Find the area of segments shaded in the figure, if PQ = 24cm, PR = 7cm, and QR is the diameter of the circle with center O. [Take $\mathrm{\pi }=\frac{22}{7}$]

Solution:

PQ=24cm, PR =7 cm

We know that any angle made by the diameter QR in the semicircle is 90°.

∴∠RPQ=90°

In right-angled ∆RPQ

RQ² = PQ² + PR² [By Pythagoras theorem]

RQ²=24²+7²

RQ²=576+49  

RQ²=625

RQ=√625cm

RQ=25cm

The radius of the circle (OQ)= RQ/2 ​= 25/2 = 12.5​cm

Area of right ∆RPQ=1/2​×RP×PQ

Area of right ∆RPQ=1/2​×7×24=7×12 = 84cm².

Area of semicircle=πr²/2 

=​ ​$\frac{22}{7}\times \frac{25}{2}\times \frac{25}{2}\times \frac{1}{2} = \frac{6875}{28} {\mathrm{cm}}^2$

Area of the shaded region = Area of the semicircle - Area of right ∆ RPQ

= $\frac{6875}{28}-84 =\frac{6875-2352}{28}=161.54{\mathrm{cm}}^2$

Hence, the area of the shaded region = 161.54 cm².