Question

In the figure, arcs are drawn by taking vertices A,B,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 area of shaded region.

Answer 1

hope it may helps....
please mark it as brainliest.....

Answer 2

Answer:

Area of all sectors is,

$\frac{275}{7}\, cm^{2}$

and,

Area of remaining part is,

$25(\sqrt{3}-\frac{11}{7})\,cm^{2}$

Step-by-step explanation:

We have been provided that,

Triangle ABC is an Equilateral triangle.

Side of triangle is = 10 cm

The arcs are drawn from each vertices of the triangle.

We get three sectors intersecting at D, E and F on the sides BC, CA and AB respectively.

Now,

AB = BC = CD = 10 cm.

The area of sector AEF = Area of sector BDF = Area of sector CDE

So,

Area of sector AEF is given by,

$Area=\pi r^{2}(\frac{\theta}{360})Area=\frac{22}{7}(5)^{2}(\frac{60}{360})\\Area=\frac{22\times 25}{7\times 6}\\Area=\frac{275}{21}\,cm^{2}$

Therefore, the area of the all the shaded region that is the area of all the sectors is given by,

$Area=3\times \frac{275}{21}\\Area=\frac{275}{7}\, cm^{2}$

Now,

Area of Equilateral triangle is,

$Area=\frac{\sqrt{3}}{4}a^{2}\\Area=\frac{\sqrt{3}}{4}(100)\\Area=25\sqrt{3}\,cm^{2}$

Therefore, the area of the remaining portion is,

Remaining area = Area of triangle ABC - Area of all the sectors

$Remaining\,area=25\sqrt{3}-\frac{275}{7}\\Remaining\,area=25(\sqrt{3}-\frac{11}{7})\,cm^{2}$