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. $8c{m}^2$
• B. $2.9c{m}^2$v
• C. $2c{m}^2$
• D. $4c{m}^2$

Answer 1

The correct option is C

$2c{m}^2$

Given that, three circles are drawn as shown above. Join the centres of each circle as shown.

Radius of each circle is 3.5 cm,

$\Rightarrow$ $AB=2\times Radiusofcircle$

$=2\times 3.5=7cm$

$\Rightarrow AC=BC=AB=7cm$

$\Rightarrow$, $\Delta ABC$ is an equilateral triangle with side 7 cm.

We know that each angle between two adjacent sides of an equilateral triangle is ${60}^{\circ }$.

$\therefore$ Area of sector with angle ${60}^{\circ }$ $=\frac{60}{{360}^{\circ }}\times \pi r^2$

So, total area of all 3 sectors = 3 $\times$ $\frac{60}{{360}^{\circ }}\times \pi r^2$

$=3\times \frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \times (3.5{)}^2$

$=\frac{1}{2}\times \frac{22}{7}\times 3.5\times 3.5$

$=11\times \frac{5}{10}\times \frac{35}{10}=\frac{11}{2}\times \frac{7}{2}$

$=\frac{77}{4}=19.25c{m}^2$

Area of $\Delta ABC=\frac{\sqrt{3}}{4}\times Sid{e}^2$
$=\frac{\sqrt{3}}{4}\times 7^2=21.21$

Area of the shaded region

= Area of the equilateral triangle - Area of three sectors

$=21.21-19.25=1.96c{m}^2$

Hence, the required area enclosed between these circles is $2c{m}^2\left(approx\right)$