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