Question

In figure $ABC$ is a right-angled triangle right-angled at $A$. Semicircle are drawn on $AB,ACandBC$ as diameters. Find the area of the shaded region.

• A. $6$sq. unit
• B. $9$sq. unit
• C. $8$sq. unit
• D. $5$sq. unit

Answer 1

The correct option is A $6$sq. unit

Area of shaded region = area of semicircle $AB$ + area of semi-circle $AC$ - (area of semicircle on side $BC$) + area of $\Delta ABC$

Area of semicircle AB = $\frac{\pi r^2}{2}=\frac{22}{7\times 2}\times \frac{3}{2}\times \frac{3}{2}=3.54$ sq. units

Area of semicircle AC = $\frac{\pi r^2}{2}=\frac{22}{7\times 2}\times 2\times 2=6.28$ sq. units

Area of semicircle BC = $\frac{\pi r^2}{2}=\frac{\pi }{2}\times {\left(\frac{BC}{2}\right)}^2$

Now, $BC=\sqrt{4^2+3^2}=\sqrt{25}=5$

$\therefore$ Area of semi-circle BC = $\frac{22}{7\times 2}\times {\left(\frac{5}{2}\right)}^2=9.82$ sq. units.

Area $\Delta ABC=\frac{1}{2}\times AB\times AC=\frac{1}{2}\times 3\times 4=6$ sq. units

Therefore, area of shaded region$=$ $3.54+6.28-9.82+6$

$=$ $6$ sq. units