Question

An equilateral triangle had $7cm$ long sides. Equal circles were drawn about the three vertices of the triangle so that each circle touched the other two. Calculate the area of space enclosed between these circles

• A. $1.967c{m}^2$
• B. $1.900c{m}^2$
• C. $1.867c{m}^2$
• D. $1.800c{m}^2$

Answer 1

The correct option is A $1.967c{m}^2$

As it is clearly seen from the figures

radius of circle=$\frac{AB}{2}=\frac{7}{2}=3.5cm$

As angles of equilateral are 60${}^{\circ }$ each

Area of Sector of circle=$\frac{\pi r^2\times 60}{360}=\frac{22}{7}\times \frac{49}{4}\times \frac{1}{6}=\frac{77}{12}c{m}^2=6.41c{m}^2$

Area of equilateral triangle =$\frac{\sqrt{3}}{3}{a}^2=\frac{\sqrt{3}}{4}\times 7^2=21.214c{m}^2$

Area of remaning part=Area of triangle-Area of 3 sector of circle

=$21.2145-3\times 6.41=1.97c{m}^2$