Question

Find the area of the shaded region in figure ,if PQ=24 cm , PR= 7 cm and O is the centre of the circle.

• A. $100.98c{m}^2$
• B. $161.54c{m}^2$
• C. $101.54c{m}^2$
• D. None of these

Answer 1

Answer: (B)

$161.54c{m}^2$

PQ$=24cm$ ,PR $=7$ cm

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

$\therefore \angle RPQ=90°$

In right angled $\Delta RPQ$

$R{Q}^2=P{Q}^2+P{R}^2$ [By pythagoras theorem]

$RQ²=24²+7²$

$RQ²=576+49$

$RQ²=625$

$RQ=\surd 625cm$

$RQ=25cm$

radius of the circle $\left(OQ\right)=\frac{RQ}{2}=\frac{25}{2}cm$

Area of right $\Delta RPQ=\frac{1}{2}\times Base\times height$

Area of right $\Delta RPQ=\frac{1}{2}\times RP\times PQ$

Area of right $\Delta RPQ=\frac{1}{2}\times 7\times 24=7\times 12=84cm²$

Area of right $\Delta RPQ=84cm²$

Area of semicircle$=\frac{\pi r²}{2}$

$=\frac{22}{7}\times \frac{25}{2}\times \frac{25}{2}\times \frac{1}{2}$

$=\frac{11\times 25\times 25}{28}$

$=\frac{6875}{28}c{m}^2$

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

$=\frac{6875}{28}-84$

$=\frac{6875-2352}{28}$

$=\frac{4523}{28}=161.54c{m}^2$

Hence, the area of the shaded region = 161.54 cm²