**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 = 90^{o}

AB^{2} = AO^{2} – OB^{2} [By Pythagoras theorem]

=) AB^{2 }= (17)^{2} – (8)^{2}

=) AB^{2 }= 289 – 64

=) AB^{2 }= 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 = 90^{o}

OA^{2 }= AB^{2} + OB^{2} [By Pythagoras theorem]

=) OB^{2} = OA^{2 }– AB^{2}

=) OB^{2} = (25)^{2} – (24)^{2}

=) OB^{2} = (25+24) (25-24) [∴a^{2} – b^{2} = (a + b) (a – b)]

=) OB^{2} = 49 x 1

=) OB^{2} = 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.

AO^{2} = OP^{2} + PA^{2} [By Pythagoras theorem]

=) PA^{2} = AO^{2} – OP^{2}

=) PA^{2} = (6.5)^{2} – (2.5)^{2}

=) PA^{2} = 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.

AD = AF = x

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

∴ AD = 7 cm

**(5)** Given,

PA and PB are tangents to a circle with center O

Here, OA = OB

OA ⊥ AP and OA ⊥ BP

∠OBP = ∠OAP = 90^{o}

∠OBP + ∠OAP

= 90^{o }+ 90^{o}

= 180^{o}

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

AOBP is a quadrilateral

A, O, B, and P are contingent

**(6)**

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

∴ ∠OCA = ∠OCB = 90^{o}

Now, in △OCA and △OCB

∠OCA = ∠OCB = 90^{o}

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

=) AD = 3 cm

∴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

OP^{2} = OA^{2} + PA^{2}

= (6)^{2} + (10)^{2}

= 36 + 100

= 136 cm^{2}

In △OBP

BP^{2} = OP^{2} – OB^{2}

= 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,

PO^{2} = 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,

TO^{2} = TP^{2} +PO^{2}

=) (TR + OR^{2}) = TP^{2} +PO^{2}

=) (y + 1.8)^{ 2} = x^{2} + (3)^{2}

=) y^{2 }+ 3.6y + 3.24 = x^{2} + 9

=) x^{2} – y^{2} = 3.6y + 3.24 – 9

=) x^{2} – y^{2} = 3.6y – 5.76 ———————- (ii)

x^{2} – y^{2} = 3.6y – 5.76 in equation (ii)

y^{2} + 5.76 – y^{2} = 3.6y – 5.76

=) 3.6y = 5.76 + 5.76

=) y = 11.52/3.6 = 3.2 cm

Y = 3.2 in equation (i)

x^{2} = (3.2)^{ 2} + 5.76

=) x^{2} = 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 = 180^{o} [adjacent interior angles]

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

∠DGO = 90^{o}

∴ 90^{o} + ∠PQR = 180^{o}

=) PQR = 180^{o} – 90^{o} = 90^{o}

Similarly ∠POR = 90^{o}

∴ ∠POR + ∠POQ = 90^{o} + 90^{o} = 180^{o}

∴ 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= 90^{o} 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

Now, Ad = 23 cm

=) 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 = 90^{o}

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

Now, in △APB

∠BAP + ∠APB + ∠PBA = 180^{O}

=) ∠BAP = 180^{O} – ∠APB – ∠PBA

=) ∠BAP = 180^{O} – 90^{O} – 30^{O} = 60^{O}

OP = OA [radii]

∠PAB = ∠OPA = 60^{o}

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

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

=) 60^{o} + 60^{o} + ∠AOP = 180^{o}

=) ∠AOP = 60^{o}

All angles of △OPA equals to 60^{o}

∴△OPA is an equilateral triangle

∴OP = OA = PA

∠OPT = 90^{o}

Now, ∠OPA + ∠APT = 90^{o}

=) 60^{o} + ∠APT = 90^{o}

∴∠APT = 30^{o}

Again, ∠PAB + ∠PAT = 180^{o}

=) 60^{o} + ∠PAT = 180^{o}

=) ∠PAT = 120^{o}

In △APT

∠APT + ∠PAT + ∠PTA = 180^{O}

=) 30^{O} + 120 + ∠PTA = 180^{O}

=) ∠PTA = 30^{O}

∴∠APT = ∠PTA = 30^{O}

∴AT = PA

IN △ABP

Sin ∠ABP = AP/BA

=) Sin30^{O} = AT/BA [AT = PA]

=) ½ = AT/BA

∴BA: AT = 2:1