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Chapter 10 Theorems Related To cyclic Quadrilateral Exercise 10.1

 

Application 1. Look at the picture of two circles with centre O given below and let us write by calculating the value of x° in each case.

Solution:

 

 

 

Read and Learn More WBBSE Solutions For Class 10 Maths

Application 2. ABCD is a cyclic quadrilateral and O is the centre of that circle. If <COD 120° and <BAC 30°, let us write by calculating the value of <BOC and <BCD.

 

 

Application 3. Sides AD and AB of a cyclic quadrilateral ABCD beside are produced to the points E and F respectively. If <CBF = 120°, let us write by calculating the value of CDE.

Solution: In the figure, ∠BAD= ∠CBF=120º

∠CDE = 180º-120º

= 60º

Application 4. ABCD is a cyclic quadrilateral. Sides AB and DC, when produced, meet at point Q. If ZADC = 85° and BPC = 40°, let us write by calculating the value of ZBAD and COD.

Solution: For cyclic quadrilateral ABCD, 

exterior ∠PBC = ∠ADC = 85° 

∴ In ΔBPC, ∠BCP = 180° – (85° + 40°) = 55°

Again, ∠BAD = exterior∠BCP = 55°

∴ In ΔCQD, ∠CQD + ∠DCQ = 85°

∴ ∠CQD = 85° – ∠DCQ 

= 85° – ∠BCP 

= 55°

∴ ∠BAD = 55° 

and ∠CQD = 30°

Again, ∠BAD = exterior ∠BCP = 55°

∴ In ΔCQD, ∠CQD + ∠DCQ = 85°

∴ ∠CQD = 85° – ZDCQ 

= 85° – ∠BCP 

= 55°

∴ ∠BAD = 55° 

and ∠CQD = 30°

Application 5. Let us prove that a cyclic parallelogram is a rectangular 

Solution :

Given: ABCD is a cyclic parallelogram.

To prove: Quadrilateral ABCD is a rectangular picture.

Proof: ABCD is a parallelogram.

∴∠ABC =∠ADC

Again, ABCD is a cyclic quadrilateral.

∴ ∠ABC + ∠ADC = 180°

So, 2∠ABC= 180°. 

∴ ∠ABC = 90°

One angle of the parallelogram is a right angle. So, ABCD is a rectangular picture.

 

 

Question 1. In the picture beside two diagonals of quadrilateral PQRS in- intersect each other at point X in such a way that PRS = 65° and ZRQS = 45°. Let us write by calculating the value of SQP and ZRSP. 

Solution: ∠SQP = ∠PRS = 65° (on the same segment)

∠RPS = ∠RQS = 40° (on the same segment)

∠RSP = 180° – (65° +45°)

= 180° – 110°

= 70°

 

 

2. Side AB of a cyclic quadrilateral ABCD is produced to the point X and by measuring we see that XBC= 82° and ∠ADB = 47°. Let us write by measuring the value of <BAC.

Solution: Find ∠BAC.

As ABCD is a cyclic quadrilateral.

.. ∠ADC + ∠ABC = 2 rt. angles

Again, ∠ABC +∠XBC= 2 rt. angle 

.. ∠ADC + ∠ABC = ∠ABC + <XBC 

∠ADC = ∠XBC = 80° (given)

.. ∠BDC + ∠BDA = ∠ADC

∠BDC + 47° = 80° [∠BDA = 47° (given)]

.. ∠BDC = 82°-470°

= 35°

.. ∠BDC & ∠BAC are in the same segment.

.. <BAC = ∠BDC = 35°

.. <BAC 35°

 

 

Question 3. Two sides PQ and SR of a cyclic quadrilateral PQRS are produced to meet at the point T. O is the centre of the circle. If ∠POQ = 110°,∠QOR = 60°, and ∠ROS = 80°, let us write by measuring ∠RQS and ∠OTR.

Solution: PQ & RS, the two sides of the cyclic quadrilateral, meet at T, when produced. 

O is the centre of the circle.

∠POQ=110°, 

∠QOR = 60°, 

& ∠ROS = 80°

To find ∠RQS &∠QTR.

On the same segment SR, ∠RQS is the angle on the circumference &

∠ROS =1/2 

∠ROS =1/2 x 80° = 40°

.. We know, ∠POQ + ∠QOR + ∠ROS + ∠SOP = 360°

..∠SOP= 360° – (∠POQ+ ∠QOR + ∠ROS)

= 360°-(110° +60° +80°) 

= 360° – 250° 

= 110°

or, ∠SOP = 110°, 

∠PQR = ∠PQS + ∠SQR 

= 55° + 40° = 95° 

.. ∠PQR + ∠RQT = 180°

∠RQT 180° – ∠PQR 180° – 95° = 85°

In ΔQOR, OQ = OR (same radiii)

∴ ∠OQR = OR

Again, ∠OQR + ∠ORQ+∠QOR = 180°

or, 2∠OQR +60° = 180°

∠OQR = 1/2 (180° 60°) 60° = ∠ORQ

Again, In ΔORS, OR = OS (same radii)

∴ ∠ORS = ∠OSR

∠ORS + ∠OSR +∠SOR= 180°

2∠ORS+80° = 180°

∠ORS = ∠OSR =1/2(180° 80°)= 50°

∠SRQ = ∠ORS + ∠ORQ = 50° + 60° = 110° 

∠QRT = 180°-∠SRQ 

= 180°-110°-70°

In ΔQTR, ∠QTR + ∠QRT+ ∠RQT = 180° 

∠QTR = 180° (∠QRT + ∠RQT)

= 180° (70° 85°) = 25°

∴ ∠RQS = 40° &∠QTR = 25°

 

 

Question 4. Two circles intersect each other at points P and Q. Two straight lines through points P and Q intersect one circle at points A and C respectively and the other circle at points B and O respectively. Let us prove that AC || BD.

Solution: To prove AC || BD.

Join A, C; B, D & P, Q.

Proof ACQP is a cyclic quadrilateral.

:. ∠PAC + ∠PQC = 2 rt. angles

Again, PQ meets CD at Q

∴∠PAC+ ∠PQD = 2 rt. angles

∴ ∠PAC + ∠PQC =∠PQC +∠PQD 

∠PAC = ∠PQD

As PQDB is a cyclic quadrilateral.

∴ ∠PBD + ∠PQD = 2 rt. angles

or, ∠PBD +∠PAC = 2 rt. angles (∠PQD = ∠PAC)

.. AB cuts AC & BD, and the sum of the internal angles of the same sides of the intercepts is equal to two right angles.

.. AC || BD

 

 

Question 5. I drew a cyclic quadrilateral AQCD and the side BC is produced to point E. Let us prove that the bisectors of ∠BAD and ∠DCE meet in the circle.

Solution: ABCD is a cyclic quadrilateral. 

Side BC is produced to E & the bisector of ∠BAD cuts the circle at F.

Join C, F.

To prove, CF straight line bisects DCE.

Proof: DCF & ∠DAF are in the same segment.

∴ ∠DCF = ZDAF

AF bisects ∠BAD.

∴ ∠DAF = 1/2  ZBAD

∴ ∠DCF =∠DAF = ∠BAD———(1)

ABCD is a cyclic quadrilateral. 

∴ ∠BAD + 2BCD = 2 rt. angles ——-(2)

BE meets DC at C.

∴ ∠DCE + ∠BCD = 2 rt. Angles ——(3)

From (2) & (3),

∠BAD + ∠BCD = ∠DCE + ∠BCD

or ∠BAD = ∠DCE ———-(4)

∴ ∠DCE= 2∠DCF

∴CF is the bisector of ∠DCE.

 

 

Question 6. Mohit drew two straight lines through any point X exterior to a circle to intersect the circle at points A, B, and points C, D respectively. Let us prove logically that ΔXAC and ΔXBD are equiangular.

Solution: From an external point X, two straight lines are drawn, which cut the circle at A, B, and C, D. 

To prove that two angles of each of the ΔXAC & ΔXBD are equal.

i.e., ΔXAC & ΔXBD are equiangular.

Join A, C & B, D.

ABCD is a cyclic quadrilateral

∴ ∠ABD + ∠ACD = 2 rt. angles——-(1)

AC meets DX at C

∴ ∠ACD + ∠ACX 2 rt. angles———–(2)

∠ABD+∠ACD = ∠ACD +∠ACX

∠ABD = ∠ACX = <XBC

Again, as ΔBCD is a cyclic quadrilateral,

∠BDC + <BAC = 2 rt. angles

& ∠BAC + ∠CAX = 2 rt. angles. 

∠BDC + ∠BAC = ∠BAC + ∠CAX . 

∠BDC=∠CAX or, ∠BDX = <CAX

Now, in Δ XAC & ΔXBD,

∠XCA = ∠XBD; 

∠XAC = <BDX 

& ∠AXC = ∠BXD (Same angle)

∴ΔXAC & ∴XBD are equiangular.

 

 

Question 7. I drew two circles that intersect each other at points G and H. Now I drew a straight line through point G which intersects two circles at points P and Q and the straight line through point H parallel to PQ intersects the two circles at points R and S. Let us prove that PQ = RS.

Solution: Two circles intersect each other at G & H. A straight line passing through G cuts the circles at P & Q. 

And a straight line passing through H cuts the circles at R & S. To prove PQ = RS.

Join P, R; G, H & Q, S.

PRHG is a cyclic quadrilateral.

∴ ∠RPG + ∠RHG = 2 rt. angles.

Again, GH meets RS at H.

∴ ∠RHG + ∠GHS = 2 rt. angles

∴ ∠RPG+∠RHG = ∠RHG + ∠GHS

∴ ∠RPG = ∠GHS——–(1) 

Again, as GHSQ is a cyclic quadrilateral. 

∴ ∠GQS+ ∠GHS = 2 rt. angles. 

and ∠GHS + ∠GHR = 2 rt. angles. 

∠GQS +∠GHS = ∠GHS + ∠GHR 

∠GQS = ∠GHR——(2)

Adding (1) & (2).

<RPG + ∠GQS = <GHS + ∠GHR

or, ∠RPG+∠GQS = 2 rt. angles

i.e., straight line PQ cuts PR & QS, & the sum of the internal angles on the same side of the intercept are equal to 2 rt. angles.

.. PR || QS & PQ || RS (given).

PQSR is a parallelogram.

.. PQRS Proved.

 

 

Question 8. I drew a triangle ABC of which AB = AC and E is any point on BC produced. If the circumcircle of ABC intersects AE at point D. Let us prove that ∠ACD =∠AEC. 

Solution: In AABC, AB = AC, E is a point on produced BC.

AE cuts the circumcircle of AABC at D.

To prove, ∠ACD = ∠AEC

Join C,D.

Proof: ABCD is a cyclic quadrilateral.

∠ABC + ∠ADC = 2 rt. angles.

Again, straight-line CD meets AE at D. 

∴ ∠ADC + ∠CDE = 2 rt. angles. 

∴ ∠ABC + ∠ADC= ∠ADC + ∠CDE 

∴ ∠ABC = CDE

In ΔABC, AB = AC, 

∴ ∠ABC = ∠ACB 

∴∠ACB = ∠CDE

Again, side EC of CDE is produced.

External vDCB = ∠CDE+∠CED

or, ∠ACB+∠ACD = ∠CDE + ∠AEC

or, ∠CDE+∠ACD = ∠CDE + ∠AEC [ ∠ACD = ∠CDE]

∴∠ACD = ∠AEC. Proved.

 

 

Question 9. ABCD is a cyclic quadrilateral. Chord DE is the external bisector of BDC. Let us prove that AE (or produced AE) is the external bisector of <BAC. 

Solution: ABCD is a cyclic quadrilateral, and chord DE is the extra 

To prove, AE (or produced AE) is the external bisector of <BAC. 

Proof: DE is the external bisector of ∠BDC.

∴ ∠AED = ∠ACD = ∠ABC (angles on the same segment) 

Again, <BAC = ∠BDC (angles on the same segment) [as DE is A ∠BDC]

.. AE is the external bisector of BAC.

 

 

 

Question 10. BE and CF are perpendicular on sides AC and AB of t Let us prove that four points B, C, E, and F are on the same circle 

Solution: InΔABC, BE & CF are two perpendiculars on the side To prove points B, C, E, and F are on the same circle.

That prove also AAEF & AABC are equiangular.

Join E, F.

Proof: AS 1 BE 1 AC & CF 1 AB

∴ ∠BEC = 1 rt angle & ZBFC = 1 rt. angle

∴ BEC = ∠BFC = 1 rt. angle

∴Points B, C, E, and F are on the same circle.

Now, inΔAEF &ΔABC,

∠FAE=∠BAC (Same angle)

 

 

Question 11. ABCD is parallelogram. A circle passing through points A and B intersects the sides AD and BC at points E and F respectively. Let us prove that the four points E, F, C, and D are concyclic.

Solution: ABCD is a parallelogram. 

A circle passing through points A & B intersects the sides AB & BC at E & F respectively. 

To prove that the points E, F, C, and D are concyclic. 

Join E, F.

Proof: ABCD is a parallelogram.

∴ ∠BAD + ∠ADC = 2 rt. angles.

Again, ABEF is a cyclic quadrilateral.

∠BAE+∠BFE = 2 rt. angles

∴ ∠EFC =∠BAE

∠EFC = <BAD

But∠BFE+∠CFE = 2 rt. angles.

∴∠EFC + ∠ADC = 2 rt. angles.

∠EFC + ∠EDC = 2 rt. angles. 

∴∠EFCD is a cyclic quadrilateral. 

.. Points E, F, C,D are concyclic

 

 

Question 12. ABCD is a cyclic quadrilateral. The two sides AB and DC are produced to meet at point P and the two sides AD and BC are produced to meet at point R. The two circumcircles of ABCP and ACDR intersect at point T. Let us prove that the points P, T, and R are collinear.

Solution: To prove that points P, T, and R is collinear.

Joint P, T;

T, R & C, T.

In BCP & ADP,

∠BPC = ∠APD; ∠PBC =∠ADP

& ∠BCP = ∠PAD

∴ΔBCP & ΔAPD are equiangular.

i.e., they are similar.

Now, in ΔPCT & ΔRCT,

∠PCT = ∠CRT,

CT common & ∠RCT = ∠TPC

∴ΔPCT ≅ ΔRCT

∴<PTC =∠RTC (corresponding angle)

As CT is the common side of∠PTC & ∠RTC

∴TR & TP are on the same straight line.

∴ P, T, and R are collinear.

 

 

Question 13. If point O is the ortho of triangle ABG, let us prove that it is the incentre of ΔDEF.

Solution: To prove O is the incentre of the ΔDEF.

In ΔBDO & ΔAEO.

∠BOD = ∠AOE, ∠BDO = ∠AEO = 90°

& Rest ∠BDO = Rest ∠EAO

ΔBDO ≅ ΔΑΕΟ,

∴ OD = OE

Similarly, ΔBOD ≅ ΔAFO

∴ OD = OF 

∴ OD =OE = OF

∴ O is the incentre of the ΔDEF.

 

 

Question 14. I drew a cyclic quadrilateral ABCD such that AC bisected <BAD. Now produced AD to a point E is such a way that DE AB. Let us prove that CE = CA.

Solution: To prove CE = CA

Join B, D.

Proof: Diagonal AC bisects ZBAD.

∴<BAC = ∠DAC

Again, ∠DBC= <DAC

As these angles are on the same segment DC, 

Again, ∠BDC= <BAC

As on the same segment BC,

∴ ∠DBC= <BDC

∴ BC= DC.

