Question

Three circles each of radius $3.5$cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Answer 1

Given

Three circles each of radius $3.5$ cm

On joining the centers of the three circles,
Step $1$: FInd the radius of three circles:

$AB=BC=CA=2\left(radius\right)=7cm$

Therefore, triangle $ABC$ is an equilateral triangle with each side $7$cm.
Step $2$: Find the area of triangle.

∴ $Areaofthetriangle=(\frac{\surd 3}{4})\times a^2$, where a is the side of the triangle.

$=(\frac{\surd 3}{4})\times 72$

$=\left(\frac{49}{4}\right)\surd 3c{m}^2$

$=21.2176c{m}^2$
Step $3$: Find the sector angle :

$Centralangleofeachsector=\varnothing =60°(60\pi /180)$

$=\pi /3radians$
Step $4$: Find the area of sector:

We know that $areaofeachsector=(1/2)r^2\theta$

$=(1/2)\times {(3.5)}^2\times (\pi /3)$

$=12.25\times (22/(7\times 6\left)\right)$

$=6.4167c{m}^2$

$Totalareaofthreesectors=3\times 6.4167=19.25c{m}^2$
Step $5$: FInd area of enclosed between three circles:

∴ $Areaenclosedbetweenthreecircles=AreaoftriangleABC–Areaofthethreesectors$

$=21.2176–19.25$

$=1.9676c{m}^2$

Hence, the required area enclosed between these circles is $1.9676c{m}^2$