**Download NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2**

**1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is**

**(A) 7 cm****(B) 12 cm****(C) 15 cm****(D) 5 cm**

Let **P** be the point of contact and **O** be the centre of the circle with radius **r**∴ PQ=24 cm

OQ=25 cm

∵ Tangent and radius are perpendicular to each other at point of contact

∴ In △OPQ, ∠P=90°

OP^{2} + PQ^{2} = OQ^{2} (By Pythagorean Theorem)

r^{2} + 24^{2} = 25^{2}r^{2} = 25^{2} – 24^{2}r^{2} = (25 – 24)×(25 + 24)

r^{2 }= 49

r = 7 cm

Therefore, the correct option is (A)

**2. In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ=110°, then ∠PTQ is equal to**

**(A) 60°****(B) 70°****(C) 80°****(D) 90°**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠TPO = ∠TQO = 90°

∵ Sum of angles of a quadrilateral = 360°

∴ In quadrilateral TPOQ

∠TPO + ∠POQ + ∠OQT + ∠QTP = 360°

90° + 110° + 90° + ∠QTP = 360°

290° + ∠QTP = 360°

∠QTP = 360° – 290°

∠QTP = 70°

Therefore, correct option is (B)

**3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to**

**(A) 50°****(B) 60°****(C) 70°****(D) 80°**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠PAO = ∠OBP = 90°

∵ Sum of angles of a quadrilateral = 360°

∴ In quadrilateral PAOB

∠PAO + ∠AOB + ∠OBP + ∠BPA = 360°

90° + ∠AOB + 90° + 80° = 360°

260° + ∠AOB = 360°

∠AOB = 360° – 260°

∠AOB = 100°

In △OPA and △OPB

∠PAO = ∠OBP = 90°

PO = PO [Common]

AO = OB [Radius]

∴ △OPA ≅ △OPB (RHS Congruency Rule)

⇒ ∠AOP = ∠BOP = ∠AOB = x 100° = 50°

Therefore, correct option is (A)

**4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel**

**Given:** **PQ** and **RS** be the two tangents drawn at the end points of the diameter **AB** of circle with centre **O**

**To Prove: PQ **and **RS **are parallel to each other

**Proof:**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠PBO = ∠QBO = 90°

∠RAO = ∠SAO = 90°

⇒ ∠RAO = ∠QBO

∠PBO = ∠SAO

∵ Alternate angles are equal

∴ PQ and RS are parallel to each other

**5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.**

**Given:** A circle with **O** as the centre of the circle and **AB** is a tangent to the circle with point of contact at **P**

**To Prove: **Perpendicular to the tangent **AB **passes through the centre **O**

**Proof (by contradiction):**

Let the perpendicular to the tangent AB at P does not pass through O but passes through a point O’, not lying on the line joining O and P

∴ ∠O’PB = 90° **…(i)**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠OPB = 90° **…(ii)**

From equation (i) and (ii)

∠O’PB = 90° = ∠OPB

This is possible only if O’, O and P lie on the same line

It is contradiction to the earlier assumption

Therefore, O’ lies on the line joining O and P

Hence, the perpendicular to the tangent AB at P passes through O

Therefore, the perpendicular at the point of contact to the tangent to a circle passes through the centre

**6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.**

Let **B** be the point of contact and **O** be the centre of the circle with radius **r**∴ BA = 4 cm

OA = 5 cm

∵ Tangent and radius are perpendicular to each other at point of contact

∴ In △OAB, ∠B=90°

OB^{2} + BA^{2} = OA^{2} (By Pythagorean Theorem)

r^{2} + 4^{2} = 5^{2}r^{2} = 5^{2} – 4^{2}r^{2} = 25 – 16

r^{2 }= 9

r = 3 cm

Therefore, the radius of the circle is 3 cm.

**7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.**

Let **AC **be the chord of the bigger circle and is a tangent to the smaller circle with point of contact **B**, both circles have common centre **O**∴ OA = 5 cm

OB = 3 cm

∵ Tangent and radius are perpendicular to each other at point of contact

∴ In △OBA, ∠B=90°

OB^{2} + BA^{2} = OA^{2} (By Pythagorean Theorem)

3^{2} + BA^{2} = 5^{2}BA^{2} = 5^{2} – 3^{2}BA^{2} = 25 – 9

BA^{2 }= 16

BA = 4 cm

AC = 2×AB (Perpendicular to the chord from the centre bisects it)

AC = 2×4 = 8 cm

Therefore, the length of the chord of larger circle which touches the smaller circle is 8 cm.

**8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC**

The points **P**, **Q**, **R** and **S** are the point of where the circle is touched by the sides **AB**, **BC**, **CD** and **DA**, respectively.

