Question

Please any give the

Answer 1

Given , PQR is an isosceles triangle inscribed in a circle . if PQ = PR = 25 cm, and QR = 14 cm.
then calculate the radius of the circle to the nearest cm.

solution :
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let r be the required radius of circle and D be the mid point on QR.

since, QR = 14 cm and PD is perpendicular to QR.

then ,QD = QR /2 = 14 / 2 = 7 cm

QD = DR = 7 cm

from ∆ PDR ,
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by Pythagoras theorm , we get

(PR)^2 = ( DR )^2 + ( PD )^2

(25 )^2 = ( 7 )^2 +( PD)^2

(PD)^2 = 625 - 49 = 576 cm

PD = √576 = 24 cm

OP = OR = r ( radius )

since, PD = 24 cm, and OP = r

then, OD = PD - OP = 24 - r

Solve for ' r ' :
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now in ∆ ODR ,

by Pythagoras theorem, we get

(OR )^2 = ( DR )^2 + ( OD )^2

r^2 = ( 7)^2 + ( 24 - r )^2

r^2 = 49 + 576 + r^2 - 48r

r^2 - r^2 = 625 - 48r

0 = 625 - 48r

48r = 625

r = 13.02 cm

therefore, radius of circle = 13.02 cm

Answer : radius = 13.02 cm

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