Question

In the given figure, $ACB$ is the largest triangle that can be inscribed in a semicircle of radius $r$. If two semicircles are drawn considering $AC$ and $BC$ as diameters, then the area of the shaded region is

• A. $\frac{1}{2}{r}^2$
• B. $\frac{1}{2}{r}^2(\pi -2)$
• C. $r^2$
• D. $r^2(\pi -2)$

Answer 1

The correct option is C $r^2$

Given that $\Delta ACB$ is the largest triangle that can be inscribed in a semicircle of radius $r$.
$\therefore \angle ACB={90}^{\circ }$ and $AC=BC!$ (As $\Delta ACB$ is the largest triangle


Join $C$ to the mid-point of $AB$, i.e. at $O$.
$\therefore OC=OA=OB$ is the radius of the semicircle.
$\Rightarrow \Delta AOC\cong \Delta BOC$ (By SSS congruency rule)
$\Rightarrow \angle AOC=\angle BOC$ (By CPCT)
As, $\angle AOC+\angle BOC={180}^{\circ }$ (Linear pair)
$\Rightarrow \angle AOC=\angle BOC={90}^{\circ }$
Consider $OA=OB=OC=r$

$\therefore A{C}^2=A{O}^2+O{C}^2$
(By Pythagoras Theorem)
$=r^2+r^2$
$\Rightarrow AC=r\sqrt{2}$
Thus, Area of semicircle with diameter
$AC=\frac{1}{2}\times \pi \times {\left(\frac{AC}{2}\right)}^2$
$=\frac{\pi }{2}\times {\left(\frac{r\sqrt{2}}{2}\right)}^2$
$=\frac{\pi r^2}{4}$

Similarly, area of semicircle with diameter $BC=\frac{\pi r^2}{4}$ $(\because AC=BC)$

Area of right triangle $ACB=\frac{1}{2}\times AB\times OC$
$=\frac{1}{2}\times 2r\times r$ $(\because AO=OB=OC=\frac{AB}{2}=r)$
$=r^2$
Area of semicircle with $AB$ as diameter $=\frac{1}{2}\times \pi \times {\left(\frac{AB}{2}\right)}^2$
$=\frac{\pi }{2}\times {\left(\frac{2r}{2}\right)}^2=\frac{\pi r^2}{2}$
$\therefore$ Area of shaded region = Area of semicircles with diameter $AC$ and $CB+$ Area of right $\Delta ACB-$ Area of semicircle with $AB$ as diameter
$=\frac{\pi r^2}{4}\times 2+r^2-\frac{\pi r^2}{2}$
$=r^2$
Hence, the correct answer is option c.