(x^{-1}=\frac{1}{x}\)

9. \(x^{-n}=\frac{1}{x^n}\)

10. x^m=Yⁿ ,n ≠ 0 ⇒ x= y when x and y positive

11. x^m=Yⁿ  ⇒ m = n when x>n and x ≠1

12. x^m = y ∴ \(x=y^{\frac{1}{m}}\)

Read and Learn More WBBSE Solutions For Class 9 Maths

Class 9 Math Solution West Bengal Board Chapter 2 Laws Of Indices: Exercise 2.1

 

Question 1. Let us find out the values of the following:

1. \((\sqrt[5]{8})^{\frac{5}{2}} \times\left(16^{\frac{-3}{2}}\right)\)

Solution:

 

2. \(\left\{(125)^{-2} \times(16)^{\frac{-3}{2}}\right\}^{\frac{-1}{6}}\)

Solution:

 

WBBSE Class 9 Math Book Solution

3. \(4^{\frac{1}{3}} \times\left[2^{\frac{1}{3}} \times 3^{\frac{1}{2}}\right] \div 9^{\frac{1}{4}}\)

Solution: \(4^{\frac{1}{3}} \times\left[2^{\frac{1}{3}} \times 3^{\frac{1}{2}}\right] \div 9^{\frac{1}{4}}\)

 

 

 

Question 2. Let us simplify:

1. \(\left(8 a^3 \div 27 x^{-3}\right)^{\frac{2}{3}} \times\left(64 a^3 \div 27 x^{-3}\right)^{-2}\)

Solution:

WBBSE Class 9 Math Book Solution

2. \(\left\{\left(x^{-5}\right)^{\frac{2}{3}}\right\}^{\frac{-3}{10}}\)

Solution:

 

 

3. \(\left[\left\{\left(2^{-1}\right)^{-1}\right\}^{-1}\right]^{-1}\)

Solution:

WB Board Class 9 Math Solution

4. \(\sqrt[3]{a^{-2}} \cdot b \times \sqrt[3]{b^{-2}} \cdot c \times \sqrt[3]{c^{-2}} \cdot a\)

Solution:

 

5. \(\left(\frac{4^{m+\frac{1}{4}} \times \sqrt{2.2^m}}{2 \cdot \sqrt{2^{-m}}}\right)^{\frac{1}{m}}\)

Solution:

 

West Bengal Class 9 Maths Solutions

\(\left(2^{\frac{5 m+2-2+m}{2}}\right)^{\frac{1}{m}}\)

\(=\left(2^{\frac{6 m}{2}}\right)^{\frac{1}{m}}=2^3=8\)

 

6. \(9^{-3} \times \frac{16^{\frac{1}{4}}}{6^{-2}} \times\left(\frac{1}{27}\right)^{-\frac{4}{3}}\)

Solution:

West Bengal Class 9 Maths Solutions

7. \(\left(\frac{x^a}{x^b}\right)^{a^2+a b+b^2} \times\left(\frac{x^b}{x^c}\right)^{b^2+b c+c^2} \times\left(\frac{x^c}{x^a}\right)^{c^2+c a+a^2}\)

Solution:

 

West Bengal Class 9 Maths Solutions

Question 3. Let us arrange in ascending order:

1. \(5^{\frac{1}{2}}, 10^{\frac{1}{4}}, 6^{\frac{1}{3}}\)

Solution:

 

Class 9 Math Solution West Bengal Board

2. \(3^{\frac{1}{3}}, 2^{\frac{1}{2}}, 8^{\frac{1}{4}}\)

Solution:

 

3. \(2^{60}, 3^{48}, 4^{36}, 5^{24}\)

Solution:

Class 9 Math Solution West Bengal Board

Question 4. Let us prove:

1. \(\left(\frac{a^q}{a^r}\right)^p \times\left(\frac{a^r}{a^p}\right)^q \times\left(\frac{a^p}{a^q}\right)^r=1\)

Solution: L.H.S\(\left(\frac{a^q}{a^r}\right)^p \times\left(\frac{a^r}{a^p}\right)^q \times\left(\frac{a^p}{a^q}\right)^r\)

Ganit Prabha Class 9 Solutions

= R.H.S Proved.

