Question

Three circles are placed on a plane in such a way that each circle just touches the other two, each having a radius of $10$ cm. Find the area of the region enclosed by them.

Answer 1

$\Rightarrow$ Here, $AB=BC=AC=10+10=20cm$

$\therefore$ $△ABC$ is equilateral triangle.

$\therefore$ $\angle A=\angle B=\angle C={60}^o$

$\Rightarrow$ Let $a=20cm$

$\Rightarrow$ Area of equilateral $△ABC=\frac{\sqrt{3}}{4}\times a^2$

$\Rightarrow$ Area of equilateral $△ABC=\frac{\sqrt{3}}{4}\times (20{)}^2$

$\therefore$ Area of equilateral $△ABC=\frac{400\times \sqrt{3}}{4}=173.20c{m}^2$

$\Rightarrow$ Here, $\theta ={60}^o$ and $r=10cm$

$\Rightarrow$ Area of $3$ sectors $=3\times \frac{\theta }{{360}^o}\pi r^2$

$\Rightarrow$ Area of $3$ sectors $=3\times \frac{{60}^o}{{360}^o}\times \frac{22}{7}\times (10{)}^2$

$\therefore$ Area of $3$ sectors $=3\times 52.38=157.14c{m}^2$

$\Rightarrow$ Required area = Area of equilateral triangle - Area of $3$ sectors.

$\Rightarrow$ Required area $=173.20-157.14=16.06c{m}^2$