## In this post, you will find the NCERT solutions for class 10 maths ex 10.2. These solutions are based on the latest syllabus of NCERT Maths class 10.

**NCERT Solutions for Class 10 Maths Ex 10.2**

**In Q. 1 to 3, choose the correct option and
give justification.**

**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) 24.5 cm**

**Solution: (A)** We have, ∠OPQ = 90°

[Since the tangent at any point of a circle is perpendicular to the radius through the point of contact]

In right-angled ∆OPQ, we have

OQ^{2
}=^{ }OP^{2 }+ PQ^{2} [By Pythagoras theorem]

(25)^{2 }=^{ }OP^{2
}+ (24)^{2}

625
= OP^{2} + 576

OP^{2 }= 625 – 576 = 49

OP
= 7 cm

**2. In the 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°**

**Solution: (B) **We
have, ∠POQ = 110°, ∠OPT
= 90° and ∠OQT = 90°

[Since the
tangent at any point of a circle is perpendicular to the radius through
the point of contact.]

In
quadrilateral OPTQ, we have

∠POQ
+ ∠OPT
+ ∠OQT
+ ∠PTQ
= 360° [Angle sum property of a
quadrilateral]

110°
+ 90° + 90° + ∠PTQ
= 360°

290° + ∠PTQ = 360°

∠PTQ
= 360° – 290°

∠PTQ
= 70°

**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°**

**Solution: (A) **We have, ∠APB = 80°, ∠OAP
= 90° and ∠OBP = 90°

[Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.]

In
quadrilateral OAPB, we have

∠APB
+ ∠OAP
+ ∠OBP
+ ∠AOB
= 360° [Angle sum property of a
quadrilateral]

80°
+ 90° + 90° + ∠AOB
= 360°

260° + ∠AOB = 360°

∠AOB
= 360° – 260°

∠AOB
= 100°

Now, ∠POA
= ½ ∠AOB [Since the line OP bisects the ∠AOB and ∠APB]

∠POA
= ½ × 100°

∠POA
= 50°

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

**Solution:
Given**: PQ is a diameter
of a circle with centre O and two tangents AB and CD are drawn at the points P
and Q, respectively.

**To Prove**: AB || CD

**Proof**: Since AB is a tangent to the circle
at P and OP is the radius through the point of contact.

Therefore, ∠OPA = 90° ………. (i)

[Since the
tangent at any point of a circle is perpendicular to
the radius through the point of contact.]

Since CD is
a tangent to the circle at Q and OQ is the radius through the point of contact.

Therefore, ∠OQD = 90° ………. (ii)

From
equations (i) and (ii), we have ∠OPA = ∠OQD
……….
(iii)

But
∠OPA
and ∠OQD
form a pair of alternate angles which are equal.

Hence, AB
|| CD

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

**Solution: **We know that the tangent at any point
of a circle is perpendicular to the radius through the point of contact and the
radius essentially passes through the centre of the circle, 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.**

**Solution:
**Since the tangent
at any point of a circle is perpendicular to
the radius through the point of contact.

Therefore, ∠OPA = 90°

OA^{2
}= OP^{2 }+ AP^{2} [By Pythagoras theorem]

(5)^{2
}= (OP)^{2 }+ (4)^{2}

25
= (OP)^{2 }+ 16

OP^{2 }= 9

OP
= 3 cm

Hence,
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.**

**Solution: **Let O be the common centre of the two
concentric circles.

Let
AB be a chord of the larger circle which touches the smaller circle at a point
P.

Join
OP and OA.

We
have, ∠OPA
= 90°

Then, OA^{2} = OP^{2} + AP^{2} [By Pythagoras theorem]

(5)^{2
}= (3)^{2 }+ AP^{2}

25
= 9 + AP^{2}

AP^{2}= 16

AP
= 4 cm

Since the
perpendicular from the centre of a circle to a chord bisects the chord,
therefore, AP = BP = 4 cm.

Thus, AB
= AP + BP

= AP + AP = 2AP

= 2 × 4 = 8 cm

Hence,
the length of the chord of the larger circle is 8 cm.

**8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove
that:**

**AB + CD = AD + BC **

**Solution: **Since the tangents from an external
point to a circle are equal.

Therefore, AP
= AS ………. (i)

BP
= BQ ………. (ii)

CR
= CQ ………. (iii)

DR
= DS ………. (iv)

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

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

AB
+ CD = (AS + DS) + (BQ + CQ)

AB
+ CD = AD + BC
Hence proved.

**9. In the 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°.**

**Solution:
Given**: 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.

**To Prove**: ∠AOB = 90°

**Construction**: Join OC.

**Proof**: ∠OPA = 90° ………. (i)

∠OCA
= 90° ………. (ii)

In
right-angled triangles OPA and OCA, we have

OA
= OA [Common]

AP
= AC [Tangents from an external point
to a circle are equal]

∠OPA
= ∠OCA [From equations (i) and (ii)]

Thus, ∆OPA ≅ ∆OCA [By RHS congruence criterion]

Therefore, ∠OAP = ∠OAC [By C.P.C.T.]

