Question

Question 12
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 $\pi =3.14$)

Answer 1

Since, ABC is an equilateral triangle

$\therefore \angle A=\angle B=\angle C={60}^{\circ }$

And AB = BC = AC = 10 cm

So, E, F and D are mid- points of the sides.

$\therefore AE=EC=CD=BD=BF=FA=5cm$

Now, area of sector CDE $=\frac{\theta \pi r^2}{360}=\frac{60\times 3.14}{360}(5{)}^2$

$=\frac{3.14\times 25}{6}=\frac{78.5}{6}=13.0833c{m}^2$

$\therefore$ Area of shaded region = 3 (Area of sector CDE)

$=3\times 13.0833$

$=39.25c{m}^2$