Again, ∠ACD = ∠ABD

As on the same segment AD, 

∴ ∠ABC = ∠ABD + ∠DBC

∠CDE = <DAC + ∠ACD

∴ ∠ABC = ∠CDE

Now, in ΔABC & ΔCED,

AB = DE (given), 

BC = DC & ∠ABC = ∠CDE

∴ ABC == ΔCED (SAS)

∴ AC = CE

 

 

Question 15. In two circles, one circle passes through the centre Q of the other circle and the two circles meet each other at point A intersecting the circle which passes through point O at the point P and the circle with centre O at point R. Joining P, B, and R, B. Let us prove that PR = PQ.

Solution: Let the straight line PB cut the circle with centre O at point C.

Join A, C.

In ΔPRB & ΔACP

∠BPR = ∠APC (same angle)

∠PBR = ∠CAP [. ARBC is a cyclic quadrilateral]

∴ ∠PBR + ∠RAC = 2 rt. angles

= ∠RAC + ∠CAP

∴ ΔPRB + ΔACP are obtuse-angled triangles, i.e., sim

∴∠PRB = ∠PBR

∴ PB = PR Proved.

 

 

Question 16. Let us prove that any four vertices of a regular pentagon are concyclic.

Solution: Let ABCDE is a pentagon.

To prove A, B, C, and E are concyclic.

Join C, E parallel to AB.

As AB || CE

∴ ∠ABC +∠EAB = 2 rt. angles

Again,∠ABC + ∠BCE 2 rt. angles

As the sum of opposite angles of the quadrilateral ABCE is a supplement

∴ Points A, B, C, and E are concyclic.

 

Chapter 10 Theorems Related To cyclic Quadrilateral Exercise 10.1 Multiple Choice Question

 

Question 1. In beside O is the centre of the circle and AB is the diameter. ABCD is a cyclic quadrilateral. <BAC is

1. 50°
2. 60°
3. 30°
4. 40°

Solution: ABC 180° = 120° = 60°

∴∠ACB = 90° (semicircle angle)

<BAC = 180° – (90° +60°) 

= 180° – 150° 

= 30°

Answer. 3. 30°

 

Question 2. In beside O is the centre of the circle and AB is the diameter. ABCD is a cyclic quadrilateral. The value of ∠BCD is

1. 75°
2. 105°
3. 115°
4. 80°

Solution: ∠ACB 90° (semicircle angle)

∠ADC = 180° – 65° 

= 115°

∠ACD = 180° – (115° 40°) 

=180° – 155° 

= 25°

∠BCD = (90° +25°) 

= 115°

Answer. 3. 115°

 

 

Question 3. In beside O is the centre of the circle and AB is the diameter. ABCD is a cyclic quadrilateral in which AB || DC and if BAC = 25° then the value of <DAC is

1. 50°
 2. 25°
3. 130°
4. 40°

Solution: ∠ACB = 90° (semicircle angle)

∴∠ABC = 180° – (90° +25°)

= 180° – 115° 

= 65°

∠DCA alternate ∠CAB = 25°

∠ADC = 180° – 65° 

= 115°

∠DAC = 180° – (115° +25°) 

= 180° – 140° 

= 40°

Answer. 4. 40°

 

 

Question 4. Besides ABCD is a cyclic quadrilateral. BA is produced to the point F. If AE || CD, ∠ABC= 92° and  FAE = 20°, then the value of ∠BCD is

1. 20°
2. 88°
3. 108°
4. 72°

Solution:∠ADC = 180° – 92° 

= 88°

∠DAF = ADC = 88° [AE || CD]

∠DAE = 88° + 20° 

= 108°

∠BAD = 180° – 108° 

= 72°

∠BCD = 180° – 72° 

= 108°

Answer. 3.108°

 

 

Question 5. Besides two circles intersect each other at points C and D. Two straight lines through points D and C intersect one circle at points E and F respectively. If DAB = 75°, then the value of DEF is

1. 75°
2. 70°
3. 60°
4. 105°

Solution: Join D, C.

ZDCF = ZBAD = 75°

ZDEF = 180°- 75° 

= 105°

Answer. 4. 105°

 

 

Chapter 10 Theorems Related To cyclic Quadrilateral Exercise 10.1 True Or False

 

1. The opposite angles of a cyclic quadrilateral are complementary.

False

2. If any side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to the interior opposite angle. 

True

Chapter 10 Theorems Related To cyclic Quadrilateral Exercise 10.1  Fill In The Blanks

 

1. If the opposite angles of a quadrilateral are supplementary then the vertices of the quadrilateral will be Concyclic. 

2. A cyclic parallelogram is a Rectangle.

3. The vertices of a cyclic square are Concyclic.

 

Chapter 10 Theorems Related To cyclic Quadrilateral Exercise 10.1 Short Answer

 

Question 1. In the beside two circles with centres P are Q intersect each other at points B and C. ACD is a line segment. If ∠ARB = 150°, ZBQD = x°, then let us find the value of x.

Solution: BCD 150°

.. Reflex ∠BQD = 2 x 150° 

= 300°

.. x° = zBQD 

= 360° – 300° 

= 60°

.. x = 60°

 

 

Question 2. In the beside two circles intersect at the points P and Q. If∠QAD = 80° and PDA = 84°, then let us find the value of ∠QBC and BCP.

Solution.∠QPC = <DAQ = 80°

∠QBC 180° – 80° = 100°

∠PQB = ∠ADP = 84°

∠BCP = ∠AQD  = 180° – (100° + 60°) 

= 20°

 

 

Question 3. If ∠BAD = 60°, ∠ABC = 80°, then let us find the value of ∠DPC and ∠BQC.

Solution. ∠ADQ = ∠ADC= 180° 80° = 100°

∠DPC = ∠APB 

= 180° – (80° +60°) 

= 40°

∠BQC = ∠AQD 

=180° – (100° + 60°) 

= 20°

 

 

4. In beside O is the centre of the circle and AC is its diameter. If∠AOD = 140° and CAB = 50°, let us find the value of BED.

Solution. ∠BOC =180° – 80° =100°

<BEC = 1/2 <BOC = 1/2 x 100° = 50°

.. OB = OC 

∠OCB =∠OBC = 180°-100°/2

= 80°/2

= 40°

∠BCF =(40° + 10°) = 50°

.. BED (50° + 50°) = 100°

 

 

Question 5. Beside O is the centre of the circle and AB is its diameter. If∠AOD = 140° and CAD = 50°, let us find the value of ∠BED.

Solution. ∠BOD = 180°-140° 

= 40°

.. OB = OD

∠OBD = ∠ODB = 140°/2

= 70°

∠DBE = 180°-70° 

= 110°

∠ACF = 180°-70° 

= 110°

∠BED = ∠AEC = 180° – (110° +50°) 

= 180° – 60° 

=20°

 

 

 

 

 

Chapter 9 Quadratic surd Exercise 9.3

 

Question 1:

1. If m + 1/m = √3, let us calculate the simplified value of (i) m2 + 1/m2 and  (ii) m3 + 1/m3.

Solution (1):  m2 + 1/m2 = (m + 1/m)2 – 2.m.1/m

=(√3)²-2

=3-2 

= 1 

Read and Learn More WBBSE Solutions For Class 10 Maths

Solution (2): m3 + 1/m3 = (m+1/m)3 -3.m.1/m(m+1/m)

 =(√3)²-3√3

= 3√3-3√3 

=0 Ans.

 

2. Let us show that√5+√3 / √5-√3 – √5 – √3 / √5+ √3 = 2√15.

Solution: √5+√3 / √5-√3 – √5 – √3 / √5+ √3 = 2√15

L.H.S.= √5+√3 / √5-√3 – √5-√3 / √5+√3.

= (√5+√3)²-(√5-√3)² / (√5-√3) (√5+√3)

= (5+3+2√15)-(5+3-2√15) / (√5)²-(√3)²

= 8+2√15-8+2√15 /5-3

=4√15 / 2

= 2√15 R.H.S.

Question 2.

1. √2 (2+ √3) / √3(√3+1)  – √2 (2-√3) /√3(√3-1)

Solution : √2 (2+ √3) / √3(√3+1)  – √2 (2-√3) /√3(√3-1)

= √2 / √3 [2+√3 / √3+1 – 2-√3 / √3-1]

= √2 (2√3 −2+3−√3)-(2√3+2−3−√3) / (√3)2-(1)2

= √2 / √3[2√3-2+3-√3-2√3-2+3+√3 / 3-1]

= √2 / √3 x 6-4/2

= √2 / √3 x 2/2

= √2 x √3 / √3.√3

=√6/3

2. 3√7 / √5+ √2 – 5√5 / √2 + √7 + 2√2 / √7 + √5

Solution: 3√7 / √5+ √2 – 5√5 / √2 + √7 + 2√2 / √7 + √5

= 3√/7(√5 – √2) / (√5+√2)(√5+√2) –  5√5(√7-√2) / (√7+√2)(√7-√2) + 2√5/(√7+√5)(√7-√5)

= (√5+√2) √5-√2) (√7 + √2) √7 – √2) + (√7+ √5 √7-√5)

= 3√7(√5-√2)/(√5)2-(√2)2 – 5√5(√7-√2) / (√7)2-(√2)2 + 2√2 / (√7)2-(√5)2

= 3√7(√5-√2) /5-2 – 5√5(√7-√2)/7-2 + 2√2(√7-√5)/7-5

= 3√2(√5-√2)/3 – 5√5(√7-√2)/5 + 2√2(√7-√5)/2

=√35-√14-√35+√10+√14-√10

= 0

3. 4√3 / 2 -√2 – 30/4√3 – √18 – √18/3-√12

 

4. 3√2/√3+√6 – 4√3/√6+√2 + √6/√2+√3

 

 

Question 3. If x = 2, y = 3 and z = 6 let us write by calculating the value of 3√x/√y+√z – 4√y/√z+√x + √z/√x+√y

 

 

Question 4. If x = √7+√6 let us calculate the simplified values of

1. x – 1/x

Solution: x-1/x = (√7+√6)+(√7-√6)

= √7+√6 – √7 +√6

=2√6

 

2. x + 1/x

Solution: x – 1/x

= (√7+√6) + (√7-√6)

=√7+√6+√7-√6

2√7

 

3. x²+1/x²

Solution: x²+1/x²

=(x+1/x)² – 2.x.1/x

=(2√7)² – 2.1

= 28-2

= 26

4. x³ +1/x³

Solution: x³+1/x³

= (x+1/x)³ – 3.x.1/x(x+1/x)

=(2√7)³-3.1.2√7

=8 x 7√7 – 6√7

= 50√7

Question 5. x+√x2-1/x-√x2-1+ x-√x2-1/x+√x2-1 If the simplified value is 14, let us write by calculating what is the value of x.

 

Question 6. If a= √5+1 / √5-1 and b= √5-1/√5+1, let us following expressions

1. a²+ab+b/a²-ab+b²

 

 

2. (a-b)³/(a+b)³

Solution: (a-b)³/(a+b)³

= (√5)³/(3)

=5√5/27

 

3. 3a²+5ab+3b²/3a²-5ab+3b²

 

4. a³+b³/a³-b³

 

 

Question 7. If x = 2 + √3, y = 2-√3, let us calculate the simplified value of:

Solution : x = 2+√3

∴ 1/x = 1/2-√3

= 2-√3/(2+√3)(2-√3)

=2-√3/4-3

=2-√3/1

2-√3

Again, y=2-√3

∴ 1/y = 1/2- √3

=(2+√3)/(2+√3)(2-√3)

=2+√3/4-3

=2+√3/1

2+√3

1. x-1/x

Solution: x-1/x

=(2+√3) – (2-√3)

= 2+√3-2+√3

=2√3

2. y²+1/y²

Solution: y²+1/y²

=(y+1/y)²-2.y.1/y

={(2-√3)+(2+√3)}²

=(2-√3+2+√3)²-2

={(4)²-2)}

= 16-2

= 14

3. x³-1/x³

Solution: x³-1/x³

= (x-1/x)³+3.x.1/x(x-1/x)

=(2√3)³+3.1.2√3

=8.3.√3+6√3

=24√3+√3

=30√3

4. xy+1/xy

Solution: xy+1/xy

=x.y

=(2+√3)(2-√3)

=(2)²-(√3)²

= 4-3

= 1

∴ xy +1/xy

1+1/1

=1+1

=2

5. 3x²-5xy+3y²

Solution: 3x²-5xy+3y²

=3x²-6xy+3y²+xy

=3(x²-2xy+y²)+xy

=3(x-y)²+xy

=3{(2+√3)-(2-√3)}²+1

=3 x 4 x3 +1

=36+1

=37

Question 8. If x = √7+√3/√7-√3 and xy=1, let us show that x²+xy+y²/x²-xy+y² = 12/11

 

 

Chapter 9 Quadratic surd Exercise 9.3 Multiple Choice Question

Question 1. If x=2+√3, the value of x + 1/x is

1. 2
2. 2√3
3. 4
4. 2-√3

Solution:  .. 1/x

=1/ 2+√3

=1x(2-√3)/(2+√3)(2-√3)

=2-√3/4-3

2-√3/1

2-√3

.. x+1/x = 2+v3+2-√3

=4

Answer. 3. 4

Question 2. If p + q = √13 and p-q= √5 then the value of pq is

1. 2
2. 18
3. 9
4. 8

Solution: We know,

pq = (p+q-p-q)/4 

=(√13)2-(√5)2/4

= 13-5/4

=8/4

=2

Answer. 1. 2

Question 3. If a + b = √5 and a-b=√3, the value of (a2 + b2) is

1. 8
2. 4
3. 2
4. 1

Solution: a2+ b2= (a+b)2+(a-b)2/2

= (√5)2+(√3)2/2

= 5+3/2

=8/2

=3

Answer. 1. 8

Question 5. If we subtract √5 from √125, the value is

1. √80
2. √120
3. √100
4. none of these

Solution: √125-√5 

= √5x5x5

= √5x5x5-√5-√5

=4√5

= √16×5

= 80

Answer. 1. √80

Question 6. The product of the bracketed terms (5 -√3), (√3 -1), (5+ √3), and (√3+1) is

1. 22
2. 44
3. 2
4. 11

Solution: (5-√3) (5+√3) (√3-1) (√3+1)

= {(5)2- (√3)2} {(√3)2- (1)}

=(23) x (3-1)

=22 x 2 

= 44

Answer. 2. 44

Chapter 9 Quadratic surd Exercise 9.3 True Or False

 

1. √75 and √147 are similar surds.

Solution: √75 & √5x5x3

=5√3 & √147

= √7x7x3 

=7√3

Answer. True

2.√π is a quadratic surd. 

Answer. False

 

Chapter 9 Quadratic surd Exercise 9.3 Fill In The Blanks

 

1. 5√11 is an Irrational number (rational/irrational)

2. Conjugate surd of (√3-5) is   -√3-5.

3. If the product and sum of two quadratic surds is a rational number, then the surds are Irrational surd.

 

Chapter 9 Quadratic surd Exercise 9.3 Short Answers

Question 1. If x=3+2√2, let us write the value of x + 1/x

Solution: 1/x =1/3 +2√2

1x(3-2√2) / (3+2√2) (3-2√2)

=3-2√2/9-8

=3-2√2

=3+2√2+3-2√2=6

Question 2. Let us write which one is greater between (√15+ √3) and (√10+ √8). 

Solution: Now, (√15+√3)2=(√15)2+(√3)2+2.√15.√3

= 15+ 3 + 2√45

= 18+ 2√45

& (√10 + √8)2 = (√10)2+(√8)2 +2. √10 √8 

= 10 +8 +2√80 

= 18+2√80

As 2√80 is greater than 2√45.

