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:
Given
In the figure aside, ABCD and ABEF are two parallelograms.
∵ 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 this by saying 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: 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 length 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:
Given
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.
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:
Given
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:
Given
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:
Given
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.
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:
Given
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:
Given
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:
Given
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:
Given
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:
Given
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:
Given
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:
Given
In the parallelogram ABCD,∠A: ∠B 3:2.
∵ 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:
Given
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.
∵ 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:
Given
The equilateral triangle AOB lies within the square ABCD.
∵ 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:
Given
In the square ABCD, M is a point on AD so that ∠CMD=30°. The diagonal BD intersects CM at the point P.
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:
Given
In the rhombus ABCD, the length of the side AB is 4 cm and ∠BCD=60°.
In ΔBCD, ∠BCD = 60°
∴ ∠CBD + ∠BDC = 180°-60° = 120°
∵ BC=CD
∴ ∠ BDC= ∠CBD = \(\frac{120^{\circ}}{2}\) = 60°
∴ ΔBCD is an equilateral triangle.
∴ BC = CD = BD
∵ AB = 4 cm
∴BC= 4 cm
∴ BD 4 cm