# RS Aggarwal Solutions Chapter 12 Circles Exercise 12A Class 10 Maths

 Chapter Name RS Aggarwal Chapter 12 Circles Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 12BMCQ Exercise Related Study NCERT Solutions for Class 10 Maths

### Exercise 12A Solutions

1. Find the length of tangent drawn to a circle with radius 8 cm form a point 17 cm away from the center of the circle.

Solution

Let O be the center of the given circle.

Let P be a point, such that

OP = 17 cm

Let OT be the radius, where

OT = 5 cm

Join TP, where TP is a tangent.

Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.

∴ OT ⊥ PT
In the right ∆OTP, we have:

OP2 = OT2 + TP2 [By Pythagoras’ theorem]

∴ The length of the tangent is 15 cm.

2. A point P is 25 cm away from the center of a circle and the length of tangent drawn from P to the circle is 24 cm. Find the radius of the circle.

Solution

Draw a circle and let P be a point such that OP = 25 cm

Let TP be the tangent, so that TP = 24 cm

Join OT where OT is radius.

Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.

∴ OT ⊥ PT

In the right ∆OTP, we have:

OP2 = OT2 + TP2 [By Pythagoras’ theorem]

∴ The length of the tangent is 7 cm.

3. Two concentric circles are of radii 6.5 cm and 2.5 cm. Find the length of the larger circle which touches the smaller circle.

Solution

We know that the radius and tangent are perpendicular at their point of contact

In right triangle AOP

AO2 = OP2 + PA2

⇒ (6.5)2 = (2.5)2 + PA2

⇒ PA2 = 36

⇒ PA = 6 cm

Since the perpendicular drawn from the center bisects the chord.

∴ PA = PB = 6 cm

Now, AB = AP + PB = 6 + 6 = 12 cm

Hence, the length of the chord of the larger circle is 12 cm.

4. In the given figure, a circle inscribed in a triangle ABC, touches the sides AB, BC and AC at points D, E and F Respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, find the length of AD, BE and CF.

Solution

We know that tangent segments to a circle from the same external point are congruent.

Now, we have

AD = AF, BD = BE and CE = CF

Now, we have

AD = AF, BD = BE and CE = CF

Now, AD + BD = 12 cm ...(1)

AF + FC = 10 cm

⇒ AD + FC = 10 cm ....(2)

BE + EC = 8 cm

⇒ BD + FC = 8 cm

AD + BD + AD + FC + BD + FC = 30

⇒ 2(AD + BD + FC) = 30

⇒ AD + BD + FC = 15 cm ...(4)

Solving (1) and (4), we get

FC = 3 cm

Solving (2) and (4), we get

BD = 5 cm

Solving (3) and (4), we get

∴ AD = AF = 7 cm, BD = BE = 5 cm and CE = CF = 3 cm

5. In the given figure, a circle touches all the four of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Solution

Let the circle touch the sides of the quadrilateral AB, BC, CD and DA at P, Q, R and S respectively.

Given, AB = 6 cm, BC = 7 cm and CD = 4 cm

Tangents drawn from an external point are equal.

∴ AP = AS, BP = BQ, CR = CQ and DR = DS

Now, AB + CD (AP + BP) + (CR + DR)

⇒ AB + CD = (AS + BQ) + (CQ + DS)

⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

⇒ AD = (AB + CD) – BC

⇒ AD = (6 + 4) – 7

∴ The length of AD is 3 cm.

6. In the given figure, the chord AB of the larger of the two concentric circles, with center O, touches the smaller circle at C. Prove that AC = CB.

Solution

Construction: Join OA, OC and OB

We know that the radius and tangent are perpendicular at their point of contact

∴ ∠OCA = ∠OCB = 90°

Now, In ∆OCA and ∆OCB

∠OCA = ∠OCB = 90°

OA = OB (Radii of the larger circle)

OC = OC (Common)

By RHS congruency

∆OCA ≅ ∆OCB

∴ CA = CB

7. From an external point P, tangents PA and PB are drawn to a circle with center O. If CD is the tangent to circle at a point E and PA = 14 cm, find the perimeter of ∆PCD.

Solution

Given, PA and PB are the tangents to a circle with center O and CD is a tangent at E and

PA = 14 cm

Tangents drawn from an external point are equal.

∴ PA = PB, CA = CE and DB = DE

Perimeter of ∆PCD = PC + CD + PD

= (PA – CA) + (CE + DE) + (PB – DB)

= (PA – CE) + (CE + DE) + (PB – DE)

= (PA + PB)

= 2PA (∵ PA = PB)

= (2 × 14) cm

= 28 cm

∴ Perimeter of ∆PCD = 28 cm

8. A circle is inscribed in a ∆ABC touching AB, BC and AC at P, Q and R respectively. If AB = 10 cm, AR = 7 cm and CR = 5 cm, find the length of BC.

Solution

Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle

Tangents drawn to a circle from an external point are equal.

∴ AP = AR = 7 cm, CQ = CR = 5 cm

Now, BP = (AB – AP)

= (10 – 7)

= 3 cm

∴ BP = BQ = 3 cm

∴ BC = (BQ + QC)

⇒ BC = 3 + 5

⇒ BC = 8

∴ The length of BC is 8 cm.

9. In the given figure, PA and PB are the tangents segments to a circle with centre O. Show that the points A, O, B and P are concyclic.

Solution

Here, OA = OB

And OA ⊥ AP, OA ⊥ BP (Since tangents drawn from an external point are perpendicular to the radius at the point of contact)

∴ ∠OAP = 90°, ∠OBP = 90°

∴ ∠OAP + ∠OBP = 90° + 90° = 180°

∴ ∠AOB + ∠APB = 180° (Since, ∠OAP + ∠OBP + ∠AOB + ∠APB = 360°)

Sum of opposite angle of a quadrilateral is 180°.