(√10+√8)2 > (√15+√3)2

(√10+√8) is greater than (√15+√3).

 

Question 3. Let us write two quardratic surds whose product is a rational number. 

Solution: (5+2√6) & (5-2√6).

Question 4. Let us write what should be subtracted from √72 to get √32.

Solution. Required number = √72-√32 = √6x6x2 – √4x4x2 

=6√2 -4√2 

= 2√2 Ans.

 

Question 5. Let us write the simplified value of (1/√2+1 + 1/√3+ √2 + 1/√4+ √3)

Solution: 1(√2-1)/(√2+1) (√2-1) + 1(√3-√2)/(√3 + √2) (√3-√2)+1(√4-√3)/(√4+ √3)(√4-√3)

=√2-1/2 -1 +  √3-√2/3-2 +  √4-√34-3

=√2-1+√3-√2+√4-√3

= √4-1

=2-1

= 1.

 

 

 

 

 

Chapter 9 Quadratic surd Exercise 9.2

 

Question 1:

1. Let us find the product of 3  and √3.

 

 

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2. Let us write what should be multiplied by 2√2 to get the product 4.

Solution: Required Number = 4/2√2

= 2 x 2 /2√2

=2/√2

=2.√√2/√2.√2

=2√2/2

=√2

 

3. Let us calculate the product of 3√5 and 5√3. 

Solution: 3√5×5√3

=3×5 x√5 × √3 

= 15√15

4. If √6 x √15 =x√10, then let us write by calculating the value of x.

Solution: If √6 x √15 

= x√10

or, √90 = x√10

or, √9x√10=x.√10

∴ x = √9 = 3

5. If (√5+ √3) (√5-√3) = 25-x2 is an equation, then let us write by calculating the value of x.

Solution: If (√5+√3) (√5-√3)

= 25 – x²

Or, (√5)²-(√3)²

= 25-x2

or, 2=25-x²

or, x² = 25-2 = 23

∴ x = ±√23

Question 2:

1. √7 x √14

Solution: √7x√14 = √7x√7×2 = √7× √7×√2

= 7√2


2. √12 x 2√3

Solution: √12×2√3 = √2×2×3×2√3 = 2√3×2√3 =2x2x√3.√3

= 4 x 3 

=12


3. √5 x √15 x √3

Solution: √5x√15x√3

= √15x√15 √15×15

= 15

4. √2x (3+ √5)

Solution: √2x(3+√5) = 3√2+√2 × √5 = 3√2+√10

5. (√2+ √3) (√2 – √3)

Solution: (√2+√3)(√2-√3) = (√2)-(√3) 2-3-1


6. (2√3 +3√2) (4√2 + √5)

Solution:

(2√3+3√2) (4√2+√5)

=8√6+2√15+12√4+3√10

=8√6+2√15+12.2+3√10

=8√6+2√15 +24+3√10


7. (√3+1) (√3-1) (2-√3) (4+2√3)

Solution : (√3+1)(√3-1) (2-√3) (4+2√3)

= {(√3)2 -(1)2} (2−√3) (4+2√3)· 

=(3-2) (2-√3) × 2(2+√3). 

=2×2×(2-√3) (2+√3) 

= 4x {(2)² – (√3)²}

= 4x (4-3)

= 4 × 1 

= 4 

Question 3:

1. If √x is the rationalizing factor of √5, let us write by calculating the smallest value of x (where x is an integer).

Solution: x = √5

 

2. Let us calculate the value of 3 √2 ÷ 3.

Solution: 3√2 ÷ 3 

= 3√2/3

=√2

3. Let us write which smallest factor we should multiply with the denominator to rationalize the denominator of 7 ÷ √48.

Solution: 7÷ √48 

= 7/√48

= 7/√4x4x3

= 7/4√3

Required smallest factor = √3

 

4. Let us calculate the rationalizing factor of (√5+2) which is also its conjugate surd. 

Solution: (√5+2)

The conjugate surd of (√5+2) is 2- √5.

5. If (√5+ √2) ++ √7 = 1/7(√35 + a), let us calculate the value of a.

Solution: (√5+√2) ++√7 = 1/7(√35+a)

If √5+ √2 /7 =√35+a/7

=(√5+√2)√7 / √7 x √7 = √35+a/7

Or, √35+ √14/7 = √35+a/7

Or, √35+ √14 = √35+a

∴ a = √14

 

6. Let us write a rationalizing factor of 5/√3-2 which is not its conjugate surd.

Solution: 5/√3-2

The rationalizing factor of the denominator of 5/√3-2 is (√3+2).

 

Question 4. Let us write the conjugate surds of mixed quadratic surds (9-4√3) and (-2 -√7). 

Solution: Conjugate surd of (9-4√5) is (9+4√5)

& conjugate surd of (-2-√7) is (-2+√7).

5. Let us write two conjugate surds of each of the mixed quadratic surds given below.

1. √5+ √2

Solution: √5+√2

Two conjugate surds of √5+ √2 are (√5-√2) & (√5+√2).

 

2. 13+ √6

Solution: 13+ √6

Two conjugate surds of 13+ √6 are (13-√6) & (-13+ √6). 

3. √8-3

Solution: √8-3

Two conjugate surds of √8-3 are (- √8-3) & (√8+3).

 

4. √17-√15

Solution: √17-√15

Two conjugate surds of √17-√15 are (√17 + √15) & (-√17 – √15).

 

6. Let us rationalize the denominators of the following surds 

1. 2√3+3√2 / √6

Solution: 2√3+3√2/√6 

=(2√3+3√2) x √6/ √6 x √6 

= 2√18+3√12/6

=2×3√2+3×2√3 / 6

=6√2+6√3 / 6

= 6(√2+√3)/6

=√2 + √3

2. √2-1+√6 / √5

Solution: √2-1+√6/√5

 = (√2-1+ √6)x√5 / √5 x √5

= √10-√5+√30 / 5

3. √3-1 / √3+1

Solution: √3-1 / √3+1

= (√3+1)(√3+1) / (√3-1)x(√3+1)

= 3+1+2√3 / 3-1

=4+2√3 / 2

=2(2+√3) / 2

= (2 + √3) 

4. 3+√5 / √7-√3

Solution: 3+√5 / √7-√3 

= (3+√5) (√7 + √3) / (√7-√3) (√7 + √3) 

= 3√7+√35+3√3+ √15 / (√7)²-(√3)²

= 3√7+√35+3√3+ √15 / 7-3

= 3√7+√35+3√3+√15 / 4

5. 3√2+1 / 2√5-1

Solution: 3√2+1 / 2√5-1

(3√2+1)x(2√5+) / (2√5-1)(2√5+1) 

= 6√10+3√2+2√5+1 / (2√5)² – (1)²

=6√10 +3√2+2√5+1 / 4×5-1

6√10+3√2+2√5+1 / 19

6. 3√2+2√3 / 3√2-2√3

Solution: 3√2+2√3 / 3√2- 2√3

= (3√2+1)x(3√2+2√3) / (3√2-2√3) (3√2+2√3)

=9×2+6√6+6√√6+4×3 / (3√2)²-(2√3)²

= 18+12√6+10 / 9×2-4×3

30+12√6 / 18-12

 6(5+2 5+2√3) / 6

=5+2√3.

7. Let us divide first by second and rationalize the divisor.

1. 3√2 + √5, √2+1

Solution : 3√2+√5 / √2+1

= (3√2+√5)x(√2-1) / (√2+1)x(√2-1)

= 3×2-3√2+√10-√5 /(√2)² – (1)²

= 6-3√2+√10-√5 / 2-1

=6-3√2+√10-√5 Ans.

2. 2√3-√2, √2-√3

Solution:

= 2√3-√2 – √2-√√3

=(2√3-√2)x(√2+√3) / (√2-√3)x (√2+√3)

= 2√6+2×3-2-√6 / (√2)-(√3)

= √6+6-2 / -1

= √6+4 / -1
=  -(√6+4)

 

3. 3+√6, √3+√2

Solution:  3+√6  / √3+√2

= (√3+√6) (√3-√2) / (√3+√2) (√3-√2) 

= 3√3-3√2+√18-√12 / (√3)²-(√2)²

= 3√3-3√2+3√2-2√3 / 3-2

=√3

Question 8: 

1. 2√5+1 / √5+1 – 4√5-1 / √5-1

Solution: 2√5+1 / √5+1 – 4√5-1 / √5-1

= (2√5+1)(√5-1) – (4√5-1√5+1)  / √5+1 √5-1

= (2×5-2√5+√5-1)-(4×5+4√5-√5-1) / (√5)²-(1)²

= 10-√5-1-20-3√5+1 / 5-1

= -10-4√5 / 4 

= 2(-5-2√5) / 4

= (-5-2√5) / 2

2. 8+3√2 / 3+√5 – 8-3√2 / 3-√5

Solution: 8+3√2 / 3+√5 – 8-3√2 / 3-√5

=(8+3√2 3-√5)-(8-3√2)(3 -+√5) / (3+√5) (3-√5)

=(24-8√5+9√2-3√10)-(24+8√5-9√2-3√10) / (3)²-(√5)²

=24-8√5+9√2-3√10-24-8√5+9√2+3√10 / 9-5

18√2-16√5 / 4

2(29√2-8√5) / 2

= 9√2-8√5 / 2

 

Application 1: 3√20+2√28+ √12 / 5√45+2√175+√75

Solution: 3√20+2√28+√12 / 5√45+2√175+√75

= 3x√2x2x5+2√2x2x7 +2√2x2x3 / 5√3x3x5 +2√5x5x7 + √5x5x3

= 6√5+4√7+4√3 / 15√5+10√7+5√3 

= 2(3√5+2√7+√3) / 5(3√5+2√7+√3)

= 2/5

 

Application 34: 5 / √2+√3 – 1/√2-√3

Solution: 5/√2+√3 – 1/√2-√3

= 5(√2-√3)-1(√2+√3) / (√2+√3)(√2-√3 )

= 5√2-5√3-√2-√3 / (√2)² – (√3)²

= 4√2-6√3 / 2-3

= 4√2-6√3 / -1

=6√3-4√2 

Application 36. If x =  √3+√2  let us calculate the simplified value of (x – 1/x) (x³ – 1/x³) and (x² – 1 / x²)

Solution: If x = √3+√2, find the values of (x – 1/x) (x³ – 1/x³) and (x² – 1 / x²)

1/x = 1 / √3-√2

∴ x – 1/x = (√3+√2)-(√3-√2)

=√3+ √2−√3+ √2 

=2√2

=(√3+√2)+(√3-√2)

=√3+ √2+√3-√2 

=2√3

Now, x³ + 1/x³

=(x + 1/x)³ – 3.x.1/x(x+1/x)

=(2√3)³-3×2√3 

=8×3√3-6√3

=24√3-6√3 

= 18√3

&  x²- 1/x² = (x + 1/x)(x – 1/x)

=2√3 ×2√2 

= 4√6 

 

 

 

 

Chapter 9 Quadratic surd Exercise 9.1

 

Application 1. I write by understanding 4 pure quadratic surds and 4 mixed quadratic surds.

Solution: 4 pure quadratic surds are √3,- √5

4 mixed quadratic surds are 2-√3; 2+ √6 , 3/2 – √10 , 3+ √5

Application 2. Are √4, √25 quadratic surds?

Solution: Apparently √4. √25 are in the form of surds but they are not surds. 

Rational number, √4 = 2 and √25 = 5.

I apply Sreedhar Acharyya’s formula for solving the equation x2 – 2ax + (a2-b2) = 0. 

We see that the roots are a + √b and a- √b, 

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both of which are mixed surds, where b is a positive rational number which is not a square number of any rational number.

Application 3. What type of number do we get by the addition, subtraction, multiplication, division, and square of the two numbers 8 and 12?

Solution: 8+ 12 = 20 (Integer)

8-12= -4 (Integer)

8 x 12 = 96 (Integer)

8/12 = 2/3 (Rational)

Application 4. We write similar surds in a specific place from the following quadratic surds. 

√45, √80, √147, √180 and √500

Solution:  √45, √80, √147, √180 and √500

√45

= √9×5 

= 3√5

√80

= √16×54 

= 4√5

√147

= √7x7x3

= 7√3

√180

=√6x6x5 

= 6√5

√500

= √2x2x5x5x5 

=2×5√5

=10√5

Application 5. Let us write the similar surds among the quadratic surds √48,

√27, √20 and √75

Solution: √48,√27, √20 and √75

√48
= √2x2x2x2x3
= 4√3

√27
= √3x3x3
= 3√3

√20
=√2×2×5
=2√5

√75
= √5x5x3
=5√3

√48. √27 √75 are similar surds.

Application 6. Let us write by calculating the value of (√12+√45) and (√2-√8) and see whether they can be expressed in pure quardratic surds.

Solution:

√2+ √8
= √2 + √2×2×2

= √2+2√2

= 3√2

√2-√8 

= √2-√2x2x2

= √2-2√2

=-√2

(√2-√8)  is a pure quadratic surd.

Application 7. Let us write by calculating the sum of √12,-4√3 and √3 

Solution: (√12)+(-4√3) +6√3

= 2√3 −4√3 +6√3

= 4√3.

Application 8. (9-2√5) + (12+7√5)   

Solution: (9-2√5) + (12+7√5)

=9+12-2√5

= 21 +5√5

Application 9. I write any other two quadratic surds whose sum is a rational number. 

Solution: (6+√7)+(6-√7)=6+6+ √7-√7 = 12

Question 1. Let us write the following numbers in the form of the product of rational and irrational numbers.

1. √175

Solution: √175

=√5x5x7

= 5√7

2. 2 √112

Solution: 2√112

= 2.√4x4x7

=2×4√√7=8√7

3. √108

Solution: √108 

= √2x2x3x3x3

=2×3√3 =6√3

4. √125

Solution: √125

= √5x5x5

= 5√5

5. 5√√119

Solution: 5√119

= 5√7×17

= 5√119

Question 2. Let us show that √108-√75 = √3

Solution: √108 – √75 = √3

L.H.S

= √108 – √75 

= √6x6x3 – √5x5x3

= 6√3-5√3 

= √3 

R.H.S.

Question 3. Let us show that √98+ √8-2√32 = √2

Solution: √98 + √8-2√32 = √2

L.H.S= √98 + √8 -2√32 

= √7x7x2 + √2x2x2 -2√4x4x2 

=7√2 +2√2 -2×4√2

=9√2-8√2 

= √2 

R.H.S.

Question 4. Let us show that 3 √48-4√75+ √192 = 0

Solution: 3√48 -4√75 + √192 =0

L.H.S.

= 3√48-4√75 + √192

= 3√4x4x3 -4√5x5x3 + √8x8x3

=3×4√3-4×5√3 +8√3 

= 12√3-20√3 +8√3 

=20√3-20√3 

= 0 

R.H.S.

Question 5. Let us simplify: √12 + 18+ √27 – √32

Solution: √12+ √18+ √27-√32

=√2x2x3 + √3x3x2 + √3×3×3 – √4x4x2

=2√3 +3√2 +3√3-4√2

=2√3 +3√3 +3√3 -4√2

=5√3 – √2

 

Question 6.

1. Let us write what should be added with √5+ √√3 to get the sum 2√5. 

Solution: Required number = 2√5 – (√5+√3)

=2√5-√5-√3

=√5-√3

2. Let us write what should be subtracted from 7-√3 to get the sum of 25.

Solution: Required number = (7-√3)-(3+√3)

=7-√3-3-√3 

=7-3-√3 

=4-2√3.

3. Let us write the sum of 2+√3, √3+ √5, and 2+ √7. 

Solution: Required sum = 2 + √3 + √3 + √5 +2+√7

=2+2 + √√3 + √√3 + √5 + √7

=4+2√3 +√5+√7.

