Question

Let $AB$ be a line segment of length $2$. Construct a semicircle $S$ with $AB$ as diameter. Let $C$ be the midpoint of the arc $AB$. Construct another semicircle $T$ external to the triangle $ABC$ with chord $AC$ as diameter. The area of the region inside the semicircle $T$ but outside $S$ is

• A. $\frac{\pi }{2}$
• B. $\frac{1}{2}$
• C. $\frac{\pi }{\sqrt{2}}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B $\frac{1}{2}$

As C is the mid point of arc ACB,

Arc length $AC$=$CB$ and radius is same for both arc,So

$\angle CAD=\angle CBD$

It is also clear that $\angle ACB={90}^o$ as angle on circumference of half-circle.

So $\angle CAD=\angle CBD={45}^o$

it means $AD=DC=DB=1$

$A{C}^2=A{D}^2+D{C}^2=1+1=2$

$AC=\sqrt{2}$ which is the diameter of arc T.

Area of shaded region $=$ area of semicircle $T$$-$area of $\Delta ADC$

$=\frac{1}{2}\Pi (\frac{AC}{2}{)}^2-\frac{1}{2}AD\times DC$

$=\frac{1}{2}\Pi (\frac{\sqrt{2}}{2}{)}^2-\frac{1}{2}1\times 1$

$=\frac{\pi }{4}-\frac{1}{2}$

The area of the region inside the semicircle TT but outside SS is

$=$ area of semicircle $T$$-$area of shaded region

$=\frac{\pi }{4}-(\frac{\pi }{4}-\frac{1}{2})$

$=\frac{1}{2}$

Option B is correct.