Question

In the given figure, arcs are drawn by taking vertices $A,B$ and $C$ of an equilateral triangle of side 10 cm, to intersect the sides $BC,CAandAB$ at their respective mid-points $D,EandF$. Find the area of the shaded region.$(Use\pi =3.14)$

• A. $29.25c{m}^2$
• B. $39.25c{m}^2$
• C. $49.25c{m}^2$
• D. $59.25c{m}^2$

Answer 1

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

Since $\Delta$ABC is an equilateral triangle.


$\therefore \angle A=\angle B=\angle C={60}^{\circ }$
Area of the sector, $AFEA=\frac{\theta }{360}\times \pi r^{2}c{m}^2$
$=(\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi (5{)}^2)c{m}^2$
$=\frac{25}{6}\pi c{m}^2$
$\because$ Areas of all three sectors are equal.
$\therefore$ The total area of shaded region $=3\left(\frac{25}{6}\pi \right)c{m}^2$
$=\frac{25\times 3.14}{2}$
$=39.25c{m}^2$