4. Let us subtract (-5+3 √11) from (10+ √11) and let us write the value of the subtraction.

Solution: Required subtraction = (10-√11)-(-5+3√11)

= 10- √11+5-3√11 

= 15-4√11

5. Let us subract (5+ √2+ √7) from the sum of (-5+ √7) and (√7 + √2) and find the value of the subtraction.

Solution: Required value of subtraction = (-5+√7) + (√7+√2) – (5+√2+√7)

=-5+ √7 + √7 + √2-5-√2-√7 

=-10+ √7

6. I write two quadratic surds whose sum is a rational number. 

Solution: Two quadratic surds whose sum is a rational number, 

5+√3:5-√3.

Application 1. Let us write by calculating the product of (3+ √7 √5) and (2√2-1) 

Solution: (3+ √7-√5) x (2√2-1)

=6√2-3+2√14 – √7-2 √10-√5

Application 2. Let us write two rationalizing factors of √7. 

Solution: √7 & 2√7

Application 3. Let us see what will be the rationalizing factor of (5+ √7).

Solution : (5+√7)

= (5+√7)x (5-√7)

= (5)²- (√7)²

= 25-7

=18 [ (a+b) (a-b) = a2-b2]

Again, (5+√7)x(5+7)

=(√7 +5) (√7 -5)

=(√7)²- (5)²

=-18

Application 4. Let us write two rationalizing factors of 7-√3 

Solution: 7-√3

=(7+√3); (-7-√3)

Application 5. Let us see the rationalizing factors of (√11 – √6).

Solution: (√11-√6) (√11+√6)

=(√11)²-(√6)²

= 11-6

=5

Again, (√11-√6) √11-√6) = [(√11-√6) (√11+√6)]

= [11 – 6]

=-5

Application 6. Let us write two rationalizing factors of (√15+ √3)

Solution : (√15+√3)

(√15-√3): (-√15+√3)

Application 7. Let us write the conjugate guards of the following mixed and pure surds

1. 2+√3

Solution: b

=2-√3

2. 5-√2

Solution: 5-√2

=5+√2

3. √5-7

Solution: √5-7

=-√5+7 

4. √11 + 6

Solution: √11 + 6

= (6-√11)

5. √5

Solution: √5

= – √5

Application 8. Let us rationalize the denominator of

1. 4√5 / 5/√3

Solution:4√5 / 5√3

=4√5.√3 / 5√3.√3

=4√3/5×3

= 4√3 / 15

2. √6/3√7

Solution: √6/3√7

= 3√7/√6

= 3√7x√6 /√6x√6

= 3√42 /6

= √42/2

Application 9. Let us rationalize the denominator of 

1. (4+2√3)+(2-√3)

Solution: 4+2√3 / 2-√3

=(4+2√3) (2+√3) /(2-√3)2+√3)

=8+4√3 +4√3 +6/(√2)²-(√3)²

=14+8√3/4-3

= 14+8√3

2. (√5+ √3) + (√5 – √3)

Solution: √5+√3/√5-√3

=(√5+√3)(√5+√3)/(√5-√3)(√5+√3)

= 5+2√5.√3+3 /(√5)²-(√3)²

=8+2√15/ 5-3 

=2(24+√15)/2

= 4+ √15.

 

 

 

Ganit Prakash Class 10 Chapter 8 Solutions

Chapter 8 Right Circular Cylinder Exercise 8.2

 

Question 1. Let us look at the picture of the solid below and answer the following

1. Solid has    surfaced.

Answer. 3

2. Number of the curved surfaces is    and number of plane surface is                                                              

Answer. 1, 2.

Read and Learn More WBBSE Solutions For Class 10 Maths

Question  2. Let us write the names of five solid objects of my house, the shapes of which are right circular cylinder.

Solution: Glass, 

Pipe, 

Tube light, 

Pencil, 

Gas cylinder.

 

Question  3. The length of diameter of a drum made of steel covered with a lid is 28 cm. If a 2816 sq cm steel sheet is required to make the drum, let us write by calculating the height of the drum.

Solution: Let the height = h cm

& radius of the drum = 28/2 = 14 cm.

According to the problem,

2π(14+ h) x 14 = 2816

or, 2 x 22/7 x 14 (14+ h) = 2816

or, 14+h=2816 x 7/2 x 22 x 14 

= 32

∴ h=32-14

18 cm.

∴Height of the drum = 18 cm.

Ganit Prakash Class 10 Chapter 8 Solutions

Question  4. Let us write by calculating how many cubic decimetres of concrete materials will be necessary to construct two cylindrical pillars, each of whose diameter is. 5.6 decimetres and height are 2.5 meters.

Solution: Height of each pillar = h = 2.5 m = 25 dcm.

Radius of each pillar = (r) = 5.6/2 = 2.8 dcm.

The volume of plaster materials required to cover the two pillars

= 2 × π  x (2.8)2 × 25 cu dcm.

= 2x 22/7 x 28/10 x 28/10 25 

= 56 x 22

= 1232 cu dcm.

The curved surface area of two pillars = 2 x π x 2.8 x 25 sq dcm.

= 2 x 2 x 22/7 x 28/10 

= 880 Sq dcm. 

= 8.80 sqm.

Total cost at the rate of Rs. 125 per sqm for the two pillars

= Rs. 8.8 x 125

= Rs. 1100. 

 

Question  5. If a gas cylinder for fuel purposes having a length of 7.5 dcm and the length of an inner diameter of 2.8 dcm carries 15.015 kg of gas, let us write by calculating the weight of the gas of per cubic dcm.

Solution: Weight of gas in the cylinder 15.015 kg = 15015. gm.

The radius of the cylinder = 2.8/2

= 1.4 dcm.

∴The volume of the cylinder = (1.4)2 x 7.5 cu dcm.

=22/7 x 14/10 x 14/10 x 75/10

=462/10

= 46.2 cu dcm.

∴ Weight of 1 cu dcm gm = 15015/46.2gm 

= 325 gm.

Ganit Prakash Class 10 Chapter 8 Solutions

Question  6. Out of three jars of equal diameter and height, 2/3 part of the first, 5/6 part of the second, and 7/9 part of the third were filled with dilute sulphuric acid. The whole of the acid in the three jars was poured into a jar of 2.1 dcm diameter, as a result the height of acid in the jar becomes 4.1 dcm. If the length of the diameter of each of the three equal jars is 1.4 dcm, let us write by calculating the height of the three jars.

Solution: Let the height of each jar = h dcm.

According to problem,(2/3 + 5/6 + 7/9) x h = π(2.1/2)2 x 4.1

Or,12+15+14/18 x h

= 22/7 x 21/20 x 21/20 41/10

.. h = 21/7 x 21/20 x 21/20 x 41/10 c 18/41 

4.05 dcm.

∴ The height of each jar is 4.05 dcm.

Ganit Prakash Class 10 Chapter 8 Solutions

Question  7. The total surface area of a right circular pot open at one end is 2002 sq cm. If the length of diameter of base of the pot is 14 cm, let us write by calculating how many liters of water the drum will contain.

Solution: Let the height of the cylinder = h cm.

& the radius of the cylinder =14/2 = 7 cm.

According to the problem,

or, 2π x 7 x h + л(7)2 = 2002

or, 2 x 22/7 x 7 x h+ 22/7 x 7 x 7 = 2002

or, 44h+ 154 = 2002

∴ 44h = 2002 – 154 = 1848

∴ h= 1848/44

= 42

∴ Height of the cylinder = 42 cm. = 4.2 dcm.

The radius of the cylinder = 7 cm. = 0.7 dcm.

∴ The volume of water in the cylinder = 7(0.7)2 x 4.2 cu dcm.

= 22/7 x 7/10 x 7/10 x 42/10 

=6468 / 1000

= 6.468 cu dcm.

=6.468 litre [1 cu dcm = 1 litre]

Maths Class 10 WBBSE Solutions

Question  8. If a pump set with a pipe of 14 cm diameter can drain 2500 meters of water per minute, let us write by calculating how much kilometer of water that pump will drain per hour. [1 liter = 1 cubic dcm.]

Solution: Radius of the pipe = 14/2 = 7 cm = 0.7 dcm.

Height of the pipe = 2500 m = 25000 dcm. 

The cross-sectional area of the mouth of the pipe

= 22/7 x 72 sq cm. = 154 Sq cm. 1.54 sq dm.

The volume of water irrigated by the pipe in one minute

= (1.54 x 25000) cu dcm 

= 38500 cu dcm

∴ The volume of water irrigated by the pipe in one hour

= (38500 x 60) cu dcm 

= 2310000 cu dcm.

Now, 1 cu dcm = 1 liter

∴ The pump set can irrigate 2310000 liters

= 2310 Kilolitre in 1 hour.

 

Question  9. There is some water in a long gas jar of 7 cm in diameter. If a solid right circular cylindrical pipe of iron having 5 cm length and 5.6 cm diameter be immersed completely in that water, let us write by calculating how much the level of water will rise. 

Solution: Let the water level will rise by h cm.

According to the problem,

π(7/2)² x h = π(5.6/2)² x 5

49/4 h = 28/10 x 28/10 x 5 

H = 28 x 28 x 5 / 10 x 10 x 49 x 4 

= 32/10

= 3.2

∴ The water level will rise by 3.2 cm.

Maths Class 10 WBBSE Solutions

Question  10. If the surface area of a right circular cylindrical pillar is 264 sq meters and the volume is 92 cubic meters, let us write by calculating the height and length of the diameter of this pillar.

Solution: Let the radius of the pillar = rm & height = h m.

According to 1st condition, 2πth = 264 ——– (1)

According to 2nd condition, πr²h = 924———(2)

πr²h/2πrh = 924/264

Or, r = 7

2πrh 2 x 22/7 x 7 h = 264

H = 264/44 

= 6

∴ Diameter of the pillar = 2 x 7 = 14 m

& height of the pillar = 6 m.

 

Question  11. A right circular cylindrical tank of 9 meters in height is filled with water. Water comes out from there through a pipe having a length of 6 cm diameter with a speed of 225 meters per minute and the tank becomes empty after 2 hr. 24 minutes, let us write by calculating the length of the diameter of the tank.

Solution: Let the radius of the tank = R m

& radius of the pipe = 6/2 

= 3 cm.

= 0.03m.

According to the problem,

π R² x9= 36 (0.03)∴ x 225       (Where radius of pipe = R)

R² = 36 x 3/100 x 3/100 x 225 x 1/9

= 81/100

∴ R = 9/10

Diameter of the tank = 2 x 9/10

= 1.8 m.

 

Question  12. The curved surface area of the right circular cylindrical log of wood of uniform density is 440 sq dcm. If I cubic dcm of wood weighs 1.5 kg and the weight of the log is 9.24 quintals. Let us write by calculating the length of the diameter of the log and its height. 

Solution: Weight of the wooden log = 9.24 quintal = 924 kg.

∴ The volume of the log = 924 / 1.5 cu dcm 616 cu dcm.

Let the radius of the log = r dcm

& height of the log = h dcm.

According to 1st condition, 2лth = 440———-(1)

 According to 2nd condition, лr²h = 616———(2)

= лr²h/2лrh = 616/440 

Or, r/2 = 616/440

R = 2 x 616 / 440

= 28/10

Or, 2 x 22/7 x 28/10 

h = 100/4

= 25

∴ Diameter of the log = 2 x 28/10 

= 56/10

= 5.6

Height of the log = 25. dcm.

Maths Class 10 WBBSE Solutions

Question  13. The lengths of the inner and outer diameter of a right circular cylindrical pipe open at two ends are 30 cm and 26 cm respectively and the length of the pipe is 14.7 metres. Let us write by calculating the cost of painting its all surfaces with coal tar at Rs. 2.25 per dcm.

Solution: Internal radius of the pipe = 26/20

= 1.3 dcm.

The external radius of the pipe = 30/20

= 1.5 dcm.

Height of the pipe 14.7 m = 147 dcm.

The total surface area of the pipe

= [27(1.5+ 1.3) 147+ 2π((1.5)2- (1.3)2)] sq dcm..

= [2π x 2.8 × 147 + 2 x 2.8 x 0.2] sq dcm.

= 2π x 2.8(147+ 0.2)

= 2x 22/7 x 28/10 x 147.2 sq dcm.

= 17.6 x 147.2 2590.72 sq dcm.

Total cost for painting with coal tar

= Rs. (2.25 x 2590.72)= Rs. 5829.12.

 

Question  14. The height of a hollow right circular cylinder, open at both ends, is 2.8 metres. If the length of the inner diameter of the cylinder is 4.6 dcm and the cylinder is made up of 84.48 cubic dcm of iron, let us calculate the length of the outer diameter of the cylinder. 

Solution: Let the external radius = r dcm. of the cylinder & the height of the cylinder = 2.8 m = 28 dcm.

The internal radius of the cylinder = 4.6/2

= 2.3 dcm.

According to the problem,

л(r² (2.3)²) x 28 = 84.48

or,22/2(r2-5.29) × 28 = 84.48

r² – 5.29 = 84.48/88 = 0.96

R² = 5.29

= 5.29 + 0.96

= 6.25

External diameter = 2 x 2.5 5 dcm.

 

Question  15. Height of a right circular cylinder is twice its radius. If the height would be 6 times its radius, then the volume of the cylinder would be greater by 539 cubic dcm, let us write by calculating the height of the cylinder.

Solution: Let the height of the cylinder = h dcm & radius = r dcm.

∴ h = 2r

Volume of the cylinder = r2 x 2r= 2πr3 

If height h=6r, volume = r2 x 6г = 6πr3 

According to the problem, 6r²r³ = 539

4πr³ = 536

Or, 4 x 22/4 x r³ =539

∴ r³ = 539 x 7/88

=(7/3)³

r =7/2

2r = 7

∴ Height 2r= 7dcm.

Maths Class 10 WBBSE Solutions

Question  16. A group of fire brigade personnel carried a right circular cylindrical tank filled with water and pumped out water at a speed of 420 metres per minute to put out the fire in 40 minutes by three pipes of 2 cm in diameter each. If the diameter of the tank is 2.8 meters and its length is 6 metres, then let us calculate 

1. what volume of water has been spent in putting out the fire and 

2. the volume of water that still remains in the tank.

Solution: Radius of the tanker = 2.8/2

= 1.4

= 14 dcm.

Length of the tanker = 6 m = 60 dcm.

The volume of the tanker = π(14)² x 60 cu dcm.

Radius of each pipe = 2/2 

= 1 cm

= 1/10 dcm.

Length of the pipe = 420 m = 4200 dcm.

Quantity of water ejected in 40 minutes by 3 pipes

= 40 x 3 x π(1/10)² x 4200 cu dcm.

= 40 x 3 x 22/7 x 1/10 x 1/10 x 4200 cu dcm.

= 15840 cu dcm.

∴ Quantity of water left in the tanker

= (3696015840) cu dcm.

= 21120 cu dcm.

 

Question  17. It is required to make a plastering of sand and cement with 3.5 cm thick, sur- rounding four cylindrical pillars, each of whose diameter is 17.5 cm.

1. If each pillar is of 3-meter height, let us write by calculating how many cubic dcm of plaster materials will be needed.

2. If the ratio of sand and cement in the plaster material be 4: 1, let us write how many cubic dcm of cement will be needed. 

Solution: Length of each pillar = 3 m = 30 dcm.

Internal radius of each pillar = 17.5/2

175/20 cm.

= 7/8 dcm.

The external radius of each pillar = (7/8 + 75/10) dcm.

