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

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 = 180O [By angle sum property]

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

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

=) 2∠PAB = 130O

=) ∠PAB = 130O/2 = 65O

Here, AP is a tangent to the circle

∴∠OAP = 90O

∠PAB + ∠OAB = 90O

=) 65O + ∠OAB = 90O

=) 65 + ∠OAB = 90

=) ∠OAB = 90O – 65O = 25O

(3)

Join OT and OQ

OT = OQ [radius]

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

Now, in quadrilateral OQPT

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

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

=) ∠QOT = 360o – 250o = 110o

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 110o

= 55o

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

∴ EA = EC for the circle having center O1 and ED = EB for the circle having center O2

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 = 90o

Now, ∠OPQ = ∠OPT – ∠QPT

=) ∠OPG = 90o – 70o

∴∠OPQ = 20o

OP = OQ [Radii of the same circle]

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

Now, in △POQ,

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

=) ∠POQ = 180O – 20O – 20O = 140O

∴∠POQ = 140O

(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

OA2 = OC2 + AC2

=) AC2 = OA2 – OC2

=) AC2 = 52 – 32

=) AC2 = 252 – 92

=) AC2 = 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 = 90o

And ∠OQB = 90O

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 = 120O

=) ∠PRO = 60O

In △POR,

Cos60o = 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 = 90o

Now, in quadrilateral AOBP,

∠PAO + ∠PBO + ∠AOB + ∠APB = 360

=) 90o + 90o + ∠AOB + ∠APB = 360o

=) ∠AOB + ∠APB = 360o – 90o – 90o = 180o

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

∴AOBP is a cycle quadrilateral

(12)

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

∴∠OPA = 90o

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,

AO2 = OP2 + PA2

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

=) 25 = OP2 + 16

=) OP2 = 25 – 16

=) OP2 = 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 = 90o

Now, ∠OPQ = ∠OPT – ∠QPT

= 90o – 60o

= 30o

OP = OQ [radii of the same circle]

∠OPQ = ∠OPQ = 30o

In △OQP

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

=) 30o + 30o + ∠POQ = 180o

=) ∠POQ = 180o – 30o – 30o

∴∠POQ = 120o

Now, ∠POQ + reflex∠POQ = 360o

=) reflex∠POQ = 360o – POQ

= 360o – 120o

= 240o

∴reflex∠POQ = 240o

∴∠PRQ = ½ X reflex angle POQ

= ½ X 240o

= 120o

(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 = 180o [angle sum property of a triangle]

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

=) 2∠PAB = 180o – 60o

=) ∠PAB = 120o/2 = 60o

Here, PA is a tangent,

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

∴ ∠OAP = 90o

Now, ∠OAB = ∠OAP – ∠PAB

=) ∠OAB = 90o – 60o

∴ ∠OAB = 30o


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