# RS Aggarwal Class 10 Math Twelfth Chapter Circles Exercise 12A Solution

### RS Aggarwal Class 10 Math Twelfth Chapter Circles Exercise 12A Solution

Exercise 12A RS Aggarwal Class 10 Math Book Solution for CBSE English medium Students. For any problem to Solveout this Chapter 12A Solution please comment us below.

(1) Here, AB is a tangent of a circle with centre O and OB is a radius of the circle OB = 8cm Join

AO = 17cm

Now in triangle AOB, ∠B = 90o

AB2 = AO2 – OB2 [By Pythagoras theorem]

=) AB2 = (17)2 – (8)2

=) AB2 = 289 – 64

=) AB2 = 225

=) AB = Ö225

=) AB = 15

∴ AB = 15 cm

∴ The length of the tangent = 15 cm

(2) Here, AB is a tangent of a circle with centre O and OB is a radius of the circle.

AO = 25cm and AB = 24cm

Now, in triangle △AOB, ∠B = 90o

OA2 = AB2 + OB2 [By Pythagoras theorem]

=) OB2 = OA2 – AB2

=) OB2 = (25)2 – (24)2

=) OB2 = (25+24) (25-24) [∴a2 – b2 = (a + b) (a – b)]

=) OB2 = 49 x 1

=) OB2 = 49

∴ OB = Ö49

∴ OB = 7cm

∴ The radius of the circle is 7 cm

(3) We know that the radius and tangent one perpendicular at their point of contact in right triangle AOP.

AO2 = OP2 + PA2 [By Pythagoras theorem]

=) PA2 = AO2 – OP2

=) PA2 = (6.5)2 – (2.5)2

=) PA2 = 42.25 – 6.25

=) PA = Ö36

=) PA = 6

∴ PA = 6 cm

Here AB is a chord of the larger circle we know the perpendicular drawn from the center bisects the chord.

∴ AB = 2PA

=) AB = 2 x 6

∴ AB = 12

∴ The length of the chord of the larger circle is 12cm

(4) We know that tangent drawn from external point to a circle are equal.

BD = BE = y

CF = CE = Z

Given, AB = 12cm

BC = 8CM AND

AC = 10 cm

Now,

AB = AD + DB = x + y = 12 cm ———- (i)

BC = BE + EC = y + z = 8 cm ———— (ii)

And AC = AF + CF = x+ z = 10 cm ——- (iii)

Adding the three equations, we get

x + y + y + z + x + z = 12 + 8 + 10

=) 2x + 2y + 2z = 30

=) 2 (x + y + z) = 30

=) x + y + z = 30/2

=) x + y + z = 15

=) 12 + z = 15

=) z = 15 – 12

=) z = 3

∴ CF = 3 cm

Z = 3 in equation (ii)

Y + z = 8

=) y +3 = 8

=) y = 8 -3

=) y = 5

∴ BE = 5 cm

Z = 3 in equation (iii)

X + 3 = 10

=) x = 10 – 3

(5) Given,

PA and PB are tangents to a circle with center O

Here, OA = OB

OA ⊥ AP and OA ⊥ BP

∠OBP = ∠OAP = 90o

∠OBP + ∠OAP

= 90o + 90o

= 180o

We know, sum of opposite angles in a quadrilateral is 180o

A, O, B, and P are contingent

(6) We know the radius and tangent are perpendicular at their print of contact.

∴ ∠OCA = ∠OCB = 90o

Now, in △OCA and △OCB

∠OCA = ∠OCB = 90o

OA = OB

OC = OC (common)

△OCA ≅ △OCB [By RHS congruency]

∴ CA = CB

(7) Given, tangents PA and PB are down to a circle with center O and CD is the tangent to the circle at a point E and PA = 14cm

The tangent 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 – CE + CE + DE + PB – DE

= PA + PB

= PA + PA [∵ PA = PB]

= 2PA

= 2 X 14 [∵ PA = 14cm]

= 28 cm

∴Perimeter of △PCD = 28 cm

(8) Given, A circle is inscribed in a △ABC,

Touching AB, BS and AC at P, Q and R respectively

AB = 1 cm, AR = 7cm and CR = 5cm

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

∴AP = AR = 7cm, CQ = CR = 5cm

Now, BP = (AB –AP) = (10 – 7) cm = 3cm

∴BP = BQ = 3 cm

BC = 3 + 5

BC = 8

∴The length of BC is 8cm

(9) Here, ABCD is a quadrilateral. A circle touches the sides AB, BC, CD and Ad at P, Q, R, S respectively. AB = 6 cm CD = 4 cm and BC = 7 cm

We know the length of tangents  drawn from on external point to the circle are equal

AP = AS

BP = BQ

CQ = CR

DR = DS

Now, AB + CD = (AP + PB) + (CR + DR)

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

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

∴AB + CD = AD + BC

=) AD = AB + CD – BC

=) AD = 6 + 4 – 7

∴The length of the side AD is 3 cm.

(10) Given, triangle ABC is and isosceles triangle and AB = AC

We know, the triangle drawn from an external point to the circle are equal.

∴AR = AQ

BR = BP ——- (i)

CP = CQ ——– (ii)

Now, AB = AC [Given]

=) AR = BR = AQ + CQ

=) AQ + BR = AQ + CQ [∵AR = AQ]

BR = AQ + C – AQ

=) BR = CQ

=) BP = CP [from (i) and (ii)]

∴P bisects base BC

(11) Given, O is the centre of the two concentric circles of radii OA = 6cm and OB = 4cm AP and PB are tangents to the outer and inner circle respectively and PA = 10 cm.