Hence, A, O, B and P are concyclic.

10. In the given figure, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that point of contact P bisects the base BC.

Solution

We know that tangent segments to a circle from the same external points are congruent

Now, we have

AR = AO, BR = BP and CP = CQ

Now, AB = AC

⇒ AR + RB = AQ + QC

⇒ AR + RB = AR + OC

⇒ RB = QC

⇒ BP = CP

Hence, P bisects BC at P.

11. In the given figure, O is the centre of the two concentric circles of radii 4 cm and 6 cm respectively. AP and PB are tangents to the outer and inner circle respectively. If PA = 10 cm, find the length of PB upto one place of the decimal.

Solution

Given, O is the center of two concentric circles of radii OA = 6 cm and OB = 4 cm.

PA and PB are the two tangents to the outer and inner circles respectively and PA = 10 cm.

Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.

∴ ∠OAP = ∠OBP = 90°

∴ From right-angled OAP, OP2 = OA2 + PA2

∴ The length of PB is 10.9 cm.

12. In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BC and DC into which BC is divided by the point of contact D, are of lengths 6 cm and 9 cm respectively. If the area of ∆ABC = 54 cm2 then find the length of sides AB and AC.

Solution

Construction: Join OA, OB, OC, OE ⊥ AB at E and OF ⊥ AC at F.

We know that tangent segments to a circle from me same external point are congruent

Now, we have

AE = AF, BD = BE = 6 cm and CD = CF = 9 cm

Now,

Area (∆ABC) = Area (∆BOC) + Area(∆AOB) + Area (∆AOC)

⇒ 54 = 1/2 × BC × OD + 1/2 × AB × OE + 1/2 × AC × OF

⇒ 108 = 15 × 3 + (6 + x) × 3 + (9 + x) × 3

⇒ 36 = 15 + 6 + x + 9 + x

⇒ 36 = 30 + 2x

⇒ 2x = 6

⇒ x = 3 cm

∴ AB = 6 + 3 = 9 cm and AC = 9 + 3 = 12 cm

13. PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.

Solution

Let TR = y and TP = x

We know that the perpendicular drawn from the center to me chord bisects it.

∴ PR = RQ

Now, PR + RQ = 4.8

⇒ PR + PR = 4.8

⇒ PR = 2.4

Now, in right triangle POR

By Using Pythagoras theorem, we have

PO2 = OR2 + PR2

⇒ 32 = OR2 + (2.4)2

⇒ OR2 = 3.24

⇒ OR = 1.8

Now, in right triangle TPR

By using Pythagoras theorem, we have

TP2 = TR2 + PR2

⇒ x2 = y2 + (2.4)2

⇒ x2 = y2 + 5.76 ...(1)

Again, In right triangle TPQ

By using Pythagoras theorem, we have

TO2 = TP2 + PO2

⇒ (y + 1.8)2 = x2 + 32

⇒ y2 + 3.6y + 3.24 = x2 + 9

⇒ y2 + 3.6y = x2 + 5.76 ...(2)

Solving (1) and (2), we get

x = 4 cm and y = 3.2 cm

∴ TP = 4 cm

14. Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

Solution

Suppose CD and AB are two parallel tangents of a circle with centre O

Construction: Draw a line parallel to CD passing through O i.e., OP

We know that the radius and tangent are perpendicular at their point of contact.

∠OQC = ∠ORA = 90°

Now, ∠OQC + ∠POQ = 180° (co-interior angles)

⇒ ∠POQ = 180° - 90° = 90°

Similarly, Now, ∠ORA + ∠POR = 180° (co-interior angles)

⇒ ∠POQ = 180° - 90° = 90°

Now, ∠POR + POQ = 90° + 90° = 180°

Since, ∠POR and ∠POQ are linear pair angles whose sum is 180°

Hence, QR is a straight line passing through center O.

15. In the given figure, with center O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, B = 90° and DS = 5 cm then find the radius of the circle.

Solution

We know that tangent segments to a circle from the same external point are congruent

Now, we have

DS = DR, AR = AQ

⇒ AR + RD = 23

⇒ AR = 23 – RD

⇒ AR = 23 – 5 [∴ DS = DR = 5]

⇒ AR = 18 cm

Again. AB = 29 cm

⇒ AQ + QB = 29

⇒ QB = 29 – AQ

⇒ QB = 29 – 18 [∵ AR = AQ = 18]

⇒ QB = 11 cm

Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ

Hence, BQOP is a square.

We know that all the sides of square are equal.

Therefore, BQ = PO = 11 cm

16In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If PBT = 30°, prove that BA : AT = 2 : 1.

Solution

AB is the chord passing through the center

So, AB is the diameter

Since, angle in a semicircle is a right angle

∴ ∠APB = 90°

By using alternate segment theorem

We have ∠APB = ∠PAT = 30°

Now, in APB

∠BAP + ∠APB + ∠BAP = 180° (Angle sum property of triangle)

⇒ ∠BAP = 180° - 90° - 30° = 60°

Now, ∠BAP = ∠APT + ∠PTA (Exterior angle property)

⇒ 60° = 30° + ∠PTA

⇒ ∠PTA = 60° - 30° = 30°

We know that sides opposite to equal angles are equal

∴ AP = AT

In right triangle ABP

sin ∠ABP = AP/BA

⇒ sin 30° = AT/BA

⇒ 1/2 = AT/BA °

∴ BA : AT = 2 : 1