Question

In the adjoining figure, O is the centre of the circle. If PQ = 24 cm and PR = 7 cm, find the area of the shaded region.

Answer 1

<QPR =90°......theorem

∆PQR is a equilateral triangle.

by PGT

QR = 25 cm

area of ∆ PQR = 1/2*24*7

= 12*7

= 84 cm^2

area of semicircle = 1/2*πr^2

=1/2*22/7*25/2*25/2

$= \frac{25 \times 22 \times 25 }{2 \times 2 \times 7} \\ = \frac{11 \times 625}{14} \\ = 491.07$

shaded area = 491.07-84

= 407.07 cm^2

hope it helps you

Answer 2

Explanation:

Hope it helps you ...............