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.

• A. $3.932c{m}^2$
• B. $19.251c{m}^2$
• C. $21.217c{m}^2$
• D. $1.966c{m}^2$

Answer 1

Answer: (D)

$1.966c{m}^2$

Given, $A,B$ and $C$ are the centres of three circles touching each other at $P,Q\&R$. All the circles have equal radii $r=3.5$ cm.

To find out: The area enclosed by the circles.

Solution:

We join $AB,BC$ and $CA$. Since $AB$ joins the centres of two circles touching each other, we have $AB=$ sum of their radii $=2r=2\times 3.5$ cm $=7$ cm.

Similarly $BC=7$ cm and $CA=7$ cm.

$\therefore \Delta ABC$ is equilateral with sides $=a=7$ cm.

So area $\Delta ABC=\frac{\sqrt{3}}{4}\times a^2=\frac{\sqrt{3}}{4}\times 7^{2}c{m}^2=\frac{49\sqrt{3}}{4}\times c{m}^2=21.217c{m}^2$.

Also each angle of $\Delta ABC={60}^o$.

Now $AQ\&AP$ are the radii of the same circle of which $PQ$ is an arc.

$\therefore APQ$ is a sector of central angle $=\theta ={60}^o$ and radius $=r=3.5$ cm.

$\therefore$ Area of sector $APQ=\frac{\theta }{{360}^o}\times \pi \times r^2=\frac{{60}^o}{{360}^o}\times \frac{22}{7}\times {3.5}^{2}c{m}^2=6.417c{m}^2$.

Since all the sectors have same radii and central angle they are equal in areas.

$\therefore$ Area of three sectors $=3\times 6.417c{m}^2=19.251c{m}^2$.

$\therefore$ Area of enclosed region $=$ area .$\Delta ABC-$ Area of three sectors $=(21.217-19.251)c{m}^2=1.966c{m}^2$