= (7/8 + 35/100)

= (7/8 + 7/20)dcm.

= 35+14 / 40

= 49/40 dcm.

1. Plaster material is required for 4 pillars

= 4 π {(49/40)² – (7/8 )²} x 30 cu dcm.

= 4x 22/7 {(49/40 + 7/8)(49/40 – 7/8)} x 30 cu dcm.

= 4 x 22/7 x 84/40 x 14/40 x 30 cu dcm.

= 2772/10

= 277.2 cu dcm.

2. Volume of cement required = 1/5x 277.2 cu dcm. 

= 55.44 cu dcm.

School Mathematics Class 10 Solutions

Question 18. The length of the outer and inner diameter of a hollow right circular cylinder are 16 cm and 12 cm respectively. Height of the cylinder is 36 cm. Let us calculate how many solid cylinders of 2 cm radius and 6 cm length may be obtained by melting this cylinder.

Solution: External radius of the cylinder = 16/2

= 8 cm.

Internal radius of the cylinder = 12/2

= 6 cm.

Height of the cylinder = 36 cm.

The radius of the solid cylinder = 2/2 

= 1 cm.

Height of the cylinder = 6 cm.

Volume of hollow cylinders = π{(8)²- (6)²} x 36 cu cm.

The volume of x no. of solid cylinders = x. π(1)². 6 cu cm.

According to the problem,

2. π(1)². 6=π{(8)² – (6)²) x 36

or, x. π. 1.6 = πx (64-36) x 36

or, x = 28 x 36 / 6

= 168

∴ No. of solid cylinders = 168.

 

Multiple Choice Question

Question 1. If the lengths of radii of two solid right circular cylinders are in the ratio 2:3 and their heights are in the ratio 5: 3, then ratio of their lateral surface areas is

1. 2:5
2. 7:7
3. 10: 9
4. 16:9

Solution: Ratio of area of curved surfaces

=2π ×2×5:2π × 3 × 3 

=10:9

Answer. 3. 10: 9

School Mathematics Class 10 Solutions

Question 2. If the lengths of radii of two solid right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5: 3, then the ratio of their volumes is

1. 27: 20
2. 20:27
3. 40:9
4. 9:4

Solution: Ratio of volumes of the cylinder

=π(2)² × 5: π(3)² x 3 = 20:27

Answer. 2. 20:27

Question 3. If volumes of two solid right circular cylinders are the same and their heights are in the ratio 1 2, then the ratio of lengths of radii is

1. 1:√2
2. √2:1
3. 1:2
4. 2:1

Solution: Ratio of the radius of the cylinder = √2:1

Answer. 2. √2:1

4. In a right circular cylinder, if the length of radius is halved and height is doubled, volume of the cylinder will be

1. Equal
2. Double
3. Half
4. 4 times

 

 

Question 5. If the length of the radius of a right circular cylinder is doubled and the height is halved, the lateral surface area will be

1. Equal
2. Half
3. Double
4. 4 times

 

School Mathematics Class 10 Solutions True Or False

1. The length of a right circular drum is r cm and the height is h cm. If half part of the drum is filled with water then the volume of water will be r2h cubic cm.

False

2. If the length of the radius of a right circular cylinder is z unit, the numerical value of volume and surface area of the cylinder will be equal for any height.

True

 

Fill In The Blanks

1. The length of a rectangular paper is units and the breadth is b units. The rectangular paper is rolled and a cylinder is formed whose perimeter is equal to the length of the paper. The lateral surface area of the cylinder is lb sq unit.

2. The longest rod that can be kept in a right circular cylinder has a diameter of 3 cm and height of 4 cm, then the length of the rod is 5  cm.

3. If the numerical values of volume and lateral surface area of a right circular cylinder are equal then the length of the diameter of the cylinder is 4 units.

School Mathematics Class 10 Solutions

Chapter 8 Right Circular Cylinder Exercise 8.2 Short Answers

Question 1. If the lateral surface area of a right circular cylindrical pillar is 264 sq meters and the volume is 924 cubic meters, let us write the length of the radius of the base of the cylinder.

Solution: Let the radius & height of the pillar = r m & h m respectively.

According to 1st condition, 2rh = 264——–(1)

According to 2nd condition, r2h = 924 ——-(2)

(2) ÷ (1),

= πr²h/2πrh

= 924/264

Or, r/2 = 7/2

R = 7

∴ Radius = 7 cm.

 Question 2. If the lateral surface area of a right circular cylinder is c square units, the length of the radius of the base is r units and the volume is cr/v cubic units, let us write the value of 

Solution. Let height = h unit.

Cr/v=  2πrhxr / πr²h

= 2

 Question 3. If the height of a right circular cylinder is 14 cm and the lateral surface area is 264 sq cm, let us write the volume of the cylinder.

Solution. Let the height of the cylinder = r cm.

∴ 2πгh = 264 

or, 2 x 22/7 x r x 14= 264

r=3

∴ Volume = π(3)² x 14 cu cm 

= 22/7 x 9 x 14 cu cm = 396 cu cm.

 Question 4. If the heights of two right circular cylinders are in the ratio of 1: 2 and prim- meters are in the ratio of 3: 4, let us write the ratio of their volumes.

Solution. The Radius & height of the two cylinders are r unit & R unit & heights are h unit & 2h unit respectively.

∴ 2πr : 2πR 3:4 

∴ r: R 3:4 

h =3R / r

The ratio of volume = (r)² h: (R)²

2h = (3R/4)²: 2R²

Or, 9R/16 : 2R2 = 9:32

 

Question 5. The length of the radius of a right circular cylinder is decreased by 50% and height is increased by 50%, let us write by how much percent of the volume will be changed.

 

 

 

 

 

West Bengal Board Class 10 Math Book Solution In English

Chapter 7 Theorems Related To Angles In A Circle Exercise 7.1

Question 1. In the adjoining figure, ∠AMB, formed by the circular arc ÁPB, is the angle, and ∠ANB, formed by the circular arc is the front angle of the circle.

 

Solution: In the adjoining figure, ∠AMB formed by the circular arc APB is a cyclic angle and ∠ANB formed by circular arc AQB is the front angle of the circle.

Read and Learn More WBBSE Solutions For Class 10 Maths
Question 2. ∠SLT in the adjoining figure at point L by the chordis front angle. Again, since the point L lies on the circle, so angle formed by the chord ST is a front angle in the segment.Again, ∠SLT is formed by the circular arc is front angle of the circle. The four angles in the segment of a circle with center O in the adjoining figure are,, and

Solution: ∠SLT in the adjoining figure at the point L by the chord ST is the front angle.

Again, since point, L lies on the circle, so angle SLT formed by the chord ST is a front angle in the segment.

 

————–(1)

 

Again, ∠SLT formed by the circular arc SNT is a front angle of the circle. The four angles in the segment of a circle with centre O in the adjoining figure are

∠ADB, ∠AEB, ∠ACB and ∠ASB.

 

—————–(2)

 

We see in no. (1), (2) circles the angle at the center formed by the arc AQB

West Bengal Board Class 10 Math Book Solution In English

is ZAOB and an angle in the segment is ZAPB. But in no. (3) circle, the angle at the center formed by the arc ASB is ZAOB and the angle in the segment formed by the arc ASQ is ZAPQ.

 

 

Class 10 WBBSE Math Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5

 

Question 1. In AABC, ZB is a right angle. A circle drawn taking AC as diameter intersects AB at the point P; let us write the correct information from the followings:

1. AB > AD

2. AB = AD

3. AB < AD.

Solution: We know the angle in a semicircle angle = 90°

∴ ∠ADC = 90° (given)

But ∠ABC 90°

It is possible that points B & point D are the same points. 

Read and Learn More WBBSE Solutions For Class 10 Maths

∴ AB = AD

 

 

Question 2. Let us prove that the circle drawn with any one of the equal sides of an isosceles triangle as diameter bisects the unequal side.

Solution: In Δ ABC, AB = AC.

A circle with center O and AB as the diameter is drawn.

The circle cuts BC at D.

To prove D bisects BC.

Proof: Join A, B.

As AB is the diameter,

.. ∠ADB is one right angle.

i.e., AD ⊥ BC & ∠ADC = 90°

Now, in two right-angled ΔABD & ΔACD, Hypotenuse AB Hypotenuse AC (given), And AD are common.

∴ΔABD ≅ ΔACD 

∴ BD = CD

∴ D bisects the unequal side of the isosceles triangle ABC.

 

Ganit Prakash Class 10 Solutions Pdf In English

Question 3. Sahana drew two circles that intersect each other at the points P and Q. If the diameters of the two circles are PA and PB respectively, then let us prove that A, Q, and B are collinear.

Solution: Two circles cut each other at P & Q.

If PA & PB are the diameters of the two circles, 

prove that points A, Q, and B are collinear.

Proof: Join Q, A; Q, B & P, Q.

As AP is the diameter & AQP is a semi-circle angle,

∴ ∠AQP = 1 rt. angle

Similarly, BP is the diameter.

∴∠BQP = 1 rt. angle

∴ ∠AQP+∠BQP = 2 rt. angles

Now, AQ & BQ meet at Q & the sum of the adjacent angles = 2 rt. angles

∴ AQ and BQ are on the same straight line.

∴ A Q & B are collinear.

 

Question 4. Rajat drew a line segment PQ whose midpoint is R and two circles are drawn with PR and PQ as diameters. I drew a straight line through point P which inter- sects the first circle at point S and the second circle at point T. Let us prove with the reason that PS = ST.

Solution: PQ is a straight line whose midpoint is R. Now two circles are drawn with PR & PQ as diameter.

A straight line passing through P is drawn which cuts the 1st circle at S & cuts the 2nd circle at T.

To prove PS = ST.

Join R, S & Q, T.

As PR is the diameter of the 1st circle,

∴ Semicircle angle PSR = 90°

∴ SR ⊥ PT

Similarly, PTQ = 90°

∴ QT I PT

∴ SR & QT are both perpendiculars on PT.

∴ SR || QT

In APQT, the midpoint PQ is R

and SR || OT.

∴ S is the midpoint of PT.

∴PS = ST.

Ganit Prakash Class 10 Solutions Pdf In English

 

Question 5. Three points P, Q, and R lie in a circle. The two perpendiculars PQ and PR at point P intersect the circle at points S and T respectively. Let us prove that RQ = ST.

Solution: Let P, Q & R are the three points on a circle.

Perpendiculars drawn from P on the chords PQ & PR are PS & PT, which cut the circle

at S & T respectively.

To prove RQ ST.

Proof Join R, Q; S, T; S, Q; & T, R.

Let the straight lines RT & SQ intersect at O.

∠SPQ = 1 rt. angle [as PS ⊥ PQ (given)]

∴SQ is a diameter.

Again, RPT 1 rt. angle [as PT ⊥ PR (given)]

∴RT is a diameter.

∴O is the center as the diameters SQ & RT intersect each other.

∴ OR= OQ =OT

∴In triangles ORQ & OST,

OR=OS (Radius of the same circle)

OQ= OT (Radius of the same circle)

and ∠ROQ=∠SPT (vertically opposite)

∴ΔORQ ≅ ΔOST

∴RQ = ST. Proved.

Ganit Prakash Class 10 Solutions Pdf In English

 

Question 6. ABC is an acute-angled triangle. AP is the diameter of the circumcircle of the triangle ABC; BE and CF are perpendiculars on AC and AB respectively and they intersect each other at point Q. Let us prove that BPQC is a parallelogram. 

Solution: ABC is an acute-angled triangle. AP is the diameter of the circumcircle of ΔABC.

BE & CF are the perpendiculars on the sides AC & AB, respectively.

They meet at Q.

To prove BPCQ is a parallelogram..

Proof: CF AB (given)

and ∠ABP is a semicircle angle.

∴ BP ⊥ AB

CF || BP or CQ || BP

Again, BE 1 AC (given)

and ∠ACP is an angle in a semi-circle.

∴CP ⊥ AC

∴CP || BE or CP || BQ.

∴BPCQ is a parallelogram. Proved.

Maths WBBSE Class 10 Solutions

 

Question 7. The internal and external bisectors of the vertical angle of a triangle intersect the circumcircle of the triangle at the points P and Q. Let us prove that PQ is the diameter of the circle.

Solution: ABC is a triangle, and internal & external bisectors of ∠A of ΔABC are AP & AQ, respectively, which cut the circumcircle of ΔABC at P & Q respectively.

To prove PQ is the diameter of the circle.

Proof: As AP & AQ are the internal & external bisectors of∠A.

∴ ∠PAQ = 90°

∴ PAQ is a semi-circle angle.

∴ PQ is the diameter. Proved.

 

Maths WBBSE Class 10 Solutions Question 8. AB and CD are two diameters of a circle. Let us prove that ABCD is a rectangular

Solution: Let AB & CD be two diameters of the circle with center O

Join AC, BD, AD & BC.

To prove ACBD is a rectangle.

Proof: ∠ADB = ∠ACB= 1 rt. angle

& ∠CAD = DBC = 1 rt. angle

In ACBD quadrilateral,

∠A = ∠C = ∠B = <D = 1 rt. angle

∴ACBD is a rectangle.

 

Question 9. Let us prove that if the circles are drawn having sides of a rhombus as diameter then the circles pass through a fixed point.

Solution: If circles are drawn with diameters of the sides of a rhombus, they will pass a fixed point.

Let ABCD is a rhombus. The circle with AB as diameter cuts BC or produced BC at D. Join A, D.

∠ADB= 1 rt. angle (as a semicircle angle)

∴∠AOC = 1 rt. angle

Again the circle as AC diameter will pass through point D.

∴ Two circles with AB & AC as diameters intersect each other at D.

∴Circles with diameters of the sides of a rhombus will pass through a fixed point. Proved.

 

Maths WBBSE Class 10 Solutions Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5 Multiple Choice Question

 

Question 1. PQ is a diameter of a circle with center O, and PR RQ; the value of <RPQ is 

1. 30°
2. 90°
3. 60°
4. 45°

Solution: As PR RQ, 

∴ PRQ 90° (semi-circle angle)

∴RPQ = RQP = 180°-90° / 2 

=90° /2 

Answer. 4.  45°

 

 

 

Question 2. QR is a chord of a circle and POR is a diameter of a circle. OD is perpendicular on QR. If OD = 4 cm, the length of PQ is

1. 4 cm
2. 2 cm
3. 8 cm
4. none of these

Solution: PQ = 2 x OD = 2 x 4 = 8 cm.

Answer. 3. 8 cm.

 

 

 

WBBSE Solutions Guide Class 10 Question 3. AOB is a diameter of a circle. The two chords AC and BD when extended meet at point E. If ∠COD = 40°, the value of CED is

1. 40°
2. 80°
3. 20°
4. 70°

Answer. 3. 20°

 

 

Question 4. AOB is the diameter of a circle. If AC = 3 cm, BC = 4 cm, then the length of AB is

1. 3 cm
2. 4 cm
3. 5. cm.
4. 8 cm.

Solution: AB2 = √AC2+BC2

= √32+42 

=√25 

= 5

Answer. 3. 5. cm

 

 

WBBSE Solutions Guide Class 10 Question 5. In the adjoining figure, O is centre of circle & AB is a diameter, if BCE = 20°, ZCAE= 25°, the value of ZAEC is

1. 50°
2. 90°
3. 45°
4. 20°

Solution: ACE = 90° + 20° = 110°; AEC = 180°- (110° +25°) = 45° A 

Answer. 3. 45°

 

 

Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5  True Or False

 

1. The angle in the segment of a circle that is greater than a semi-circle is an obtuse angle.

False

2. O is the midpoint of the side AB of the triangle ABC, and OA = OB = OC; if we draw a circle with side AB as diameter, then the circle passes through point C.