∵ Tangents from an external point to a circle are of equal length

∴ AP = AS …(i)

BP = BQ …(ii)

DR = DS …(iii)

CR = CQ …(iv)

Adding equations (i), (ii), (iii) and (iv)

AP + BP + DR + CR = AS + BQ + DS + CQ

(AP + BP) + (DR + CR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

**9. In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB=90°.**

Join the centre of the circle with the point of contact of the tangent **AB**

In △OPA and △OCA

∠OPA = ∠OCA = 90° (Tangent and radius are perpendicular to each other at point of contact)

AO = AO [Common]

PO = CO = Radius

∴ △OPA ≅ △OCA (RHS Congruency Rule)

⇒ ∠POA = ∠COA **…(i)**(By CPCT)

Similarly, △OQB ≅ △OCB

⇒ ∠QOB = ∠COB **…(ii)**(By CPCT)

∵ PQ is a straight line

∴ ∠QOB = 180°

∠POA + ∠COA + ∠QOB + ∠COB = 180°

Substituting values from equation (i) and (ii)

2∠COA + 2∠COB = 180°

⇒ ∠AOB = 90°

**10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.**

**Given: ****A** and **B** are the points of contact of the tangents from the external point **P** to the circle with centre **O**

**To Prove: **∠APB + ∠AOB = 180°

**Proof:**

∵ Tangent and radius are perpendicular to each other at point of contact

∴ ∠PAO = ∠PBO = 90°

∵ Sum of angles of a quadrilateral = 360°

∴ In quadrilateral PAOB

∠PAO + ∠AOB + ∠OBP + ∠APB = 360°

90° + ∠AOB + 90° + ∠APB = 360°

180° + ∠AOB + ∠APB = 360°

∠AOB + ∠APB = 360° – 180°

∠APB + ∠AOB = 180°

Hence, it is proved that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

**11. Prove that the parallelogram circumscribing a circle is a rhombus.**

**ABCD** is a parallelogram and the points **P**, **Q**, **R** and **S** are the point of where the circle is touched by the sides **AB**, **BC**, **CD** and **DA**, respectively

∴ AB = CD …(i)

BC = AD …(ii)

∵ Tangents from an external point to a circle are of equal length

∴ AP = AS …(iii)

BP = BQ …(iv)

DR = DS …(v)

CR = CQ …(vi)

Adding equations (iii), (iv), (v) and (vi)

AP + BP + DR + CR = AS + BQ + DS + CQ

(AP + BP) + (DR + CR) = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

Substituting values from equation (i) and (ii)

AB + AB = BC + BC

2AB = 2BC

AB = BC …(vii)

From equation (i), (ii) and (vii)

AB = BC = CD = DA

⇒ All sides of the parallelogram circumscribing a circle are equal

Since, a parallelogram with sides of equal length is a rhombus

Therefore, a parallelogram circumscribing a circle is a rhombus.

**12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.**

Let the sides AB and AC touch the circle at E and F respectively

Let AF = x

∵ Tangents from an external point to a circle are of equal length

∴ AE = AF = x

BD = BE = 8 cm

CF = CD = 6 cm

AB = AE + EB = x + 8

BC = BD + CD = 8 + 6 = 14 cm

CA = CF + FA = 6 + x

Semi-perimeter of △ABC (s) =(AB + BC + CA) = (x + 8 + 14 + 6 + x) = x + 14

Area of △ABC = (Heron’s Formula)

Area of △ABC =

Area of △ABC =

Area of △ABC = …(i)

Area of △ABC = ar(△AOB) + ar(△BOC) + ar(△COA)

= (AB×OE) + (BC×OD) + CA×OF)

= (x + 8)×4 + (14×4) + (x + 6)×4 (OD = OE = OF = 4 cm)

= 2×(x + 8) + 2×14 + 2×(x + 6)

= 2×(x + 8 + 14 + x + 6)

= 2×(2x + 28)

= 4(x + 14) …(ii)

From equation (i) and (ii)

48x(x + 14) = 16(x + 14)^{2} **(Both sides squared)**

Dividing by common factors

3x = (x + 14)

3x – x = 14

2x = 14

x = 7 cm

AB = x + 8 = 7 + 8 = 15 cm

BC = 14 cm

CA = x + 6 = 7 + 6 = 13 cm

Therefore, AB = 15 cm and AC = 13 cm.

**13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.**

Consider the quadrilateral ABCD circumscribing a circle with centre O. AB, BC, CD and DA touch the circle at points P, Q, R and S respectively

Join the centre of the circle to the points of contact and the vertices of the quadrilateral

In △POA and △SOA

∠OPA = ∠OSA = 90° (Tangent and radius are perpendicular to each other at point of contact)

AO = AO [Common]

OP = OS [Radius]

∴ △POA ≅ △SOA (RHS Congruency Rule)

⇒ ∠POA = ∠SOA (By CPCT) **…(i)**

Similarly, △SOD ≅ △ROD

⇒ ∠SOD = ∠ROD **…(ii)**

△ROC ≅ △QOC

⇒ ∠ROC = ∠QOC **…(iii)**

△QOB ≅ △POB

⇒ ∠QOB = ∠POB **…(iv)**

∠POA + ∠SOA + ∠SOD + ∠ROD + ∠ROC + ∠QOC + ∠QOB + ∠POB = 360°

Substituting values from equation (i), (ii), (iii) and (iv)

2∠POA + 2∠POB + 2∠ROD + 2∠ROC = 360°

2[(∠POA + ∠POB) + 2(∠ROD + ∠ROC)] = 360°

2(∠AOB + ∠COD) = 360°

⇒ ∠AOB + ∠COD = 180°

Similarly, ∠AOD + ∠BOC = 180°

Hence, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

**Download NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2**

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