NCERT Solutions for Class 10 Maths Ex 10.2

# NCERT Solutions for Class 10 Maths Ex 10.2

## 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

OQ2 = OP2 + PQ2               [By Pythagoras theorem]

(25)2 = OP2 + (24)2

625 = OP2 + 576

OP2 = 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 havePOQ = 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.]

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.]

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)

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

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°

OA2 = OP2 + AP2               [By Pythagoras theorem]

(5)2 = (OP)2 + (4)2

25 = (OP)2 + 16

OP2 = 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°

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

Then, OA2 = OP2 + AP2                       [By Pythagoras theorem]

(5)2 = (3)2 + AP2

25 = 9 + AP2

AP2= 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 ProveAOB = 90°

Construction: Join OC.

ProofOPA = 90°          ………. (i)

OCA = 90°              ………. (ii)

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

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)

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

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]

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

Then, the area of ABC =

=

=cm2

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

Thus,=

Therefore,= 28 + 2x + 12 + 2x + 16

= 4x + 56

= 4(x + 14)

Squaring both sides, we get

(x + 14) (x) (6) (8) = 16(x + 14)2

3x = x + 14

2x = 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.