Question

If PQ and PR ARE TANGENTS DRAWN FROM AN EXTERNAL POINT P WITH CENTRE O ARE INCLINED AT 90degree if length of each tangent is 5cm find radius of circle

Answer 1

in ∆PQR

(PR)^2 =(PQ)^2 +(PR)^2

(PR) =5√2

anglePOR =180-angleRPQ (angles are supplementry )

anglePOR=90°

in ∆POR

(PO)^2 + (RO)^2 =(PR)^2

2(PO)^2 =(5√2)^2

2(PO)^2. =50

(PO) =5√2/√2

PO=5cm

Answer 2

Given :

PQ and PR ARE TANGENTS DRAWN FROM AN EXTERNAL POINT P

Angle between PA & PR is 90°

PQ = PR = 5 cm

O is centre of circle

To Find :

Radius of circe

Solution :

•In triangle POQ & triangle POR

PQ = PR ( Given )

PO = PO ( common in both )

OQ = OR ( Radii of same circle )

=> Triangle POQ is congruent to Triangle POR

=> <QPO = <RPO ( CPCT)

also <QPR = 90° ( Given )

hence , <QPR = 2<QPO

<QPO = 45 °

•In Triangle PQO

PQ = 5 cm

<QPO = 45 °

•Using trigonometry

tan(<QPO) = OQ/PQ

tan45° = OQ/5

OQ = 5 cm

OQ is radius of circle

•Hence , radius of circle is 5 cm