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Question

O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O a point P is taken. From this point, two tangent PQ and PR are drawn to the circle. Then, the area of quad. PQOR is 

(a) 60 cm2   (b) 32.5 cm2  (c) 65 cm2  (d) 30 cm2


Solution

GIVEN:

PQ & PR are 2 tangents and QO & OR are 2 radius at contact point Q & R.

Angle PQO=90°

[A TANGENT TO A CIRCLE IS PERPENDICULAR TO THE RADIUS THROUGH THE POINT OF CONTACT]

By Pythagoras theorem

PQ²= OP² - OQ²
PQ² = 13²- 5² = 169- 25= 144
PQ= √ 144= 12
PQ=12cm

PQ= PR =12cm
[The Lengths of two tangents drawn from an external point to a circle are equal]

In ∆OPQ & ∆ OPR
OQ= OR (5cm) given
OP = OP ( Common)
PQ= PR( 12cm)

Hence ∆OPQ =~ ∆OPR ( by SSS congruence)

Area of ∆OPQ =Area ∆OPR

Area of quadrilateral QORP= 2×(area of ∆ OPR)

Area of quadrilateral QORP= 2× 1/2 × base × altitude

Area of quadrilateral QORP= OR× PR

Area of quadrilateral QORP=12× 5= 60 cm²
_____________________________
Area of quadrilateral QORP=60cm²
___________________________

 

Mathematics
Secondary School Mathematics X
Standard X

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