**RS Aggarwal Class 10 Math Twelfth**** Chapter Circles**** Exercise 12B**** Solution**

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

**(1)** We know that the lengths of tangents drawn from an external point to a circle are equal.

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

Now, AD = AS + DS

= AP +DR

= (AP – PB) + (CD – CR)

= (6 – PB) + (8 – CR)

= 6 – PB + 8 – CR

= 14 – BQ – CQ [∵BP = BQ and CR = CQ]

= 14 – (BQ + CQ)

= 14 – BC

= 14 – 9

= 5

∴The length of the side AD is 5 cm

**(2)** We know that tangents from an external point to a circle are equal.

∴PA = PB

∴∠PAB = ∠PBA

Now, in △ABP

∠PAB + ∠PBA + ∠APB = 180^{O} [By angle sum property]

=) ∠PAB + ∠PBA + 50^{O} = 180^{O} [∵APB = 50(given)]

=) ∠PAB + ∠PAB = 180^{O} – 50^{O} [∵PAB = PBA]

=) 2∠PAB = 130^{O}

=) ∠PAB = 130^{O}/2 = 65^{O}

Here, AP is a tangent to the circle

∴∠OAP = 90^{O}

∠PAB + ∠OAB = 90^{O}

=) 65^{O} + ∠OAB = 90^{O}

=) 65 + ∠OAB = 90

=) ∠OAB = 90^{O} – 65^{O} = 25^{O}

**(3)**

Join OT and OQ

OT = OQ [radius]

∠OTP = 90^{o} and ∠OQP = 90^{o} [∵PT and QP are tangents]

Now, in quadrilateral OQPT

∠QOT + ∠OTP + ∠OQP + ∠TPQ = 360^{o} [angle sum property of a quadrilateral]

=) ∠QOT + 90^{o} + 90^{o} + 70^{o} = 360^{o} [∠TPQ = 70^{o} (given)]

=) ∠QOT = 360^{o} – 250^{o} = 110^{o}

We knew that the angle subtended by a one as the centre is double the angle subtended by the one at any point on the remaining point of the circle.

∴∠TRQ – ½ QOT

= ½ X 110^{o}

= 55^{o}

**(4) **We know that tangents down from an external point to a circle are equal

∴ EA = EC for the circle having center O_{1} and ED = EB for the circle having center O_{2 }

Adding two equation and we get,

EA + ED = EC + EB

=) EA + EB = EC + ED [∵ED = EB]

=) AB = CD [Proved]

**(5)** We know that the radius and tangent are perpendicular at their point of contact

∴∠OPT = 90^{o}

Now, ∠OPQ = ∠OPT – ∠QPT

=) ∠OPG = 90^{o} – 70^{o}

∴∠OPQ = 20^{o}

OP = OQ [Radii of the same circle]

∴∠OPQ = ∠OQP = 20^{o} [angle opposite equal sides are equal]

Now, in △POQ,

∠POQ + ∠OPQ + ∠OQP = 180^{O} [angle sum property of a triangle]

=) ∠POQ = 180^{O} – 20^{O} – 20^{O} = 140^{O}

∴∠POQ = 140^{O }

**(6)**

Join, OA, OB, OC and draw OE⊥AB at E and OF⊥AC at F

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

Now, we have

AE = AF = x

BD = BE = 4cm and CD = CF = 3 cm

Now,

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

=) 21 = ½ x BC x OD +½ x AB x OE x ½ x AC x OF

= 21 = ½ x 7 x 2 + ½ (AE + BE) x OE + ½ (AE + CP) x OF

=) 21 = 7 + ½ (x + 4) x 2 + ½ (x + 3) x 2

=) 21 = 7 + x + 4 +x + 3

=) 21 = 14 + 2x

=) 2x = 21 – 14 = 7

∴x = 3.5 cm

∴AB = AE + BE

= x+ 4

= 3.5 + 4

= 7.5 cm

And AC = AF + CF

= x + 3

= 3.5 + 3

= 6.5 cm

**(7)**

Given, two circles with center O. AB be chord of the larger circle which touches the circle which touches the smaller circle at C.

