Question

Three equal circles, each of radius $6cm$, touch one another as shown in the figure. Find the area enclosed between them.

[ Take $\pi =3.14and\sqrt{3}$ = 1.732.]

Answer 1

Join $ABC$.

Since, all sides of $\Delta ABC$ are equal (radius is$6cm$), therefore, $\Delta ABC$ is an equilateral triangle.

Now, to find the shaded area, we have to subtract the area of the three sectors so formed from the area of the equilateral triangle $\Delta ABC$.

Therefore, Area of the equilateral triangle
$=\frac{\sqrt{3}}{4}\times Sid{e}^2=\frac{1.73}{4}\times 12\times 12[\because AB=AC=BC=6\times 2=12cm]=62.28c{m}^2$

Area of a sector of the circle $=\frac{\theta }{{360}^{\circ }}\times \pi r^2=\frac{{60}^{\circ }}{{360}^{\circ }}\times \frac{22}{7}\times 6\times 6=\frac{1}{6}\times \frac{22}{7}\times 6\times 6=18.86c{m}^2$

Therefore, area of the three sectors $=3\times 18.86=56.57c{m}^2$

Hence, area of the shaded portion $=$ Area of the triangle $–$ Area of the three quadrant

$=62.28–56.57=5.71c{m}^2$

Hence, the required area is $5.71c{m}^2$.