Question

Question 7
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 that, three circles are in such a way that each of them touches the other two now. We join centre of all three circles to each other by a line segment. Since, radius of each circles is 3.5 cm,

So, $AB=2\times Radiusofcircle$

$=2\times 3.5=7cm$

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

Which shows that, $\Delta ABC$ 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 $\angle A={60}^{\circ }$

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

So, area of each sector = 3 $\times$ Area of sector with angle A

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

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

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

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

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

$=21.2176-1925=1.9676c{m}^2$

Hence, the required area enclosed between these circles is $1.967c{m}^2\left(approx\right)$