2. \(\left(\frac{x^m}{x^n}\right)^{m+n} \times\left(\frac{x^n}{x^1}\right)^{n+1} \times\left(\frac{x^1}{x^m}\right)^{1+m}\)

Solution:

L.H.S = \(=\left(\frac{x^m}{x^n}\right)^{m+n} \times\left(\frac{x^n}{x^{\prime}}\right)^{n+1} \times\left(\frac{x^{\prime}}{x^m}\right)^{l+m}\)

= 1 = R.H.S Proved.

3. \(\left(\frac{\mathbf{x}^m}{\mathbf{x}^n}\right)^{m+n-1} \times\left(\frac{\mathbf{x}^n}{\mathbf{x}^1}\right)^{n+1-m} \times\left(\frac{\mathbf{x}^{\prime}}{\mathbf{x}^m}\right)^{1+m-n}=1\)

Solution: L.H.S \(=\left(\frac{x^m}{x^n}\right)^{m+n-1} \times\left(\frac{x^n}{x^{\prime}}\right)^{n+1-m} \times\left(\frac{x^{\prime}}{x^m}\right)^{1+m-n}\)

 

=1= R.H.S Proved.

4. \(\left(a^{\frac{1}{x-y}}\right)^{\frac{1}{x-z}} \times\left(a^{\frac{1}{y-z}}\right)^{\frac{1}{y-x}} \times\left(a^{\frac{1}{z-x}}\right)^{\frac{1}{z-y}}=1\)

Solution: L.H.S \(=\left(a^{\frac{1}{x-y}}\right)^{\frac{1}{x-z}} \times\left(a^{\frac{1}{y-x}}\right)^{\frac{1}{y-x}} \times\left(a^{\frac{1}{z-x}}\right)^{\frac{1}{z-y}}\)

\(=a \frac{1}{(x-y)(x-z)}+\frac{1}{(y-z)(y-x)}+\frac{1}{(z-x)(z-y)}\)

 

Ganit Prabha Class 9 Solutions

 

Question 5. If x+z=2y and b²=ac, then let us show that a ^y-z b ^z-x c^x-y = 1.

Solution:

 

 

Question 6. If a = xy^p-1, b = xy^q-1 ,and c = xy^r-1, then let us show that a ^q-r b ^r-p  c^p-q=1.

Solution: L.H.S. a ^q-r b ^r-p  c^p-q

=1 R.H.S Proved.

Question 7. If \(x^{\frac{1}{a}}=y^{\frac{1}{b}}=z^{\frac{1}{c}}\) and =xyz=1, then let us show that a+b+c =0.

Solution: \(x^{\frac{1}{a}}=y^{\frac{1}{b}}=z^{\frac{1}{c}}\)=k(let)

or, a ^q-r b ^r-p  c^p-q=1 k^a+b+c= k^o [ k⁰=1]
or, a+b+c=0 Proved.

Question 8. If a^x = b^y = c^z and abc = 1 then let us show that xy+yz+zx=0.

Solution:

Question 9. Let us solve:

1. 49^x=7³

Solution: 49^x=7³

2. 2^x+2+2^x-1=9

Solution: 2^x+2+2^x-1=9

Ganit Prabha Class 9 Solutions

3. 2^x+1+ 2^x+2= 48

Solution: 2^x+1+ 2^x+2= 48

\(or, 2^x=\frac{48}{6}
or, 2^x=8=2^3\)

∴x = 3

4. \(2^{4 x} \cdot 4^{3 x-1}=\frac{4^{2 x}}{2^{3 x}}\)

Solution:

 

5. 9Χ81^x = 27^2-x

Solution:

Ganit Prabha Class 9 Solutions

6. 2^5x+4 + 2⁹ = 2¹⁰

Solution: 2^5x+4 + 2⁹ = 2¹⁰

Class 9 Math Solution WBBSE

7. 6^2x+4=3^3x 2^x+8

Solution:

\(\text { or, }\left(\frac{2}{3}\right)^{x-4}=\left(\frac{2}{3}\right)^0\)

or, x-4=0
∴ x = 4 

Question 10. 

