RD Chapter 10 Circles Ex10.2 Solutions
Question - 41 : - AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in the figure. Prove that ∠BAT = ∠ACB. [NCERT Exemplar]
Answer - 41 : -
Since, AC is a diameter line, so angle in semicircle makes an angle 90°.
∠ABC = 90° [by property]
In ∆ABC,
∠CAB + ∠ABC + ∠ACB = 180°
[ sum of all interior angles of any triangle is 180°]
=> ∠CAB + ∠ACB = 180° – 90° = 90° ……….(i)
Since, diameter of a circle is perpendicular to the tangent.
i.e. CA ⊥ AT
∠CAT = 90°
=> ∠CAB + ∠BAT = 90° …….(ii)
From Eqs. (i) and (ii),
∠CAB + ∠ACB = ∠CAB + ∠BAT
=> ∠ACB = ∠BAT
Hence proved.
Question - 42 : - In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ∆ABC is 84 cm². [CBSE 2015]
Answer - 42 : -
In the given figure,
In ∆ABC is circle is inscribed touching it at D, E and F respectively.
Radius of the circle (r) = 4cm
OD ⊥ BC, then
OD = 4 cm, BD = 8 cm, DC = 6 cm
Join OE and OF
Question - 43 : - In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ. [CBSE 2015]
Answer - 43 : -
In the given figure,
AB is the diameter, AT is the tangent
and ∠AOQ = 58°
To find ∠ATQ
Arc AQ subtends ∠AOQ at the centre and ∠ABQ at the remaining part of the circle
∠ABQ = 12 ∠AOQ = 12 x 58° = 29°
Now in ∆ABT,
∠BAT = 90° ( OA ⊥ AT)
∠ABT + ∠ATB = 90°
=> ∠ABT + ∠ATQ = 90°
=> 29° + ∠ATQ = 90°
=> ∠ATQ = 90°- 29° = 61°
Question - 44 : - In the figure, OQ : PQ = 3:4 and perimeter of ∆POQ = 60 cm. Determine PQ, QR and OP.
Answer - 44 : -
In the figure, OQ : PQ = 3:4
Perimeter of ∆POQ = 60 cm
To find PQ, QR and OP
OQ : PQ = 3 : 4
Let OQ = 3x and PQ = 4x
Now in right ∆OPQ,
OP² = OQ² + PQ² = (3x)² + (4x)² = 9x² + 16x² = 25x² = (5x)²
OP = 5x
But OQ + QP + OP = 60 cm
3x + 4x + 5x = 60
=> 12x = 60
x = 5
PQ = 4x = 4 x 5 = 20 cm
QR = 2 OQ = 2 x 3x = 6 x 5 = 30 cm
OP = 5x = 5 x 5 = 25 cm
Question - 45 : - Equal circles with centre O and O’ touch each other at X. OO’ produced to meet a circle with centre O’, at A. AC is a tangent to the circle whose centre is O. O’ D is perpendicular to AC. Find the value of DO′/CO
Answer - 45 : -
Two equal circles with centre O and O’ touch each other externally at X
OO’ produced to meet at A
AC is the tangent of circle with centre O,
O’D ⊥ AC is drawn OC is joined
AC is tangent and OC is the radius
OC ⊥ AC
O’D ⊥ AC
OC || O’D
Now O’A = 1/2 A x or 1/2 AO
Now in O’AD and AOAC
∠A = ∠A (common)
∠AO’D = ∠AOC (corresponding angles)
Question - 46 : - In the figure, BC is a tangent to the circle with centre O. OE bisects AP. Prove that ∆AEO ~ ∆ABC.
Answer - 46 : -
Given : In the figure, BC is a tangent to the circle with centre O at B.
AB is diameter AC is joined which intersects the circle at P
OE bisects AP
To prove : ∆AEO ~ ∆ABC
Proof: In ∆OAE and ∆OPE
OE = OE (common)
OA = OP ‘ (radii of the same circle)
EA = EP (given)
∆OAE = ∆OPE (SSS axiom)
∠OEA = ∠OEP
But ∠OEA + ∠OEP = 180°
∠OEA = 90°
Now in ∆AEO and ∆ABC
∠OEA = ∠ABC (each 90°)
∠A = ∠A (common)
∆AEO ~ ∆ABC (AA axiom)
Hence proved.
Question - 47 : - In the figure, PO ⊥ QO. The tangents to the circle at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.
Answer - 47 : -
Given : In the figure, O is the centre of the circle
PO ⊥ QO
They tangents at P and Q intersect each other at T
To prove : PQ and OT are right bisector of each other
Proof : PT and QT are tangents to the circle
PT = QT
OP and OQ are radii of the circle and ∠POQ = 90° ( PO ⊥ QO)
OQTP is a square Where PQ and OT are diagonals
Diagonals of a square bisect each other at right angles
PQ and OT bisect each other at right angles
Hence PQ and QT are right bisectors of each other.
Question - 48 : - In the figure, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°.
Answer - 48 : -
Given : In the figure, O is the centre of the circle BCD is a tangent, CP is a chord
prove : ∠BAC + ∠ACD = 90°
Proof: ∠ACD = ∠CPA (Angles in the alternate segment)
But in ∆ACP,
∠ACP = 90° (Angle in a semicircle)
∠PAC + ∠CPA = 90°
=> ∠BAC + ∠ACD = 90°
(∠ACD = ∠CPA proved)
Hence proved.
Question - 49 : - Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines. [NCERT Exemplar]
Answer - 49 : -
Given : Two tangents PQ and PR are drawn from an external point P to a circle with centre O.
To prove : Centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
Construction : Join OR, and OQ.
In ∆POR and ∆POQ
∠PRO = ∠PQO = 90°
[tangent at any point of a circle is perpendicular to the radius through the point of contact]
OR = OQ [radii of same circle]
Since, OP is common.
∆PRO = ∆PQO [RHS]
Hence, ∠RPO = ∠QPO [by CPCT]
Thus, O lies on angle bisector of PR and PQ.
Hence proved.
Question - 50 : - In the figure, there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the lengths of PS. [CBSE 2017]
Answer - 50 : -
Construction : Join OS and OP.
Consider ∆POS
We have,
PO = OS
∆POS is an isosceles triangle.
We know that in an isosceles triangle, if a line drawn perpendicular to the base of the triangle from the common vertex of the equal sides, then that line will bisect the base (unequal side).
And PQ = PR = 5 cm
[PRT and PQS are tangents to the inner circle to the inner circle from a point P lying on the outer circle]
We have, PQ = QS
It is given that, PQ = 5 cm
QS = 5 cm
From the figure, we have
PS = PQ + QS
=> PS = 5 + 5
=> PS = 10 cm