True

 

Class 10 WBBSE Math Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5  Fill In The Blanks

1. Semicircular angle is a Right angle.

2. The angle in the segment of a circle that is less than a semicircle is an Obtuse angle.

3. The circle is drawn with the hypotenuse of a right-angled triangle as the diameter passes through the Vertices.

 

Class 10 WBBSE Math Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.5 Short Answer

 

Question 1. In isosceles triangle ABC, AB = AC; a circle drawn taking AB as diameter meets the side BC at point D. If BD = 4 cm, let us find the value of CD.

Solution. ABC is an isosceles triangle, where AB AC. A circle is drawn with AB as the diameter, the circle will cut BC at D.

If BD = 4 cm.

Find CD.

BD= CD = 4 cm.

 

Question 2. Two chords AB and AC of a circle are mutually perpendicular to each other. If AB = 4 cm, AC = 3, let us find the length of the radius of the circle.

Solution. AB & AC the two chords of a circle are perpendicular to each other. 

AB = 4 cm, AC 3 cm, 

the radius of the circle =?

Diameter (BC)=√32+42

= √25

= 5,

radius = 5/2

= 2.5 cm.

 

Question 3. Two chords PQ and PR of a circle are mutually perpendicular to each other. If the length of the radius of the circle is r cm, let us find the length of the chord QR.

Solution. PQ & PR are two chords of a circle perpendicular to each other;

if radius = r cm. find the length of the chord QR.

QR-Diameter – 2r cm [.. QPR = 1 rt. Angle]

 

Question 4. AOB is a diameter of a circle. Point C lies on the circle. If ZOBC = 60°, let us find the value of ZOCA.

Solution. AOB is a diameter of a circle. C is any point on the circle; if ZOBC= 60°, find ZOCA. 

∴ OB OC

ZOBC= ZOCB = 60°

∴ ZOCA = 90° – 60° 30°

 

Question 4. In the picture beside, O is the center of the circle and AB is the diameter. The length of the chord CD is equal to the length of the radius of the circle. AC and BD produced meet at the point P, let A us find the value of ∠APB.

Solution. OA = OC = OB = OD = CD

∴ ΔOCD is an equilateral triangle.

∴ ∠COD 60°

∠APB = 60°

 

 

Ganit Prakash Class 10 Chapter 8 Solutions

Chapter 8 Right Circular Cylinder Exercise 8.1

 

Application 1. If the length of the radius of the base of a right circular cylinder is 12 cm and the height is 21 cm, let us write by calculating its lateral surface area.

Solution: The length of the radius of the cylinder is 12/2 cm = 6 cm.

∴The lateral surface area of a cylinder is = 2 x π x 6 x 21 sq cm.

= 2 x 12/7 x 6 x 21 sq cm. 

= 792 sq cm.

Read and Learn More WBBSE Solutions For Class 10 Maths

Application 2. The perimeter of a right circular cylinder is 44 meters and the height is 14 meters, let us write by calculating its lateral surface area. 

Solution: Surface area of the cylinder = circumference of base x height

= (44 x 14) sqm = 616 sqm.

Ganit Prakash Class 10 Chapter 8 Solutions

Application 3. The base area of a closed cylindrical water tank is 616 sq. meters and the height is 21 meters. Let us write by calculating the total surface area of that tank. 

Solution: Let the length of the radius of the circular base of the water tank = r meter.

∴ Base area = πr² sq meter

By the condition, 

πr² = 616

or, 22/7 x r²  = 616

or, r²  = 616 7/22

r = √196 

= 14

Total surface area of water tank is = (2лr²  + 2лrh) sq meter [where height of cylinder = h meter]

= 2 x 22/7 x r(r + h) sq meter

= 2x 22/7 x 14(14+21) sq meter

= 3080 sqm.

 

Application 4. If the perimeter of the base of any closed cylindrical pot is 22 dm and the height is 5 dm, let us write by calculating the area that will be colored to paint the outside of that pot. 

Solution: Let radius = r dm & height = h dm.

∴2πr = 22 &h=5

2 x 22/7 x r = 22. 

r = 7/2 dm.

Ganit Prakash Class 10 Chapter 8 Solutions

Application 5. The total surface area of a right circular cylinder with one end open is 1474 sq cm. If the length of the diameter of the base is 14 cm, let us write by calculating its height. Again, if the pot would be closed at two ends, let us write by calculating what would be its total surface area

Solution: The length of the radius of the base of the right circular cylindrical pot = 14/2 cm. = 7 cm.

Let the height of the pot is h cm.

∴The total surface area of pot = area of the base + area of the lateral surface

= (22/7×72+2 22/7 x 7 x h) sq cm.

= (154 + 44h) sq cm.

By the condition, 154 + 44h = 1474

∴ h = 30

∴ The height of the pot is 30 cm.

If the pot would be closed at two ends, the surface area of that pot

= Area of the upper end surface + 1474 sq cm.

22/7 x 72 + 1474) sq cm. 

=.1628 sqcm.

 

Application 6. The length diameter of the base of a drum with a lid made of steel sheet is 4.2 dcm. If 112.20 sq dcm of steel sheet is required to make the drum, let us write by calculating the height of the drum. Again, if the price of 1 sq dcm of steel is Rs. 25, let us calculate the production cost of the drum. 

Solution: Let the height of the drum = h dcm.

The radius of the drum = 4.2/2 = 2.1 dcm.

According to the problem,

2.π x 2.1 (2.1+ h) = 112.20

or, 2x 22/7 x 22/7( 2.1+ h) = 112.20

(2.1+ h) = 1122x7x10 / 10×2×22×21

h 8.5 – 2.1 = 6.4 dcm.

.. Total surface area = 112.20 sq dcm = 1.1220 sqm.

.. Total cost Rs. 25 x 1.1220 = Rs. 28.05

Ganit Prakash Class 10 Chapter 8 Solutions

Application 7. If the length of the diameter of the base of the glass is 11.2 cm and the height is 15 cm, let us calculate the volume of water that the glass will contain.

Solution: We understand, if the length of the diameter of the base of the glass is 11.2 cm, and the height is 15 cm, the glass will contain water

= π Χ (11.2/2)2 x 15 cubic cm.

=22/7 x 56/10 x 56/10 x 15

= 1478.4 sq cm.

 

Application 8. The height of a right circular cylinder made of iron open at two ends is 42 cm. If the thickness of the cylinder is 1 cm and the length of its external diameter is 10 cm, let us calculate the volume of iron in it.

Solution: The length of the external radius of the cylinder = 10/2 cm. = 5 cm.

It is 1 cm in thickness.

.. Length of internal radius of the cylinder = (5-1) cm. = 4 cm. volume of iron

=22/7 (5²-4²) x 42 cubic cm.

= 2/7 x (5+4) (5-4) × 42 cubic cm.

[a²b² = (a + b) (a – b)]

=22/7 x 9 x 1 x 42 cubic cm.

= 22 x 9 x 6 cubic cm.

= 1188 cu cm.

 

 

 

Application 9. If we color the inside and outside of the hollow right circular cylinder (Application 8) open at two ends, let us write by calculating how much area we shall color.

Solution: The sum of the inner and outer surface area of this hollow right circular cylinder open at two ends

=(2 ×π × 5 × 21+ 2x π x 4 x 21) sq cm.

= 2 × π x 21(5+4) sq cm.

= 2 x 22/7 x 21 x 9 sq cm.

= 1188 sq cm.

 

Application 11. If the inner and outer radii of a hollow right circular cylindrical pipe of 6 meter in length are 3.5 cm and 4.2 cm respectively, let us write by calculating the volume of the iron that the pipe contains. If I cubic decimetre of iron weighs 5 kilograms, let us write by calculating the weight of the pipe.

Solution: Height of pipe = 6 m = 60 dcm.

Interval radius of pipe = 3.5/20 = 35/200 dcm.

The external radius of pipe = 4.2/20 = 42/100 dcm.

∴ Volume of Iron = π {(42/200)² – (35/200)² x 60} cu dcm.

= 22/7 x 77/200 x 7/200 x 60 cu dcm.

= 121 x 21 / 1000 cu dcm. 

= 2541/ 1000 cu dcm.

= 2.541 cu dcm.

∴ Weight of 1 cu dcm of iron 5 kg.

∴ Weight of 2.541 cu dcm of iron = (2.541 x 5) kg 

= 12.705 kg.

Ganit Prakash Class 10 Chapter 8 Solutions

Application 12. If the area of the base of a cylinder is 13.86 sq meters and the height is 8 meters, let us calculate the volume of the cylinder.

Solution: Let the length of the radius of the base of the cylinder is r. meter. 

By the condition, πr²= 13.86

Or, 22/7 r² = 1386/100 x 7/22 = 441/100

∴r = 21/10 m.

∴The volume of the cylinder is 22/7 x 21/10 x 21/10 x 8 cubic meters

= 110.88 cu cm.

 

Application 13. If the perimeter of the base of a cylinder is 15.4 cm and the height is 10 cm, let us calculate its volume. 

Solution: Let the radius of the base of the cylinder r cm & height = h cm = 10 cm.

∴Circumference = 2πr = 2 x 22/7 x r = 15.4

r = 154/10 x 7/2×22 = 49/20 cm.

∴The volume of cylinder πr2h

= 22/7 x 49/20 x 49/20 x 10 cu cm.

=3773/20

= 188.65 cu cm.

 

Application 14. If the glass of a tube light is 105 cm long and external circumference is 11 cm and it is 0.2 cm thick, let us write by calculating the volume (in cc) of the glass that will be required to make 5 such tube lights.

 

 

Application 15. Through a hole, 110 kiloliter of water enters a ship. After closing the hole a pump is connected for the drainage of water. If the length of the diameter of the pipe of the pump is 10 cm and the speed of water flow is 350 meters/minute, let us write by calculating the time that will be required by the pump to clear off all water in the ship.

Solution: The volume of water can be cleared off by the pump in 1 minute

=22/7 x 10/2 x 10/2 x1/100 x 3500 cubic dcm.

= 2750 cubic dcm = 2750 liters [1 cubic dcm. 1 liter.]

Time required to clear off 110 kilolitres of water =110000 / 2750

minutes 40 minutes.

So, the time required by the pump to clear off all water in the ship = is 40 minutes.

 

Application 16. A right circular cylindrical tank of 5 meters in height is fixed with water. Water comes out from there through a pipe having a length of diameter 8 cm at a speed of 225 meters/minute and the tank becomes empty after 45 minutes. Let us write by calculating the length of the diameter of the tank.

Solution: Let the radius of the tank = r m.

∴Radius of pipe = 8/2 = 4 cm = 1/25 m.

∴ πr² x 5 = 45 x π (1/25)² X 225

5r² = 45×225 / 25×25

∴ r² = 45 x 1 x 9 / 5 x 25 

=81/25

∴ r = 9/5 m.

∴Diameter of the tank = 2 x 9/5

= 18/5 

3.6 m.

 

Class 10 WBBSE Math Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.4

Question 1. In the adjoining figure, DBA = 40°, <BAC = 60°, and CAD = 20°; let us find the values of ZDCA and BCA. Also, let us see by calculating what the sum of BAD and ZDCB will be.

 

 

Read and Learn More WBBSE Solutions For Class 10 Maths

Solution: In the figure, DBA = 40°; <BAC = 60° & CAD = 20°.

<DCA & ∠DBA are the angles on the circumference on the same arc CD.

∴ ∠DCA =∠DBA = 40°

Now, BAD = <BAC + ∠CAD = 60° + 20° = 80°

The sum of the three angles of a triangle = 180°

∴ In ΔABD, <BAD + 2BDA +

∴ ∠BDA =180° – (BAD + ABD)

= 180 (80° + 40°)

= 180° -120° = 60°

ABD = 180°

∴ On the arc AB, the angle on the circumference ZBCA

ZBDA = 60°

∴ <BCD = ∠BCA + ∠ACD = 60° + 40° = 100°

∴ ∠BCD + ∠BAD

100° +80° = 180°

.. ∠DCA = 40°, ∠BCA = 60° & Sum of the ∠BAD + ∠BCD = 180°

Class 10 WBBSE Math Solution In English

 

 

Question 2. In the adjoining figure, AOB is the diameter of the circle and O is the center of the circle. The radius OC is perpendicular on AB. If P is any point on minor arc CB, let us write by calculating the values of <BAC and ∠APC.

Solution: As, AB LOC .. ZAOC = ZBOC = 90°

ΔAOC is an isosceles triangle.

.: AO = OC

∠OAC = ∠OCA

∠OAC+∠OCA = 90° [∠AOC = 90°] 

or, ∠OAC+∠OAC = 90°

or, 2 ∠OAC = 90°

Class 10 WBBSE Math Solution In English

Question 3. O is the orthocentre of the triangle ABC and the perpendicular AD drew on BC when extended, Intersects the circumcircle of AABC at point G; let us prove that OD = DG.

Solution: To prove, OD = DG.

Join B, G & C, G.

Proof: In the circumcircle of the ABC,

∠ACB is the angle on the circumference on arc AB.

∴ ∠ACB = ∠AGB

or, ∠ACB = ∠OGB

∠ECD = ∠OGB = ∠BGO———-(1)

Again, BEL AC & AD 1 BC.

∠OEC = ∠ODC = 90°

∠OEC+∠ODC= 90° + 90° = 180°

In the quadrilateral ODCE,

∠OEC+∠ECD +∠ODC +∠DOE = 4 right angles ∠ECD + ∠DOE

 = 4 right angles (∠OEC + ∠ODC) 4 right angles – 2 right angles

= 2 right angles ∠ECD + ∠EOD =∠EOD + ∠BOD

∴∠ECD = <BOD = ∠BOG——-(2)

From (1) & (2), BGO = BOG

BG = BO

Now in two right-angled triangles,

Hypotenuse BG = Hypotenuse BO, BD is common.

∴ ABDG ≅ ABDO. 

∴ OD = DG Proved.

WBBSE Solutions Guide Class 10

 

Question 4. I is the center of the incircle of AABC; Al produced intersects the circumcircle of that triangle at the point P, let us prove that PB = PC = PI

Solution: To prove PB = PC = PI

Join B, I; C, I; & P, B; P, C.

∴ The bisectors of the angles of a triangle meet at a point called In-centre.

∴ ∠BAI = ∠CAI = A/2, 

∠ABI = ∠CBI = B / 2

∠ACI = ∠BCI = C/2

∠PBC & PAC arc the angles of the circumcircle of

ΔABC, on the arc PC.

∴∠PBC=∠PAC

Now, ∠IBP = ∠PBC + ∠CBI = ∠PAC + ∠CBI = ∠CAI + ∠CBI

A/ 2 + B/2 ———-(1)

The external angle of the ΔABI, ∠BIP = ∠BAI + ∠ABI =

A/2 + B/2

From (i) & (ii), ∠IBP = ∠BIP. In ΔPBI, ∠IBP = ∠BIP 

∴ BP = PI——-(3)

Similarly, in∠PCI, ∠ICP = CIP 

i.e. PC = PI———-(4)

∴ From (iii) & (iv), PB

PC = PI Proved.

WBBSE Solutions Guide Class 10

Question 5. Timir drew two circles that intersect each other at points P and Q. Through the point P two straight lines are drawn so that they intersect one of the circles at points A and B, and the other circle at points C and D respectively; let us prove that ∠AQC = ∠BQD.

Solution: To prove, ∠AQC = ∠BQD.

Join A, Q, B, Q, C, Q & D, Q

Proof: In the circle with center L, ∠PCQ &∠PDQ are the angles on the same circumference.