1. The value of (0.243)^0.2 x (10)^0.6 is

(1)0.3
(2)3
(3)0.9
(4)9

Solution: (0.243)^0.2 x (10)^0.6

∴ (2)3

2. The value of \(2^{\frac{1}{2}} \times 2^{-\frac{1}{2}} \times(16)^{\frac{1}{2}}\) is

(1) 1
(2) 2
(3) 4
(4)1/2

Solution:\(2^{\frac{1}{2}} \times 2^{-\frac{1}{2}} \times(16)^{\frac{1}{2}}\)

Class 9 Math Solution WBBSE

∴(3) 4

3. If 4^x=8³, then the value of x is

(1)3/2
(2)9/2
(3)3
(4)9

Solution: 4^x=8³
or, 2^2x=2⁹
∴ x=9/2

∴ (2)9/2

4. If 20^-x= 1/7 then the value of (20)^2x is
(1) 1/49
(2)7
(3)49
(4)1

Solution: 20^-x= 1/7

∴ (3)49

5. If 4 x 5^x = 500 then the value of x^x is
(1) 8
(2)1
(3)64
(4)27

Solution: 4 x 5^x = 500

∴ (4)27

Question 11. Short answer type questions:

1. If (27)^x = (81)^y, then let us write the ratio x: y.

Solution: (27)^x = (81)^y
or, 33^x=34^y
or, 3x=4y

or, \(\frac{x}{y}=\frac{4}{3}\)

= x:y 4:3

2. If (5⁵+0.01)²– (55-0.01)²=5^x, then let us calculate the value of x and write it.

Solution: (5⁵+0.01)²– (55-0.01)²=5^x
or, 4 x 5⁵x 0.01= 5^x

or,\(4 \cdot 5^5 \cdot \frac{1}{100}=5^x\)

or, \(5^5 \times \frac{1}{25}=5^x\)

Class 9 Math Solution WBBSE

or, 5³ =5^x
∴ x = 3 .

3. If 3 x 27^x=9^x+4, then let us calculate the value of x and write it.

Solution: 3 x 27^x=9^x+4
or, 3 x (3³)=(3²)^x+4
or, 3 x 3^3x=3^2x+8
or, 3^{1+ 3x} = 3^2x+8
or,3x-2x=8-1
∴ x = 7

4. Let us find out the value of \(\sqrt[3]{\left(\frac{1}{64}\right)^{\frac{1}{2}}}\)and write it.

Solution:

5. Let us write explaining the greater value between \(3^{3^3} \text { and }\left(3^3\right)^3\)with reason.

Solution:

 

 

WBBSE Class IX Maths Solutions WBBSE Chapter 1 Real Numbers

Definitions :

1. Natural Numbers: The numbers from which objects are counted, are called natural numbers. Eg: 1, 2, 3, 4….., etc.
2. Whole Numbers: In the group of natural nos. by adding 0 (zero), the new group of numbers so formed, is called a group of whole nos. Eg. 0, 1, 2, 3, etc.
3. Integers: The group of negative numbers, zero, and natural numbers is called Integers. Eg: -4, -3, -2, -1, 0, 1, 2, 3, 4……….., etc. the numbers on the left side of zero are called negative integers and the numbers on the right side of zero are called positive integers.
4. Even Numbers: Those whole numbers which are completely divisible by 2 are called even numbers. Eg. 2, 4, 6, 8, etc. The digit on the units place of these numbers is any of 0, 2, 4, 6, and 8. Eg: 482, 940, etc.
5. Odd Numbers: Those whole numbers which are completely divisible by 2, are called Odd Numbers. Eg: 1,3, 5, 7, 9, etc.
6. Prime Numbers: Those natural nos. which are divisibly only by 1 (one) and themselves, and are not divisible by any other numbers, are called prime numbers. Eg: 2, 3, 5, 7, 11., etc.
7. Composite Numbers: Those natural numbers which are divisible except 1 and themselves by at least one number, are called composite Numbers. Eg: 4, 6, 8, 9, 10…, etc.
8. Rational Numbers: If any number can be divided in the ratio of two whole numbers, i.e., in the form of \(\frac{p}{q}\) where p and q are whole numbers and q≠0 then that number is called (Rational number) Eg. \(\frac{3}{8}, \frac{2}{3}, 10, \frac{-3}{5}\) are rational numbers.
9. Irrational number: Those numbers which cannot be expressed as the ratio of two whole numbers, are called Irrational numbers. Eg. \(\sqrt{5}, \sqrt[3]{7}, \sqrt[4]{21}\) etc.

Class 9 Math Chapter 3 WBBSE

 

Read and Learn More WBBSE Solutions For Class 9 Maths

Class 9 Mathematics West Bengal Board Chapter 1 Real Numbers Excercise 1.1

Question 1. Let us write the definition of rational numbers and also write 4 rational numbers.

