1.In the figure PQ, PR and BC are the tangents to the circle BC touches the circles at X. If PQ = 7cm find the perimeter of ∆PBC
Solution:
The perimeter of ∆PBC = PC + PB + BC
= PC + PB + BX + CX
But Cx = CR and BX = BQ [Since, tangents drawn from an external point]
Therefore, PC + PB + BQ + CR = (PC + CR) + (PB + BQ)
= PR + PQ
= 7 + 7 [Since PR = PQ tangents drawn from an external point]
= 14 cm
- Two concentric circles of radii 13 cm and 5 cm are drawn. find the length of the chord of the outer circle which touches the outer circle.
Solution:
Given, two concentric circles of radii 13 cm and 5 cm. We have to find the length of the chord of the outer circle which touches the outer circle.
Let O be the centre of the circles. AB is the tangent drawn to an inner circle through the point P.
We have AP = PB [Since the perpendicular drawn from the point of contact devices the chord equally]
In ∆OAP, ∠OPA = 90˚
AP2 + OP2 = OA2 [ Pythagoras Theorem]
AP2 + 52 = 132
AP2 + 25 = 169
AP2 = 169 – 25 = 144
AP = 12 cm
Therefore, AB = AP + PB = 12 + 12 = 24 cm
In the given ∆ABC AB = 12cm, BC = 8 cm and AC = 10 cm Find AF . BD and CE
Solution:
In ∆ABC,
AB = 12cm ; BC = 8cm ; AC = 10 cm
AB + BC + CA = 12 + 8 + 10 = 30
AD + BD + BE + CE + AF + CF = 30
But AF = AD = x [ Tangents drawn from an external point]
BE = BD = y [ Tangents drawn from an external point]
CE = CF = z[ Tangents drawn from an external point]
x + y + y + z + z + x = 30
2x + 2y + 2z = 30
x + y + z = 15 …………..(1)
AB = x + y = 12 cm ; BC = x + y = 8cm ; AC = x + z = 10cm
From (1), 12 + z = 15
⇒ z = 15 – 12 = 3 cm
From (1), x + 8 = 15
⇒ x = 15 – 8 = 7 cm
From (1), y + 10 = 15
⇒ y = 15 – 10 = 5 cm
AF = x = 10 cm
BD = y = 5 cm
CE = z = 3cm
In the given quadrilateral ABCD, BC = 38 cm, QB = 27 cm , DC = 25 cm and AD⊥DC find the radius of the circle.
Solution:
In the figure OPDS,
DS = DP [Tangents drawn from an external point]
OP = OS [ Radius of circle]
∠D = 90˚ [AD⊥DC]
Therefore, OPDS is a square.
In the figure,
BQ = BR = 27 cm [Tangents drawn from an external point]
CR = 38 – 27 = 11 cm = CS [Tangents drawn from an external point]
DS = 25 – 11 = 14 cm = DP [Tangents drawn from an external point]
Radius of the circle = OP = OS = 14 cm [OPDS is a square]
In the given figure AB = BC, ∠ABC = 68˚. DA and DB are the tangents to the circle with centre are the tangents to the circle with centre O. Calculate the measure of
(i) ∠ACB
(ii) ∠AOB
(iii) ∠ADB
Solution:
In the figure, AB = BC, ∠ABC = 68˚
(i) ∠ABC + ∠ACB + ∠BAC = 180˚
∠ACB + ∠BAC = 180˚ – ∠ABC
∠ACB + ∠ACB = 180˚ – ∠ABC [AB = BC]
2∠ACB = 180˚ – 68˚
∠ACB = 180˚ – 68˚/2 = 112˚/2 = 56˚
(ii) ∠AOB = 2∠ACB = 2 x 56˚ = 112˚
(iii) ∠ADB = 180˚ – ∠AOB
= 180˚ – 112˚
= 68˚
2. Riders based on tangent properties:
1. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD +BC.
Solution:
AP = AS = l
BS = BR = m
CR = CQ = n
DP = DQ = o
AB + CD = AS + SB + CQ + QD = l + m + n + o …………….(1)
AD + BC = AP + PD + BR + RC
l + m + n + o = l + o + m + n ……………(2)
From (1) and (2),
AB + CD = AD + BC
Tangents AP and AQ are drawn to circle with centre O, from an external point A . Prove that ∠PAQ = 2. ∠OPQ
Solution:
∠POQ + ∠PAQ = 180˚ [Angle between tangents + angle in the centre = 180˚] …………(1)
In ∆POQ , ∠POQ + ∠OPQ + ∠OQP = 180˚ [Sum of the angles of a triangle = 180˚]
Therefore,
∠POQ + 2∠OPQ = 180˚ [ Since ∠OPQ = ∠OQP] ………(2)
From (1) and (@)
∠POQ + ∠PAQ = ∠POQ + 2∠OPQ
∠PAQ = 2∠OPQ
In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that,
(a) tangent at P bisects AB at Q
(b) ∠APB = 90˚
Solution:
(i)
In the figure
QA = QP [tangents drawn from an external point] ………..(1)
QB = QP [tangents drawn from an external point] ………..(2)
From (1) and (2),
QA = QB
Therefore, P bisects AB at Q.
(ii)
In ∆APB,
∠QAP = ∠APQ = x [QA = QP]
∠QPB = ∠BPQ = y [QB = QP]
Therefore, In ∆APB x + x + y + y = 180˚ [Sum of the angles of a triangle]
2x + 2y = 180˚
x + y = 90˚
∠APB = 90˚
A pair of perpendicular tangents are drawn to a circle from an external point. Prove that length of each tangent is equal to the radius of the circle.
Solution:
PA = PB [tangents drawn from an external point]
OA = OB {Radius of the circle]
∠APB = 90˚ [Given]
∠OAP = ∠OBP = 90˚ [The sum of a quadrilateral = 360˚]
Therefore, OABP is a square
Hence, tangents PA and PB = Radius of the circle OA and OB
If the sides of a parallelogram touch a circle. Prove that the parallelogram is a rhombus.
Solution:
Let ABCD be a parallelogram
Therefore, AB||CD, AB = CD [Given]
AD||BC , AD = BC[Given]
AP = AS, PB = BQ [tangents drawn from an external point]
DS = DR, QC = RC
AB + CD = AP + PB + CR + DR
AB + CD = AS + BQ + QC + DS
In the figure if AB = AC prove that BQ = QC
Solution:
AP = AR …………….(1) [tangents drawn from an external point]
and AB = AC ……….(2)[Given]
From (1) and (2),
AB – AP = AC – AR
BP = CR
But BQ = BP and CQ = CR [tangents drawn from an external point]
Therefore, BQ = CQ.