In △OAP

OP2 = OA2 + PA2

= (6)2 + (10)2

= 36 + 100

= 136 cm2

In △OBP

BP2 = OP2 – OB2

= 136 – 16

= 120

∴BP = Ö120

= 10.9 cm

(12) Join OA, OB, OC

OE⊥AB, OD⊥BC and OF⊥AC

We know, tangents drawn from an external point to the circle are equal.

Now, we have,

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

Now,

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

=) 54 = ½ x BC x OD + ½ x AC x OF + ½ x AB x OE

=) 54 = (½ x 15 x 3) + [1/2 x (9+x) x 3] + [1/2 (6 + x) x 3]

=) 54 = ½ (45 + 27 + 3x + 18 +x 3x)

=) 108 = 45 + 27 + 6x + 18

=) 6x = 45 + 27 + 18 – 108

=) 6x = 18

=) x = 3

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

(13) Let, TR = y and TP = x

We know that the perpendicular drown from the center to the chord bisects it

∴PR = RQ

Now, PR + RQ = 4.8 [∵PR+ RQ = PQ = 4.8]

=) PR+ PR = 4.8 [∵PR = RQ]

=) 2PR = 4.8

PR = 4.8/2

PR = 2.4

Now, in right triangle POR,

PO2 = OR + PR [By Pythagoras theorem]

=) (3) = OR + (2.4)

=) OR = 9 – 5.76

OR = 3. 24

OR = 1.8

Again, in right triangle TRP,

TP = TR + PR [By Pythagoras theorem]

=) x = y + PR

=) x = y + (2.4)

=) x = y + 5.76 ————————– (i)

Again, in right triangle TPQ,

TO2 = TP2 +PO2
=) (TR + OR2) = TP2 +PO2

=) (y + 1.8) 2 = x2 + (3)2

=) y2 + 3.6y + 3.24 = x2 + 9

=) x2 – y2 = 3.6y + 3.24 – 9

=) x2 – y2 = 3.6y – 5.76 ———————- (ii)

x2 – y2 = 3.6y – 5.76 in equation (ii)

y2 + 5.76 – y2 = 3.6y – 5.76

=) 3.6y = 5.76 + 5.76

=) y = 11.52/3.6 = 3.2 cm

Y = 3.2 in equation (i)

x2 = (3.2) 2 + 5.76

=) x2 = 10.24 + 5.76

=) x = Ö16

∴ x = 4

∴TP = 4 cm

(14) Let, CD and AB two parallel tangents of a circle with center O.

Join OR and OQ and draw OP II CD.

Now, PQO + POQ = 180o [adjacent interior angles]

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

∠DGO = 90o

∴ 90o + ∠PQR = 180o

=) PQR = 180o – 90o = 90o

Similarly ∠POR = 90o

∴ ∠POR + ∠POQ = 90o + 90o = 180o

∴ QR is a straight line passing through the center O

(15) Given, O be the centre of a circle which is inscribed in a quadrilateral ABCD and the circle touches the side of quadrilateral at P, Q, R and S respectively.

OP = OQ [radius of the circle]

Here, AB = 29 cm, AD = 23 cm ∠B= 90o and DS = 5 cm

DR and DS are the tangents to the circle

∴ DR = DS = 5cm

AG and AR are tangents to the circle

AR = AG

=) AR  + RD = 23

=) AR = 23 – 5 [∵ DS = RD = 5]

AR = 18 cm

Again, AB = 29 cm

=) AQ + GB = 29

=) QB = 29 – 18 [∵ AR = AQ = 18]

∴ QB = 11 cm

All the angels of the quadrilateral BQOP are right angles and BQ = OP

BQOP is a square.

∴ BQ = PO = OQ = 11 cm

∴ Radius of the circle is 11 cm

(16) AB is the chord passing through the center.

AB is the diameter, Join OP.

Since, angle in a semi circle is a right angle.

∴ ∠APB = 90o

We name, ∠APB = ∠PAT = 30o [Alternate segment theorem]

Now, in △APB

∠BAP + ∠APB + ∠PBA = 180O

=) ∠BAP = 180O – ∠APB – ∠PBA

=) ∠BAP = 180O – 90O – 30O = 60O

∠PAB = ∠OPA = 60o

Now, ∠OPA+ ∠OAP + ∠AOP = 180o [By angle sum property]

=) 60o + ∠PAB + ∠AOP = 180o [PAB = OPA]

=) 60o + 60o + ∠AOP = 180o

=) ∠AOP = 60o

All angles of △OPA equals to 60o

∴△OPA is an equilateral triangle

∴OP = OA = PA

∠OPT = 90o

Now, ∠OPA + ∠APT = 90o

=) 60o + ∠APT = 90o

∴∠APT = 30o

Again, ∠PAB + ∠PAT = 180o

=) 60o + ∠PAT = 180o

=) ∠PAT = 120o

In △APT

∠APT + ∠PAT + ∠PTA = 180O

=) 30O + 120 + ∠PTA = 180O

=) ∠PTA = 30O

∴∠APT = ∠PTA = 30O

∴AT = PA

IN △ABP

Sin ∠ABP = AP/BA

=) Sin30O = AT/BA [AT = PA]

=) ½ = AT/BA

∴BA: AT = 2:1

Updated: September 1, 2020 — 12:43 am