Question

Find the area of the shaded region in the given figure $BC=BD=8cm$, $AC=AD=15cm$ and $O$ is the centre of the circle (Take $\pi =3.14)$

Answer 1

$\angle ADB=\angle ACB={90}^o$ [ Angle inscribe in a semicircle ]

$△ADB$ and $△ACB$ are right angled triangles.

$ar\left(△ADB\right)=\frac{1}{2}\times BD\times AD$

$=\frac{1}{2}\times 8\times 15$

$=60c{m}^2$

Similarly, $are\left(△ACB\right)=60c{m}^2$

In $△ADB,$

$\Rightarrow$ $AB=\sqrt{{15}^2+8^2}$ [ By Pythagoras theorem ]

$\Rightarrow$ $AB=\sqrt{289}$

$\Rightarrow$ $AB=17cm$

Now, $AB$ is diameter of a circle.

$\Rightarrow$ Radius $=\frac{AB}{2}=\frac{17}{2}cm$

Area of circle $=\pi r^2=3.14\times {\left(\frac{17}{2}\right)}^2=226.87c{m}^2$

Now, area of shaded region $=$ Area of circle $-$ $\left[ar\right(△ADB)+ar(△ACB\left)\right]$

$=226.87-120$

$=106.87c{m}^2$