Solution: Any number of the form \(\frac{p}{q}\) where p and q are in integers and q≠0, is called a rational number. All integers and fractions are rational numbers.
4 rational numbers are \(\frac{2}{3}, \frac{17}{5}, \frac{-1}{2}, \frac{-5}{1}\) (i.e., -5).

Question 2. Is 0 a rational number? Let us express 0 in the form of \(\frac{p}{q}\) [Where p & q are integers and q ≠0 and p & q have no common factor other than 1].

Solution: Yes, as \(0=\frac{0}{1}\)

Question 3. Let me place the following rational numbers on the Number Line:

Solution:

1. 7

 

2. -4

 

3. \(\frac{3}{5}\)

 

 

4. \(\frac{9}{2}\)

 

 

5. \(\frac{2}{9}\)

 

 

6. \(\frac{11}{5}\)

 

 

7. \(-\frac{13}{4}\)

 

 

Class 9 Math Chapter 3 WBBSE

Question 4. Let me write one rational number lying between two numbers given below and place them on the Number Line:

1. 4 and 5

Solution: \(\frac{4+5}{2}=\frac{9}{2}\)

 

 

2. 1 and 2

Solution: \(\frac{2+1}{2}=\frac{3}{2}\)

 

 

 

3. \(\frac{1}{4} \text { and } \frac{1}{2}\)

Solution: \(\frac{\frac{1}{2}+\frac{1}{4}}{2}=\frac{3}{8}\)

 

WBBSE Class 9 Math Chapter 3.2

4. \(-1 \text { and } \frac{1}{2}\)

Solution: \(\frac{(-1)+\frac{1}{2}}{2}=-\frac{1}{4}\)

 

 

 

5. \(\frac{1}{4} \text { and } \frac{1}{3}\)

Solution: \(\frac{\frac{1}{4}+\frac{1}{3}}{2}=\frac{7}{24}\)

 

 

6. -2 and -1

Solution: \(\frac{(-2)+(-1)}{2}=\frac{-3}{2}\)

 

 

Question 5. Let me write 3 rational numbers lying between 4 and 5 and place them on the Number Line.

Solution:\(\frac{1}{2}(4+5)=\frac{9}{2}, \frac{1}{2}\left(4+\frac{9}{2}\right)=\frac{17}{4}, \frac{1}{2}\left(5+\frac{9}{2}\right)=\frac{19}{4}\)

 

∴ 4 and 5 in between 3 rational numbers: \(\frac{17}{4}, \frac{9}{2}, \frac{19}{4}\)

 

 

Question 6. Let me write 6 rational numbers lying between 1 & 2 and place them on the Number Line.

Solution: Here x = 1, y = 2 and n = 6.

∴ \(\mathrm{d}=\frac{2-1}{6+1}=\frac{1}{7}\left(therefore \mathrm{d}=\frac{\mathrm{y}-\mathrm{x}}{\mathrm{n}+1}\right)\)

∴ 6 rational numbers are (x + d), (x + 2d), (x + 3d), (x + 4d), (x + 5d) and (x + 6d),

i.e., \(\left(1+\frac{1}{7}\right),\left(1+\frac{2}{7}\right),\left(1+\frac{3}{7}\right),\left(1+\frac{4}{7}\right),\left(1+\frac{5}{7}\right),\left(1+\frac{6}{7}\right)\)

WBBSE Class 9 Math Chapter 3.2

∴ \(\frac{8}{7}, \frac{9}{7}, \frac{10}{7}, \frac{11}{7}, \frac{12}{7}, \frac{13}{7}\) rational numbers lie between 1 and 2.

 

 

Question 7. Let me write 3 rational numbers lying between \(\frac{1}{5} \text { and } \frac{1}{4}\).

Solution: Here x = \(\frac{1}{5}\), y = \(\frac{1}{4}\) and n = 3.

∴ d=\(\frac{\frac{1}{4}-\frac{1}{5}}{3+1}=\frac{\frac{1}{20}}{4}=\frac{1}{20} \times \frac{1}{4}=\frac{1}{80}\)

∴ 3 rational numbers = (x + d), (x + 2d), (x + 3d)

\(\begin{aligned}
& =\left(\frac{1}{5}+\frac{1}{80}\right),\left(\frac{1}{5}+\frac{2}{80}\right),\left(\frac{1}{5}+\frac{3}{80}\right) \\
& =\frac{17}{80}, \frac{18}{80}, \frac{19}{80} \\
& =\frac{17}{80}, \frac{9}{40}, \frac{19}{80}
\end{aligned}\)

Class IX Maths Solutions WBBSE

 

 

Question 8. Let me put (T) if the statement is true and write (F) if the statement is wrong.

