Question

In fig., arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm, to intersect the sides BC, CA and AB at their respective mid-points D, E and F. The area of the shaded region is _____. (Use $\pi =3.14$).

• A. $37.25c{m}^2$
• B. $38.25c{m}^2$
• C. $39.25c{m}^2$
• D. $40.25c{m}^2$

Answer 1

The correct option is C $39.25c{m}^2$

Since,
$△ABC$ is an equilateral triangle.
$\therefore \angle A=\angle B=\angle C={60}^o$

Area of sector, AF EA
$=\frac{\pi }{360}\times \pi r^{2}c{m}^2$
$=(\frac{{60}^o}{{360}^o}\times \pi (5{)}^2)c{m}^2$
$=\frac{25}{6}\pi c{m}^2$

$\because$ Areas of all three sectors are equal.
$\therefore$ Total area of shaded region $=3\left(\frac{25}{6}\pi \right)c{m}^2$
$=\frac{25\times 3.14}{2}$
$=39.25c{m}^2$

Area of shaded region $=39.25c{m}^2$