∠PCQ =∠PDC——-(1)

Again, in the circle with center K, ∠PAQ = ∠PBQ———(2)

As these are angles on the same arc.

Adding (1) & (2),

∠PAQ + ∠PCQ = ∠PBQ+ ∠PDC———(3)

In AACQ sum of the three angles = 180°

∠CAQ + ∠ACQ + ∠AQC = 180°

or, ∠PAQ +∠PCQ + ∠AQC = 180°——–(4)

Similarly in ABDC, the sum of three angles = 180°

∴ ∠DBQ+∠BDQ+∠BQD = 180°

or, ∠PBQ + ∠PDQ+ ∠BQD = 180°——-(5)

From (4) & (5),

∠PAQ + ∠PCQ + ∠AQC = ∠PBQ + ∠PDQ + ∠BQD——–(6)

Subtracting (3) from (4),

∠AQC = ∠BQD Proved.

WBBSE Solutions Guide Class 10

 

Question 6. Two chords AB and CD of a circle are perpendicular to each other. If a perpendicular drawn to AD from the point of intersection of those two chords AB and CD are produced to meet BC at point E, let us prove that point E is the midpoint of BC.

Solution: In the circle, AB & CD are perpendicular to each other & intersect at O. The perpendicular from O on chord AD cuts the chord BC at E.

To prove, E is the midpoint of BC.

Proof: In ΔAFD, ∠AFO = 90° as OF AD

In ΔAFO, ∠FAO+∠AOF = 90°

i.e., ∠DAQ + ∠ADO = 90°—-(1)

Again, in ΔAOD, ∠AOD = 90° as AO ⊥ DO

∴∠DAQ +∠ADO = 90°——(2)

From (1) & (2),∠AOF = ∠ADO

or, ∠ADO = ∠EOB as ∠AOF ∠EOB (vertically oppo-site)

∴ ∠EOB = ∠ADO

Or, ∠EOB = ∠ADC

∴ ∠EOB = ∠ABC (Angles on the same circumference)

∴ ∠EOB = ∠OBC =∠OBE

In ΔOEB, ∠EOB =∠OBE. 

∴ BE = OE—–(3)

∠AOF+∠DOF = 90° (as AOD = 90°)

and ∠AOF+∠DAF 90° 

∴∠DOF = <DAO 

or,∠COE = ∠DÁB [∠COE = vertically ∠DOF]

∴<COE = <DCB [COE & DCB are angled on the same circumference]

i.e., ∠COE = ∠OCE

InΔOCE, ∠COE = ∠OCE

∴CE = OF ——– (iv) 

From (3) & (4), BE = CE

∴ E is the midpoint of BC.

 

 

West Bengal Board Class 10 Math Book Solution In English Question 7. If in a cyclic quadrilateral ABCD, AB = DC, let us prove that AC = BD. 

Solution: ABCD is a cyclic quadrilateral, AB = DC. To prove AC = BD. Proof: Join A, C & B, D.

Let AC & BD intersect at E.

∠CAB & ∠DAB are the angles on the same circumference.

∴∠CAB = ∠CBD

i.e.,∠EAB = ∠CDE

In ∠AEB & ∠DEC,

∠EAB = <CDE

∠AEB = ∠DEC (vertically opposite)

& AB = DC (given)

∴ΔAEB ≅ ADCB. (AAS)

∴AE = DE & BE = CE

∴ AC = AE + CE = DE+ BE = BD

∴ AC = BD.

 

Question 8. OA is the radius of a circle with the center at O; AQ is its chord and C is any point on the circle. A circle passes through the points O, A; C intersects the chord AQ at the point P; let us prove that CP = PQ.

Solution: Join O, A & O, Q.

Proof OQ & OA are two radii of the circle with center O.

In the isosceles ΔOQA,

∠OQA = ∠OAQ

Again, AO=CA is an isosceles triangle as OQ = OC (Radii of the same circle)

∴∠OQC = ∠OCQ

or, ∠OQA+∠AQC = ∠PCQ + ∠PCO

or, ∠OQA + ∠PQC = ∠PCQ + ∠PCO

∴ ∠PCO &∠PAO are the angles on the same circumference OP.

<PCO = <PAO = ∠OAQ (as ∠OQA = ∠OAQ)

∴ ∠PCO =  ∠OQA

∴ ∠OAQ + ∠PQC = ∠PCQ + ∠OQA or, <PQC = ∠PCQ

∴ CP = PQ (As APCQ is an isosceles triangle).

 

West Bengal Board Class 10 Math Book Solution In English Question 9. The triangle ABC is inscribed in a circle, and the bisectors AX, BY, and CZ of the angles <BAC, ZABC, and ZACB intersect at the points X, Y, and Z on the circle respectively, let us prove that AX is perpendicular to YZ.

Solution: Join X, Y; Y, Z; Z, X.

Let AX cuts YZ at P.

∴∠AXY + Y = ∠AXY + ∠BYX + ∠BYZ

= ∠ABY+∠BAX + ∠BCZ

=∠A/2 +  ∠B/2 +  ∠C/2

=∠A+∠B+∠C/2

= 1 rt. angle

∴ In PYX, ∠YXP + ∠PYX = 1 rt. angle

∴ ZP = 1 rt. angle

∴ AX T YZ.

 

Question 10. The triangle ABC is inscribed in a circle. The bisectors of the angles <BAC, ∠ABC, and ∠ACB intersect at the points X, Y, and Z on the circle respectively, let us  prove that in ΔXYZ, ∠YXZ = 90° – <BAC / 2

Solution: ΔABC is in a circle. Bisectors of <BAC, ∠ABC, & ∠ACB meet the circle at X, Y, and Z respectively.

To prove in ΔXYZ, ∠YXZ = 90° – 1/2 <BAC.

Proof: On the arc AY, ∠AXY = ZABY = 1/2 ZB

Similarly, on the arc AZ, ∠AXZ = ∠ACZ =1/2 ZC.

∴ Total ∠X = 1/2 ∠B + 1/2 ∠C

In ΔABC, 1/2 ∠A + 1/2 ∠B + 1/2 ∠C = 90°

1/2 ∠B + 1/2 ∠C = 90° – 1/2 ∠A 

Or, ∠X  = 90° – 1/2∠A 

∴ ∠YXZ 90° – 1/2 <BAC.

 

Question 11. A perpendicular drawn on BC from point A of AABC intersects the side BC at point D and a perpendicular drawn on side CA intersects the side CA at the point E; let us prove that four points A, B, D, and E are concyclic.

Solution: Join D, E.

Proof: AD T BC & AE T CA

∴∠ADB = ∠ADC = 90°.

∠AEB = ∠BEC = 90°

∴ External EDC = internal opposite <BAE

∴<BDE + ∠EDC = 2 rt. angles

i.e., ∠BDE+∠BAE= 2 rt. angles

∴ Opposite angles of the quadrilateral are supplementary.

∴ Opposite angles or the cyclic quadrilateral are supplementary. 

∴ A, B, D, E arc concyclic.

West Bengal Board Class 10 Math Book Solution In English

Chapter 7 Theorems Related To Angles In A Circle Exercise 7.4 Multiple Choice Question

 

Question 1. In the adjoining figure, O is the center of the circle, if ZACB = 30°, ZABC = 60°, ZDAB = 35° and ZDBC = x°, the value of x is

1. 35
2. 70
3. 65
4. 55

Solution. ∠BAC = 90°

∴ ∠BDC 90° – 35°= 55°

Answer: 4. 55°

 

Question 2. In the adjoining figure, O is the center of the circle, if ∠BAD = 65°, BDC = 45°, then the value of CBD is

1. 65°
2. 45°
3. 40°
4. 20°

Solution: ∠CAD = ∠BAD – <BAC = 65° – 45° = 20°

∴ ∠CAD = CBD = 20°

Answer: 4. 20°

 

 

Question 3. In the adjoining figure, the O is the center of the circle, if ZAEB = 110° and CBE = 30°, the value of ∠ADB is

1. 70°
2. 60°
3. 80°
4. 90°

Solution: ∠ACB = ∠ADB = 180° – (70° + 30°) = 80°

Answer: 3. 80°

Question 4. In the adjoining figure, O is the center of the circle, if∠BCD = 28°, ∠AEC = 38° then the value of ∠AXB is.

1. 56°
2. 86°
3. 38°
4. 28°

Solution: BCD = BAD = 28°

= ∠CBE 180° – (38° +  28°) 114°

∠ABX = 180° – 114° = 66°

∠AXB = 180° – (66° + 28°)= 86°

Answer: 2.  86°

 

 

Question 5. In the adjoining figure, O is the center of the circle and AB is the diameter. If AB || CD, ZABC= 25°, the value of <CED is

1. 80°
2. 50°
3. 25°
4. 40°

Solution; ∠ABC= ∠BCD = 25° (alternate angle)

∴∠CED = 90° – 2 x 25° = 40°

Answer: 4. 40°

 

 

West Bengal Board Class 10 Math Book Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.4  True Or False

 

Question 1. In the adjoining figure AD and BE are the perpendiculars on sides BC and CA of the triangle ABC. A, B, D, and E are concyclic.

Solution: In AABC, AD & BE are the perpendiculars on BC and AC, respectively points A, B, C, D, and E are concyclic.

True.

Question 2. In ABC, AB = AC, BE, and CF are the bisectors of the angles B ABC and ACB and they intersect AC and AB at points E and F respectively. Four points B, C, E, and F, are not concyclic.

Solution: In ΔABC, AB = AC; BE & CF are the bisectors of ZABC & ZACB, respectively which cut at E & F.

Points B, C, E, and F are not concyclic.

True

 

Chapter 7 Theorems Related To Angles In A Circle Exercise 7.4 Fill In The Blanks

 

1. All angles in the same segment are Equal.

2. If the line segment joining two points subtends equal angles at two other points on the same side, then the four points are Concyclic.

3. If two angles on the circle formed by two arcs are equal then the lengths of the arcs Are equal.

 

Chapter 7 Theorems Related To Angles In A Circle Exercise 7.4 Short Answer

 

Question 1. In the adjoining figure, O is the center of the circle, AC is the diameter, and chord DE is parallel to the diameter AC. If CBD = 60°; let us find the value of ZCDE.

Solution: ∠CDE = 90°- ∠CBD 90° – 60°= 30°

 

 

Question 2. In the adjoining figure, QS is the bisector of an angle /PQR, if SQR = 35° and ZPRQ 32°, let us find the value of ∠QSR.

Solution. ∠SRP = SQR = ∠PSQ = 35°

∴∠QSR =∠QPR = 180° – (35° +35° + 32°) = 78°.

 

 

Question 3. In the adjoining figure, O is the center of the circle and AB is the diameter. If AB and CD are mutually perpendicular to each other and ZADC= 50°; let us find the value of CAD.

Solution: BAD 90° = 50° = 40°

∴∠CAD = 2 x 40° = 80°.

 

 

Question 4. In the adjoining figure, O is the center of the circle and AB = AC; if ZABC = 32°, let us find the value of BDC.

Solution: ∠ABC = ∠ACB = 32°

∴ BAC = 180° (32° 32°) = 116°

∠BDC = 180° – 116° = 64°

 

 

Question 5. In the adjoining figure, BX and CY are the bisectors of the angles ABC and ZACB respectively. If AB = AC and BY = 4 cm, let us find the length of AX.

Solution. AX BY = 4 cm.

 

 

WBBSE Solutions Guide Class 10 Application 17. Let us prove that the angle in the segment of a circle that is less than a semicircle is an obtuse angle. 

Solution: In the figure, O is the center of the circle.

∴ The segment ACB is less than a semicircle.

To prove ∠ACB is an obtuse angle.

∴ ADB is a major arc.

In the reflex ∠AOB, the angle at the center standing on that arc is greater than 2 right angles.

∴ ∠ACB is an angle at the circle also standing on the same arc.

∠ACB is greater than 1 right angle.

∴ ∠ACB is an obtuse angle. Proved.

 

 

Application 19. Let us prove with the reason that the circle drawn with a hypotenuse of a right-angled triangle as diameter passes through the right angular vertex.

Solution: To prove that the circle drawn with BC as diameter passes through point A.

Proof: Let us suppose the circle does not pass through A. Then, let the circle intersects BA at point D.

∴∠BDC is 1 right angle ( angle in a semicircle is a right angle)

∴ BAC = 1 right angle (by hypothesis)

∴ <BAC = BDC.

This is impossible, if points D & A do not coincide, 

since the exterior angle ∠BDC of Δ ADC > the interior 

opposite angle ∠BAC.

∴ The circle passes through point A.

 

Maths WBBSE Class 10 Solutions

Chapter 7 Theorems Related To Angles In A Circle Exercise 7.3

 

Question 1. O is the circumcentre of the isosceles triangle ABC, whose AB = AC, the points A and B, and C are on opposite sides of the center O. If ZBOC is 100°, let us write by calculating the values of ∠ABC and ∠ABO.

 

Read and Learn More WBBSE Solutions For Class 10 Maths

 

Question 2. In the adjoining, if O is the center of the circumcircle of AABC and ZAOC = 110°; let us write by calculating the value of ZABC.

Solution: In the ∠AOC = 110°

∴ ∠ ABC = 1/2 of the angle at the center

= 1/2 x reflex AOC

= 1/2 x 250°

=125°

Maths WBBSE Class 10 Solutions

Question 3. ABCD is a cyclic quadrilateral of a circle with center O; DC is extended to point P. If ∠BCP = 108°, let us write by calculating the value of ∠BOD.

Solution: ABCD is a cyclic quadrilateral with the center of the circle O.

Side DC is produced to P such that BCP = 108°

Find ∠BOD.

∠BCD + BCP = 180°

∠BCD 180°- ∠BCP 180° – 108° = 72°

∠BOD = 2 x <BCD

= 2 x 72° = 144°

& Reflex ∠BOD = 360° – 144°

= 216°.

 

Question 4. O is the center of the circle; ∠AOD = 40° and ZACB = 35°; let us write by calculating the value of ∠BCO and ∠BOD, and answer with reason.

Solution: O is the center of the circle ∠AOD = 40° & ∠ACB = 35°.

Find ∠BCO & BOD.

Produce CD, which cuts the circle at E.

∠AOE is the angle at the center and ∠ACE is the angle at the circumference on the same arc AE.

Maths WBBSE Class 10 Solutions

∴ ∠ACE = 1/2 

AOE = 1/2 x 40° = 20°

Now, ∠BCO = ∠BCA + ∠ACO

= ∠ACB + ∠ACE

= 35° + 20°

= 55°.

Again,∠AOB is the angle at the center & ∠ACB is the angle at the circumference.

∴ ∠AOB = 2 x ∠ACB = 2 x 35° = 70°

∠BOD = ∠AOB+∠AOD

= 70° + 40°

= 110°

∴ ∠BCO = 55°

& BOD = 110°

Maths WBBSE Class 10 Solutions

Question 5. O is the center of the circle in the picture beside, if ZAPB = 80°, let us find the sum of the measures of ZAOB and COD and answer with reason.

Solution: ‘O’ is the center of the circle & ∠ADB = 80°.

To find the sum of ∠AOB + ∠COD.

Join B, C.

As ∠AOB is the angle at the center and DBC is the angle at the circumference on the same arc AB.

∴ ∠AOB = 2 x ∠ACB

Again, COD is the angle at the center and DBC is the angle at the circumference on the same arc CD.

 

 

∠COD=2x ∠CBD

∠AOB+∠COD = 2x (∠ACB + ∠CBD)———(1)

Now, in ABCP, the side CP is produced to A.

