Question:
In figure 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. Find the area of the shaded region, (use π = 3.14)
Solution:
Since, ABC is an equilateral triangle.
$\angle A=\angle B=\angle C=60^{\circ}$
and $\quad A B=B C=A C=10 \mathrm{~cm}$
So, $E, F$ and $D$ are mid-points of the sides.
$\therefore \quad A E=E C=C D=B D=B F=F A=5 \mathrm{~cm}$
Now, area of sector $C D E=\frac{\theta \pi r^{2}}{360}=\frac{60 \times 3.14}{360}(5)^{2}$
$=\frac{3.14 \times 25}{6}=\frac{78.5}{6}=13.0833 \mathrm{~cm}^{2}$
$\therefore \quad$ Area of shaded region $=3$ (Area of sector $C D E$ )
$=3 \times 13.0833$
$=39.25 \mathrm{~cm}^{2}$