Question

With vertices $A,B$ and $C$ of a triangle $ABC$ as centres, arcs are drawn with radius $6cm$ each in figure. If $AB=20cm,BC=48cm$ and $CA=52cm$, then find the area of the shaded region. (Use $\pi =3.14$)

Answer 1

Given:-

$a=BC=48cm$

$b=AC=52cm$

$c=AB=20cm$

Semi-perimeter of $△ABC=\frac{a+b+c}{2}=\frac{52+48+20}{2}=60cm$

By using Heron's formula,

Area of triangle,

$=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{60(60-48)(60-52)(60-20)}$

$=480{cm}^2$

Now,

Area of sectors $=\frac{\pi \theta }{360}{r}^2$

$=\frac{\pi r^2}{360}({\theta }_1+{\theta }_2+{\theta }_2)$

$=\frac{3.14\times 36}{360}\times \left(180\right)=56.52{cm}^2$

Therefore,

Area of shaded region $=$ Area of triangle $-$ Area of sectors

$\Rightarrow$ Area of shaded region $=480-56.52=423.48{cm}^2$