∴ External BPA = <PCB + <PBC

∠APB = ∠ACB + <DBC

= ∠ACB+∠DBC= ∠APB = 80°————-(2)

Now, from (1) & (2), ∠AOB + ∠COB = 2 x (∠ACB + ∠DBC)= 2 x 80° = 160°.

WBBSE Solutions Guide Class 10

Question 6. Like the adjoining figure, we draw two circles with centers C and D which inter- sect each other at points A and B. We draw a straight line through point A which intersects the circle with center C at point P and the circle with center D at point Q.

1. ∠PBQ = <CAD;

2. BPC = <BQD.

Solution: Two circles with centers C & D intersect each other at A & B.

A straight line passing through A cuts the circles at P & Q respectively.

To prove : 

∴ 1. ∠PBQ = ∠CAD

& 2. <BPC = <BQD.

Join A, B; A, C; P, C; A, D; D, Q; B, P; B, Q; B, C; B, D.

In the circle with center C,

∠ACP is the angle at the center &

∠ABP is the angle at the circumference on the same arc AP.

∴∠ABP =1/2 ∠ACP and CAP CPA [as, CA = CP (same radius)]

 

Now, in ∠CAP, ∠CAP + ∠CPA + ∠ACP = 180°

or,∠CAP + ∠ACP = 180°

or, ∠CAP 180° – ∠ACP

∴∠CAP 90° – 1/2 

∠ACP 90°= ∠ABP

∴ ∠CAP 90°-∠ABP——(1)

Again, in the circle with center D,

∠ADQ is the angle at the center and ∠ABQ is the angle at the circumference on the same arc AQ.

∴∠ABQ= 1/2  ∠ADQ

In ∠ADQ, DA = DQ (radii of the same circle)

∴ ∠DAQ = ∠DQA

∴ ∠DAQ + ∠DQA = 2 ∠DAQ

In∠ADQ, ∠ADQ+ ∠DAQ + ∠DQA = 180°

or, ∠ADQ+2∠DAQ = 180°

or, 2 ∠DAQ 180° – ∠ADQ

∴∠DAQ = 90° – 1/2 

∠ADQ 90° – ABQ————-(2)

∴∠DAQ = 90° – ∠ABQ

Adding (1) & (2),

∠CAP +∠DAQ = 90° – ∠ABP + 90° – ∠ABQ

=180° (∠ABP + ∠ABQ) = 180° – ∠PBQ.

or, ∠PBQ = 180° – (∠CAP + ∠DAQ)

∴∠PBQ = CAD Proved.

Again, In ∠ABC, CA = CB (Radii of the same circle)

∠CAB = <CBA

& In∠DAB, DA = DB (Radii of the same circle)

∴ ∠DAB = DBA

∴∠CAB+ ∠DAB = ∠CBA + ∠DBA

∠CAD = CBD but ∠CAD =∠PBQ (Proved before)

∴ ∠CBD = CAD =∠PBQ

∴ ∠CBD + ∠PBD =∠PAD+ ∠PBQ

or, CBP = DBQ [as BC= PC (Radii of same circle)] and BD = DQ

∴ ∠BPC = CBP and ∠DBQ = ∠BQD

∴ BPC = CBP =∠DBQ =∠BQD

∴ ∠BPC = ∠BQD Proved.

WBBSE Solutions Guide Class 10

Question 7. If the circumcentre of triangle ABC is O; let us prove that ∠OBC + <BAC = 90°.

Solution: O is the circumcentre of the triangle ∠ABC.

To prove, ∠OBC + ∠BAC = 90°

BOC = 2BAC

∴ <BAC = 1/2 ∠BOC

In ΔOBC, ∠OBC = ∠OCB

∠OBC+∠OCB + ∠BOC = 180°

∴2∠OBC+BOC = 180°

or, 2∠OBC= 180° – ∠BOC

or,∠OBC= 90° – 1/2 ∠BOC

= 90-∠BAC

∴∠OBC+∠BAC

= 90° Proved.

Question 8. Each of two equal circles passes through the center of the other and the two circles intersect each other at points A and B. If a straight line through point A intersects the two circles at points C and D, let us prove that ABCD is an equilateral triangle.

Solution: Let P & Q are the centres of two equal circles. They cut each other at A and

Join, A,P; A,Q; B,P; B,Q; A,B; P,Q.

∴AP = BP = ΔAB = AQ = PQ

∴ΔAPQ & BQ are equilateral triangles.

∴∠PAQ=  ∠APQ= ∠AQP = ∠PBQ = ∠BPQ = ∠BQP = 60°.

Now, ∠APB =∠APQ+ ∠BPQ = 60° + 60° = 120°

∠AQB = ∠AQP+∠PQB = 60° +60° = 120°

∴ ∠APB = ∠AQB = 120°

In the circle with centre P, APB is the angle at the centre & ADB is the angle at the circumference on the same arc AQB.

∴∠ADB = 1/2 

∠APB = 1/2  x 120° = 60°

∠CDB = 60°

Again, in the circle with center Q, ZAPB & ZACB are the angles at the circumference on the same arc.

∠ACB = ∠APB = 120°.

∴∠BCD = 180° – ∠ACB= 180°- 120° = 60°

∴ In ABCD, CDB = ∠BCD = 60°

Remaining DBC= 180° (∠CDB + <BCD) 

= 180° – (60° 60°) = 60°

∴ All the angles of BCD are equal.

∴ ABCD is an equilateral triangle.

Question 9. S is the center of the circumcircle of AABC and if ADI BC, let us prove that <BAD = <SAC.

 

WBBSE Solutions Guide Class 10

Question 10. Two chords AB and CD of a circle with center O intersect each other at the point P, let us prove that ∠AOD + ∠BOC = 2∠BPC. If ∠AOD and ∠BOC are supplementary to each other, let us prove that the two chords are perpendicular to each other.

Solution: Two chords AB & CD of a circle with center O,

intersect each other at P.

To prove,

∠AOD + ∠BOC = 2BPC

If ∠AOD & BOC are supplementary to each other, then prove that the chords are perpendicular to each other.

Join, O, A; O, B; O, C; O, D; & B, D.

In the circle with center O, AOB is the angle at the center & ABD is the angle at the circumference, in the same arc AKD.

∴∠ABD = 1/2 = ∠AOD

or, ∠AOD = 2 x ∠ABD

Again, in the circle with center O, ∠BOC is the angle at the center and BDC is the angle at the circumference on the same arc ZBLC.

∴∠BOC = 2 x <BDC

∴∠AOD + ∠BOC = 2(∠ABD + 2BDC) = (<PBD + BDP)——(1)

Now in the triangle PBD, Ext. ∠BPC

∴ ∠BPC = ∠PBD + 2BDP

From (1), ∠AOD +∠BOC = 2∠BPC (Proved 1st part)

Again, if ∠AOD + ∠BOC = 2 right angles

2 x ∠BPC =∠AOD +∠BOC = 2 right angles

∴∠BPC = 90°

BP PC, i.e., Two chords AB & CD are perpendicular to each other. Proved – 2nd part.

Question 11. If two chords AB and CD of a circle with center O, when produced, intersect each other at the point P, let us prove that ∠AOC – BOD = 2∠BPC.

Solution: Two chords AB and CD of the circle with center O, when produced, cut each other at P, outside the circle.

To prove,∠AOC – BOD = 2 x <BDC

Join O, A; O, B; O, C; O, D; & A, D.

Proof: In the circle with center O,∠AOC is K

the angle at the center & ZADC is the angle at the circumference on the same arc. AKC, 

∴ ∠AOC = 2∠ADC

Again, in that circle BAD is the angle at

the center & BAD is the angle at the circumference on the same arc BD,

∴ ∠BOD = 2∠BAD

∠AOC – ∠BOD=2(∠ADC-∠BAD) —–(1)

In ΔAPD, External∠ADC = ∠APD + ∠PAD = ∠APC + ∠PAD

or, ∠APC =∠ADC – ∠PAD

2(∠ADC-∠PAD) 2∠APC

∠AOC –

BOD = 2∠APC [from (1)]

 

Question 12. We drew a circle with point A of quadrilateral ABCD as the center which passes through points B, C, and D. Let us prove that CBD + /CDB = <BAD.

Solution: The circle is drawn with center A of the quadrilateral ABCD, passing through B, C, and D.

To prove, CBD + /CDB = 1/2 ∠BAD

Proof : 2∠BCD = 360 – <BAD

∴∠BCD = 180° -1/2 ∠BAD

From ABCD, ∠BDC + 4CBD = 180° – ∠BCD

or, <CBD + /CDB = 180° – (180° -1/2 ∠BAD) 

=1/2 ∠BAD Proved.

 

Question 13. O, is the circumcentre of AABC and OD is perpendicular on the side BC; let us prove that BOD = <BAC

Solution: O is the circumcentre & OD is perpendicular to BC. 

To prove,∠BOD = BAC

Join O, A; O, B; O, C; O, D &A, D. 

Proof: From ΔAOC, 

∠AOC+∠OAC+∠OCA = 180°

.. 2∠OAC = 180° – ∠AOC 90° – ∠ABC

∠BAD 90° – ∠ABC =∠OAC

.. <BAC = ∠OAC + ZOAB = ∠BOD Proved.

WBBSE Solutions Guide Class 10 Chapter 7 Theorems Related To Angles In A Circle Exercise 7.3 Multiple Choice Question

 

Question 1. In the adjoining figure, If ‘O’ is the center of the circle and PQ is a diameter then the value of x is

1. 140
2. 40
3. 80
4. 20

Solution: ∠ROQ 180° 140° = 40°

.. x = <RSQ = 1/2 x 40° = 20°

Answer: 4. 20°

 

Question 2. In the adjoining figure, if O is the center of circle, then the value of x is

1. 70
2. 60
3. 40
4. 200

Solution: ∠BOR = 360° (140° +80°) = 360° 220° = 140°

∴∠QPR = 1/2  ZQOR= 1/2 x 140° = 70°

Answer. 2. 70°

Question 3. In the adjoining figure, if O is the center of the circle and BC is the diameter then the value of x is

1. 60
2. 50
3. 100
4. 80

Solution:∠AOC 1800° – 80° = 100°

∴∠ADC = x2 = 100°/ 2

= 50°

Answer. 2. 50°

 

Question 4. If O is the circumcentre of AABC and ZOAB = 50°, then the value of ZACB is

1. 50°
2. 100°
3. 40°
4. 80°

Solution: If ∠OAB = 50°

∠AOB 180°

(50° + 50°) = 180° – 100° = 80°

∴ ∠ACB = 1/2

AOB = 1/2 x 80° = 40°

Answer. 3. 40°

Question 5. In the adjoining figure, if O is centre of circle, the value of POR is

1. 20°
2. 40°
3. 60°
4. 80°

Solution: ∠POQ = 180° –  20° = 160°

∠QOR = 180° – 80° = 100°

∠POR 160° – 100° = 60°

Answer. 3. 60°

 

Class 10 WBBSE Math Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.3 True or False

 

1. In the adjoining figure, if O is the center of the circle, then ZAOB = 2ZACD.

False

2. point O lies within the triangular region ABC in such a way that OA OB and ZAOB = 2ZACB. If we draw a circle with centre O and length of radius OA, then the point C lies on the circle. 

True

Chapter 7 Theorems Related To Angles In A Circle Exercise 7.3 Let us fill in the blanks

1. The angle at the Half center is the angle on the circle, subtended by the same arc.

2. The lengths of two chords AB and CD of a circle with centre O are equal. If APB and AQC are angles on the circle, then the values of the two angles are  Equal.

3. If O is the circumcentre of an equilateral triangle, then the value of the from angle formed by an side of the triangle is  120°.

 

Class 10 WBBSE Math Solution In English Chapter 7 Theorems Related To Angles In A Circle Exercise 7.3 Short answers

 

Question 1. In the adjoining figure, O is the center of the circle, if ∠OAB = 30°, ∠ABC = 120°,∠BCO = y° and ∠COA = x°, let us find x and y. 

Solution. ∠AOC = x° 360°- 2 x 120° = 120°

∠BCO = y°

=360° -2(120° +30° + 120°) = 360°

 

Question 2. O is the circumcentre of the triangle ABC and D is the midpoint of the side BC. If <BAC 40°, let us find the value of BOD. 

Solution. BOC = 2BAC = 2 x 40° = 80°

<BOD= (90° –  180-80 / 2)

= 90° – 50° = 40°

 

Question 3. Three points A, B, and C lie on the circle with center O in such B a way that AOCB is a parallelogram, let us calculate the value of the side ZAOC.

Solution. ∠AOC 360°-2/ABC

=1/3 x 360° = 120°

 

 

Question 4. O is the circumcentre of the isosceles triangle ABC and ABC= 120°; if the length of the radius of the circle is 5 cm, let us find the value of the side AB.

Solution.  ∠ACB = 180°-120° /2  = 30°

∴∠AOB 60°

As OA = OB,

∴ ∠OAB ∠OBA = 60°

∴ OA OB = AB = 5 cm.

 

 

 

Question 5. Two circles with centers A and B intersect each other at points C and D. The center B of the other circle lie on the circle with center A. If ZCQD = 70°, let us find the value of ZCOD.

Solution. CAD (360°-4 x 70) = 80°

∠CPD= 1/2 ZCAD = 1/2 x 80° = 40°

Example :

Each semi-circle angle is one right angle.

All the semicircle angles are equal.

∠APB =∠AQB = ∠ARB =∠ASB.

 

 

 

Class 10 WBBSE Math Solution In English Question 6. Let us prove that all angles of a circle formed by the same arc are equal.

Solution: Let in the circle with center O, ∠ACB &∠ADB are the two angles at the circumference on the same arc ∠APB. 

To prove ∠ACB = ∠ADB.

Join O, A & O, B.

As on the arc ∠APB, AOB is the angle at the center and ∠ACB

& ∠ADB is the angles on the circumference.

∴∠AOB = 2∠ACB & ∠AOB = 2∠ADB

∴ ∠ACB = ∠ADB Proved.

 

 

Question 7. Let us prove that if all angles of a circle subtended by circular arcs be equal then the lengths of the arcs are equal.

Solution: Let O is the center of the circle & the angles ∠ACB, ∠ADB, and ∠AEB are the angles on the circumference are equal,

i.e.,∠ACB = ∠ADB = ∠AEB.

To prove that the angles are on an arc of equal length.

Proof: If the angle on the circumference is half of the angle at the center then they will be on the same arc.

.. ∠AOB = 2∠ACB = 2∠ADB = 2∠AEB

.. The angles on the same arc∠APB.

 

Application 1. Let us see the figure of the circle below and let us find the value of x. O is the center of the circle.

Solution: O is the center of the circle.

∠ABC and∠ADC are the front angles on the circle formed by the minor arc /CPA.

∴∠ABC = ∠ADC= 40° ( Given that ∠ADC = .40°)

and in ΔABC, ∠ABC + ∠ACB +∠BAC = 180°

.. 40°+100° + x = 180°

or, x° 180° 140° = 40°

.. x = 40.

 

 

Class 10 WBBSE Math Solution In English Application 2. In the adjoining figure, 65°. CBD = 28°; Let us determine the values of ∠ADB,∠ABD, <BAC, ∠ACB, CAD, and∠ACD.

Solution: <BAC = <BDC = 50° Again, ∠CAD = ADB = 28°

BDC = 50°, ∠APB = 28°

In ABPC, Exterior ∠APB = ∠PBC +∠PCB.

65° = 28° + ACB;

ACB = 37°

InΔABP ∠ABP + ∠BPA + ∠PAB = 180°

∠ABP = ∠ACB = 37°

∠ABP = ∠ADB = 37°

∠ACD = ∠ABD = 65°