1. By adding, subtracting, and multiplying two integers, we get integers. 
   Solution: True
2. By dividing two integers, we get an integer.
   Answer: False

Question 9. Let me see and write what I will get by adding, subtracting, multiplying, and dividing (divisor is non-zero) two rational numbers.
Answer: Rational number:

Let two rational numbers be \(\frac{7}{10} \& \frac{19}{30}\)

By adding, \(\frac{7}{10}+\frac{19}{30}=\frac{21+19}{30}=\frac{40}{30}=\frac{4}{3}\)

By subtracting, \(\frac{7}{10}-\frac{19}{30}=\frac{21-19}{30}=\frac{2}{30}=\frac{1}{15}\)

By multiplying, \(\frac{7}{10} \times \frac{19}{30}=\frac{133}{300}\)

By dividing, \(\frac{7}{10} \div \frac{19}{30}=\frac{7}{10} \times \frac{30}{19}=\frac{7 \times 3}{1 \times 19}=\frac{21}{19}\)

∴ By adding, subtracting, multiplying, and dividing two rational numbers we will get the rational numbers.

Class IX Maths Solutions WBBSE

Class 9 Maths WB Board Chapter 1 Real Numbers Exercise 1.2

Question 1. Let us write the right or false statement from the following:

1. The sum of two rational numbers will always be rational.
   Solution:(True)
2. The sum of two irrational numbers will always be irrational.
   Solution:(False)
3. The product of two rational numbers will always be rational.
   Solution:(True)
4. The product of two irrational numbers will always be rational.
   Solution:(False)
5. Each rational number must be real.
   Solution:(True)
6. Each real number must be irrational.
   Solution:(False)

Question 2. What is meant by irrational numbers? – Let me understand. Let me write 4 irrational numbers.

Solution:
Irrational Number: Irrational Numbers are numbers which can not be expressed in the form of \(\frac{p}{q}\), where p and q are integers and q ≠ 0, i.e., a number √a (square root of a) is an irrational number if a is positive and a is not the square of a rational number.

Examples: √2: √3: √5: \(\sqrt{7 / 3}\)(all positive irrational numbers)

\(-\sqrt{3} ;-\frac{\sqrt{3}}{4}\) (Negative irrational numbers)

Class 9 Maths Chapter 1 WBBSE

Question 3. Let us write rational and irrational numbers from the following:

1. √9
2. √225
3. √7
4. √50
5. 100
6. -√81
7. √42
8. √29
9. -√1000

Solution:
Rational Numbers:
1. √9
2. √225
4. √100
6. -√81

Irrational Numbers:
3. √7
4. √50
6. √42
7. √29
8. -√1000

Question 4. Let me place √5 on the Number Line.

Solution:

 

On the number line, take OB = 2 unit & DB (on OB) = 1 unit.

∴ OD= \(\sqrt{2^2+1^2}\)=√5. With center O & radius OD an arc is drawn, which intersects the number line at P.
∴ OP = √5 unit.

Question 5. Let me place √3 on the Number Line.

Solution:

 

 

\(\begin{aligned}
& O D=\sqrt{(O B)^2+(B D)^2} \\
& =\sqrt{(\sqrt{2})^2+(1)^2} \\
& =\sqrt{2+1}=\sqrt{3}
\end{aligned}\)

Class 9 Maths Chapter 1 WBBSE

With center O & radius OD an arc is drawn which intersects the number line at Q. OQ=√3 unit.

Question 6. Let me place √5, √6, √7,-√6,-√8,-√11 on the Number Line.

Solution:

 

Let. 3, 2, 1, 0, 1, 2, 3 are on the number line ‘I’.
(1)Let OA = 2 unit, draw ABOA, Take AB = 1 unit.

\(\begin{aligned}
OB & =\sqrt{O A^2+A B^2} \text { unit. } \\
& =\sqrt{4+1} \\
& =\sqrt{5} \text { unit. }
\end{aligned}\)

 

Taking O as the center & OB as the radius, an arc is drawn that intersects the number line at P.
∴ OP = √5 units.
By placing √5 on the number line, we get the point P.

(2)Now draw BC OB and take BC = 1 unit.