∠OAC
= ½ ∠PAB ………. (iii)

Similarly, ∠OBQ = ∠OBC

∠OBC
= ½ ∠QBA
………. (iv)

Now, since XY
|| X’Y’
and a transversal AB intersects them.

Then, ∠PAB
+ ∠QBA
= 180°

[Sum of the co-interior
angles on the same side of the transversal is 180°]

½ ∠PAB
+ ½ ∠QBA
= ½ × 180° ……….
(v)

∠OAC
+ ∠OBC
= 90° [From equations (iii) and
(iv)]

In ∆AOB, we have

∠OAC
+ ∠OBC
+ ∠AOB
= 180° [Angle sum property of a
triangle]

90° + ∠AOB = 180° [From equation (v)]

∠AOB
= 90° Hence
proved.

**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.**

**Solution: **

We
have** **∠OAP
= 90° ………. (i)

and ∠OBP
= 90°
………. (ii)

Now, since OAPB
is a quadrilateral.

Therefore,
∠APB
+ ∠AOB
+ ∠OAP
+ ∠OBP
= 360° [Angle sum property of a
quadrilateral]

∠APB
+ ∠AOB
+ 90° + 90° = 360°
[From equations (i) and (ii)]

∠APB
+ ∠AOB
= 180°

Hence, ∠APB and ∠AOB are supplementary.

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

**Solution: Given**: ABCD is a parallelogram
circumscribing a circle.

**To Prove**: ABCD is a rhombus.

**Proof**: Since, the tangents from an external
point to a circle are equal.

Therefore, AP
= AS ………. (i)

BP
= BQ ………. (ii)

CR
= CQ ………. (iii)

DR
= DS ………. (iv)

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

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

AB
+ CD = (AS + DS) + (BQ + CQ)

AB
+ CD = AD + BC

AB
+ AB = AD + AD [Opposite sides
of a parallelogram are equal]

2AB
= 2AD

AB
= AD

But
AB = CD and AD = BC [Opposite
sides of a parallelogram]

Therefore, AB
= BC = CD = AD

Hence, the parallelogram
ABCD 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 (see
figure). Find the sides AB and AC.**

**Solution: **Join OE and OF. Also join OA, OB and
OC.

Since
BD = 8 cm

Therefore, BE
= 8 cm [Tangents from an external
point to a circle are equal]

Since
CD = 6 cm

Therefore, CF
= 6 cm [Tangents from an
external point to a circle are equal]

Let
AE = AF = *x *cm

Since
OD = OE = OF = 4 cm [Radii of a
circle are equal]

Therefore, semi-perimeter
of ∆ABC ==
(*x* + 14) cm

Now, Area
of ∆ABC = Area of ∆OBC + Area of ∆OCA + Area of ∆OAB

Therefore,*x* + 12 + 2*x* + 16

*x* + 56

*x* + 14)

Squaring
both sides, we get

(*x* + 14) (*x*) (6) (8) = 16(*x* + 14)^{2}

3*x* = *x*
+ 14

2*x* = 14

*x* = 7

Therefore, AB
= *x* + 8 = 7 + 8 = 15 cm

And
AC = *x* + 6 = 7 + 6 = 13 cm

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

**Solution: ****Given:** ABCD is a quadrilateral
circumscribing a circle whose centre is O.

**To prove:** **(i)** ∠AOB + ∠COD = 180° **(ii)** ∠BOC + ∠AOD = 180°

**Construction:** Join OP, OQ, OR and OS.

**Proof:** Since the tangents from an external
point to a circle are equal.

Therefore, AP = AS

BP
= BQ ………. (i)

CQ
= CR

DR
= DS

In ∆OBP and OBQ,

OP
= OQ [Radii of
the same circle are equal]

OB
= OB [Common]

BP
= BQ [From equation
(i)]

Therefore,
∆OBP ≅ ∆OBQ [By SSS congruence criterion]

So, ∠1 = ∠2 [By C.P.C.T.]

Similarly, ∠3 = ∠4, ∠5 = ∠6, and ∠7 = ∠8

Since,
the sum of all the angles around a point is equal to 360°.

Therefore, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°

∠1
+ ∠1 + ∠4 + ∠4 + ∠5 + ∠5 + ∠8 + ∠8 = 360°

2(∠1 + ∠4 + ∠5 + ∠8) = 360°

∠1 + ∠4 + ∠5 + ∠8 = 180°

(∠1 + ∠8) + (∠4 + ∠5) = 180°

∠AOB
+ ∠COD
= 180°

Similarly,
we can prove that ∠BOC
+ ∠AOD
= 180°

Hence proved.