Question

If 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 tangents pq and pr are drawn to the circle. then , the area of quadrilateral pqor is

Answer 1

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
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Area of quadrilateral QORP=60cm²
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Hope this will help you.. .

Answer 2

IVEN:

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