\(O C^2=O B^2+B C^2=(\sqrt{5})^2+1^2\) =5+1=√6 sq. unit

∴ OC = √6 units.
By placing √6 on the number line we get the point Q.
∴ OQ = √6 units.

(3)Now draw CD ⊥ OC and CD = 1 unit.

\(O D^2=O C^2+C D^2=(\sqrt{6})^2+1^2\)= 6 + 1 = 7 sq. unit

∴OD = √7 units.
By placing √7 on the number line, we get the point R.
∴OR= √7 units.

(4) Again take OE = 1 unit on the left of O, draw ES ⊥ OE, and take ES = 1 unit.

Taking O as the center & OS as the radius an arc is drawn, which intersects the number line at I.
∴ OT= – √2 unit

Now, OF = -2 unit, draw GF ⊥ OF, and take GF = 1 unit.
\(\mathrm{OG}^2=\mathrm{OF}^2+\mathrm{GF}^2=2^2+1^2\) =4+1 = 5 sq. unit
∴ OG √5 unit. Taking O as the center & OG as radius, an arc is drawn that intersects the number line at H.
∴ OH √5 units.
Now draw GK ⊥OG and take GK = 1 unit.

\(\begin{aligned}
\mathrm{OK} & =\sqrt{O G^2+\mathrm{GK}^2} \\
& =\sqrt{(\sqrt{5})^2+1^2} \\
& =\sqrt{5+1} \\
& =\sqrt{6} \text { unit. }
\end{aligned}\)

Class 9 Maths Chapter 1 WBBSE

Taking O as a center and OK as the radius, an arc is drawn that intersects the number line at M.
∴ OM = √6 units.

(5)Again draw KL ⊥ OK, take KL = OT (= √2)

\(\begin{aligned}
\mathrm{L} & =\sqrt{O K^2+\mathrm{KL}^2} \\
& =\sqrt{(\sqrt{6})^2+(\sqrt{2})^2} \\
& =\sqrt{6+2} \\
& =\sqrt{8} \text { unit. }
\end{aligned}\)

Taking O as the center and OL as the radius an arc is drawn that intersects the number line at N.
∴ON=-√8 units.

(6)From the perpendicular line on OK, cut KI = OG (√5)

∴\(\mathrm{OI}^2=\mathrm{OK}^2+\mathrm{KI}^2=(\sqrt{6})^2+(\sqrt{5})^2\) = 6 + 5 = 11
∴ OI = √11.

Taking O as centre and Ol as radius an arc is drawn which intersects the number line at J.
∴ OJ = √11 unit.

Class 9 Maths Chapter 1 WBBSE Chapter 1 Real Numbers Exercise 1.3

Question 1. Without division let’s find from the following which numbers have terminating decimals and write them.

1. \(\frac{17}{80}\)
2. \(\frac{13}{24}\)
3. \(\frac{17}{12}\)
4. \(\frac{16}{125}\)
5. \(\frac{4}{35}\)

Solution:

1. \(\frac{17}{80}=\frac{17}{2^4 \times 5}\)= Here denominator, 80 = 2⁴ x 5
   ∴ It is a terminating decimal number (Finite).
   ∴ \(\frac{17}{80}\) = 0.2125
2. \(\frac{13}{24}=\frac{13}{2^3 \times 3}\) = Here denominator 24 = 2³ x 3
   ∴ It is not a terminating decimal. It will be recurring decimal number (Infinite).
   ∴ \(\frac{13}{24}\)= 0.541666……..
3. \(\frac{17}{12}=\frac{17}{2^2 \times 3}\) = Here denominator 12 = 2² x 3
   ∴ It is not a terminating decimal. It will be a recurring decimal number. (Infinite)
   ∴ \(\frac{17}{12}\) = 1.41666…….
4. \(\frac{16}{125}=\frac{16}{5^3}\)= Here denominator 125 = 5³
   ∴It is a terminating decimal number (Finite).
   ∴\(\frac{16}{125}\)= 0.128
5. \(\frac{4}{35}=\frac{4}{5 \times 7}\) Here denominator 35 = 5 x 7
   ∴ It is a terminating decimal number (Infinite).
   ∴\(\frac{4}{35}\)=0.114285714…………

Class 9 Maths Chapter 1 WBBSE

Question 2. Let us expand each of the numbers given below into decimals and write the type of decimal expansions:

1. \(\frac{1}{11}\)
2. \(\frac{5}{8}\)
3. \(\frac{3}{13}\)
4. \(3 \frac{1}{8}\)
5. \(\frac{2}{11}\)
6. \(\frac{7}{25}\)

Solution:

1. \(\frac{1}{11}\)

2. \(\frac{5}{8}\)

 

 

3. \(\frac{3}{13}\)

 

 

 

4. \(3 \frac{1}{8}\)

 

 

Class 9 Maths Solution West Bengal Board

5. \(\frac{2}{11}\)

 

 

Class 9 Maths Solution West Bengal Board

6. \(\frac{7}{25}\)

 

 

Question 3. Let us express each of the following numbers in the form of \(\frac{p}{q}\)where p & q are integers &q≠0;

1. 0.3
2. 1.3
3. 0.54
4. 0.34
5. 3.14
6. 0.17
7. 0.47
8. 0.54
9. 0.001
10. 0.163

Solution:

1. 0.3 \(=\frac{3}{9}=\frac{1}{3}\)
2. 1.3 \(1 \frac{3}{9}=1 \frac{1}{3}=\frac{4}{3}\)
3. 0.54\(=\frac{54-5}{90}=\frac{49}{90}\)
4. 0.34\(=\frac{34}{99}\)
5. 3.14\(=\frac{314-3}{99}=\frac{311}{99}\)
6. 0.17\(=\frac{17-1}{90}=\frac{16}{90}=\frac{8}{45}\)
7. 0.47\(=\frac{47-4}{90}=\frac{43}{90}\)
8. 0.54\(=\frac{54}{99}=\frac{6}{11}\)
9. 0.001\(=\frac{001}{999}\)
10. 0.163\(=\frac{163}{999}\)

Question 4. Let’s write 4 numbers whose decimal expansions are non-terminating and non-recurring.

Solution:√2,√3,√5 and √7

Question 5. Let us write 3 different irrational numbers lying between \(\frac{5}{7} \& \frac{9}{7}\)

Solution:0.808008000800008, 0.858558555855558… 0.919119111911119………………

Class 9 Maths Solution West Bengal Board

Question 6. Let us write 2 different irrational numbers lying between \(\frac{3}{7} \& \frac{1}{11}\)

Solution:

1. 0.121221222122221…….
2. 0.373773777377779……

Question 7. Let us write the rational and irrational numbers from the following:

(1)√47
(2)√625
(3)_6.5757…
(4)1.1010010001…

Solution:

1. √47 = Irrational
2. √62525 = Rational
3. 6.5757… = Rational
4. 1.1010010001… = Irrational

Question  8. Let us place the following numbers on the number line:

1. 5.762
2. 2.321
3. 1.052
4. 4.178

Solution:

1. 5.762

 

 

Class 9 Maths Solution West Bengal Board

∴ By placing real number 5.762 on the number line we got point P.

 

2. 2.321

 

∴ By placing real number 2.321 on the number line we got point

3. 1.052

 

 

 

∴By placing real number 1.052 on the number line we got point R.

4. 4.178

 

 

∴ By placing real number 4.178 on the number line we got point T.

Class 9 Maths Solution West Bengal Board

Question 9. Let us place the two numbers 2.23 and 5.54 upto 4 decimal places on the number line.

Solution: 2.26 = 2.2626(upto 4 decimal points)

 

 

∴ By placing real number 2.2626 on the number line we got point S.
∴ 5.54 = 5.54444(Upto 4 decimal points)

 

Class 9 Maths Chapter 1 WBBSE

Question 10. Let us write two rational numbers lying between 0.2323323 and 0.212112111211112…

Solution: The required two rational numbers are 0.22 and 0.023.

Question 11. Let us write two rational numbers lying between 0.2101 & 0.2222…. or 0.2. 

Solution: 0.2 and 0.221

Question 12. Write ten true and ten false statements relating to real numbers, whole numbers, rational numbers and irrational numbers.

Solution: Ten true statements:

1. The smallest real number is 1.
2. Real numbers are infinite.
3. \(\frac{2}{3}\) is a rational number. 3
4. π is a irrational number.
5. 100 is a whole number.
6. There are infinite number of rational numbers between 2 < x < 5.
7. 0 is a whole number.
8. The product of two irrational numbers is always, an irrational number.
9. All whole numbers are real numbers.
10. The product of two rational numbers is always a rational number.