OA = 5cm an OC = 6cm

In △AOC

OA^{2} = OC^{2} + AC^{2}

=) AC^{2} = OA^{2} – OC^{2}

=) AC^{2} = 5^{2} – 3^{2}

=) AC^{2} = 25^{2} – 9^{2}

=) AC^{2} = 16

=) AC =Ö16

=) AC = 4 cm

We know, the perpendicular drawn from the centre to a chord of a circle bisects the chord

∴AB = 2AC

= 2 x 4

= 8 cm

∴The length of the chord of the larger circle is 8 cm

**(8)**

Given, O is the centre of the given circle tangent AB intersecting circle at the point Q

We know that tangent of circle is perpendicular towards at point of contact

PQ⊥AB

∠PQB = 90^{o}

And ∠OQB = 90^{O}

Above case is possible only when centre O lies on the line PQ

∴The perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

**(9)**

Join PO and QO

In △POR and △QOR

OQ = OP [Radii]

RP = RQ [Tangents from an external point are equal]

OR = OR [common]

∴△POR ≅ △QOR [By sss congruency]

∠PRO = ∠QRO = (CPC+)

Now, ∠PRO + ∠QRO = ∠PRQ

=) ∠PRO + ∠PRO = ∠PRQ [∵∠PRO = ∠QRO]

=) 2∠PRO = 120^{O}

=) ∠PRO = 60^{O}

In △POR,

Cos60^{o} = PR/OR

=) ½ = PR/OR

=) OR =2PR

=) OR = PR + PR

=) OR = PR + RQ [∵PR = RQ]

**(10)** We know that tangents from an external point to the circle are equal

AD = AF

BD = BE

And CE = CF

Given, AB = 12 cm

=) AD + BD = 12cm —- (i)

BC = 8 cm

=) BD + FC = 8 cm ——- (ii)

And AC = 10 cm

=) AF + CF = 10cm

=) AD + CF = 10 cm ——- (iii)

Adding equation (i), (ii) and (iii) we get

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

=) 2AD + 2BD + 2FC = 30

=) 2 (AD + BD + FC) = 30

=) AD + BD + FC = 15 —— (iv)

Solving (i) and (iv) we get

AD + BD + FC = 15

=) 12 + FC = 15

=) FC = 3

Solving (ii) and (iv) we get

AD + BD + FC = 15

=) AD + 8 = 15

=) AD = 7

And solving (iii) and (iv) we get

AD + BD + FC = 15

=) BD + 10 = 15

=) BD = 5

∴AD = 7cm

BD = BE = 5cm

and CF = 3cm

**(11)** We know, the radius and tangent are perpendicular at their point of contact

∠PAO = ∠PBO = 90^{o}

Now, in quadrilateral AOBP,

∠PAO + ∠PBO + ∠AOB + ∠APB = 360

=) 90^{o} + 90^{o} + ∠AOB + ∠APB = 360^{o}

=) ∠AOB + ∠APB = 360^{o} – 90^{o} – 90^{o} = 180^{o}

Since, the sum of the opposite angels of two quadrilaterals AOBP is 180^{o}

∴AOBP is a cycle quadrilateral

**(12)**

We know, the radius and tangent are perpendicular all their point of contact

∴∠OPA = 90^{o}

AB is a chord of the outer circle. We know that the perpendicular drown form the centre to a chord of a circle be sects the chord.

∴AP = PB = AB/2 = 8/2 = 4 cm

In right angle triangle AOP,

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

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

=) 25 = OP^{2} + 16

=) OP^{2} = 25 – 16

=) OP^{2} = 9

∴OP = 3 cm

∴The radius of the smaller circle is 3 cm

**(13)** We know, the radius and tangent are perpendicular at their point of contact.

∴∠OPT = 90^{o}

Now, ∠OPQ = ∠OPT – ∠QPT

= 90^{o} – 60^{o}

= 30^{o}

OP = OQ [radii of the same circle]

∠OPQ = ∠OPQ = 30^{o}

In △OQP

∠OPQ + ∠OQP + ∠POQ = 180^{o} [angle sum property of a triangle]

=) 30^{o} + 30^{o} + ∠POQ = 180^{o}

=) ∠POQ = 180^{o} – 30^{o} – 30^{o}

∴∠POQ = 120^{o}

Now, ∠POQ + reflex∠POQ = 360^{o}

=) reflex∠POQ = 360^{o} – POQ

= 360^{o} – 120^{o}

= 240^{o}

∴reflex∠POQ = 240^{o}

∴∠PRQ = ½ X reflex angle POQ

= ½ X 240^{o}

= 120^{o}

**(14)** We know that tangents from an external point to a circle equal.

∴ PA = PB

∠PAB = ∠PBA [angles opposite sides are equal]

Now, In PAB

∠APB +∠PAB +∠PBA = 180^{o} [angle sum property of a triangle]

=) 60^{o} + ∠PAB + ∠PAB = 180^{o} [∵∠PAB = ∠PBA]

=) 2∠PAB = 180^{o} – 60^{o}

=) ∠PAB = 120^{o}/2 = 60^{o}

Here, PA is a tangent,

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

∴ ∠OAP = 90^{o}

Now, ∠OAB = ∠OAP – ∠PAB

=) ∠OAB = 90^{o} – 60^{o}

∴ ∠OAB = 30^{o}