Question

Three circles each of radius 7 cm are placed in such a way that each of them touches the other two externally. The empty area enclosed between the circles is:$(\text{Take}\pi =\frac{22}{7})$

[1 mark]

• A. $88.7c{m}^2$
• B. $8.87c{m}^2$
• C. $78.7c{m}^2$
• D. $7.87c{m}^2$

Answer 1

The correct option is D $7.87c{m}^2$


We can conclude that the centres of circles form an equilateral triangle with side = 2r


$\text{Area enclosed between circles = Area of triangle - 3 (Area of sector)}$

$\text{Angle subtended by sector (θ) = 60° (∵ Equilateral triangle)}$

$\text{Area of sector}=\frac{\theta }{360°}\times \pi r^2$

$=\frac{60°}{360°}\times \frac{2}{27}\times 7^2$

$=\frac{1}{6}\times 22\times 7$

$=\frac{1}{3}\times 11\times 7$

$\text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}{a}^2$

$\text{a = Length of the side = 2r = 2(7) =14 cm}$

$\text{Area of triangle}=\frac{\sqrt{3}}{4}\times {14}^2=84.87c{m}^2$

$\therefore \text{Area enclosed between circles}=84.87-3(\frac{1}{3}\times 77)=7.87c{m}^2$