Ten false statements:

1. The product of two irrational numbers is a rational number.
2. 100 is the largest real number.
3. 0 is not a whole number.
4. \(\frac{2}{7}\) is an irrational number.
5. Real numbers are infinite.
6. \(\frac{1}{5}\) is a recurring rational number. 5
7. 0.219 is an irrational number.
8. There is only one rational number between 1 < x < 2.
9.  0 is a natural number.
10. 0.9 is an irrational number.

Question 13. Let us find how many rupees will be required to determine the values of the following expressions if Rs. 2 for one multipication and Rs. 1 for one addition is required and let us also see which law can be applied to find out the value of the expression with the least amount of money.

1. 3x²+2x+1, when x=5

2. 2x³+3x²+2x+3, when x=7 (Hints: 3×5²+2×5+1=3x5x5+2×5+1, here numbers of multipications and additions are 3 & 2 respectively. So Rs. 8 is required.∴ But if we write 3x² + 2x + 1 = x(3x+2)+ 1, by applying distributive law, then for two multipications and two additions Rs. 6 is required.).

Solution:

1. 3x² + 2x + 1 = 3 x 5²+2 x 5+1 =3x5x5+2×5+1 total cost for 3 multiplications and 2 additions = 3 Χ 2+2 x 1 = Rs. 8

Again: 3×2 + 2x + 1 = x(3x+2)+ 1 total cost for 2 multiplications and 2 additions = 2 x 2 + 2 × 1 = Rs. 6

The second law can be applied with least amount of money.

2.  2x³ + 3x² + 2x + 3 =2x( x²+ 1) + 3(x² + 1) = (2x+3) (x² + 1)

∴ Total cost for 4 mulltiplications and 2 additions will be = 4×2+2×1 = Rs. 10.

Class 9 Mathematics West Bengal Board

Question 14.

1. The decimal expansion of √5 is
(1)  A terminating decimal
(2)  A terminating or recurring decimal
(3)  Non-terminating and non-recurring decimal
(4) None of them

Solution:  (3) a non-terminating and non-recurring decimal

2. The product of two irrational numbers is
(1)  Always irrational number
(2)  Always Rational Number
(3)  Always an integer
(4)  Rational or irrational number

Solution: (4) rational or irrational number

3. π and 22/7 are
(1)  Always rational number
(2) Always Irrational number
(3) π is rational and is irrational
(4) π is irrational and 22/7 is rational

Solution:  (4) π is irrational and 22/7 is rational

4. Between two rational numbers, there exists
(1)No rational number
(2)Only one rational number
(3)Infinite numbers of rational numbers
(4)No irrational number

Solution: (3)Infinite numbers of rational numbers

5. Between two irrational numbers, there exists
(1)No rational number
(2)Only one rational number
(3)Infinite numbers of irrational numbers
(4)No irrational number

Solution: (3) infinite numbers of irrational numbers

6. The number 0 is

(1)Whole number but not integer
(2)Integer but not rational
(3)Rational but not real number
(4)Whole number, integer, rational and real number but not irrational

Solution: (4)Whole number, integer, rational and real number but not irrational

Class 9 Mathematics West Bengal Board

Question 15. Short-answer type questions :

1. Let us write a number where the sum of two irrationals is rational.
   
   Solution: Two irrational numbers are √5 and -√5.
   Sum of √5 and√5 = √5+(-√5) = √5-√5 = 0 = Rational number
   ∴ Number = 0
2.  Let us write a number where the difference of two irrationals is a rational number.
   
   Solution: Two irrational numbers are = (√7+2) and (√7-2)
   Difference of (√7+2) and (7-2)- (√7+2)-(√7-2) = 4 = Rational number
   ∴ Number = 4
3. Let us write a rational number lying between 1/7 and 2/7.
   
   Solution: \(=\frac{1}{2}\left(\frac{1}{7}+\frac{2}{7}\right)=\frac{3}{14}\)
4. Let us write an irrational number lying between \(\frac{1}{7} \text { and } \frac{2}{7}\)
   
   Solution:\(\frac{1}{7}\)= 0.142857
   \(\frac{2}{7}\)= 0285714
   ∴ \(\frac{1}{7} \text { and } \frac{2}{7}\) = 0.1515515551
5. Let us write the common fraction of the recurring decimal 0.0123
   Solution: 0.0123 \(=\frac{123-12}{10000-1000}=\frac{111}{9000}=\frac{37}{3000}\)