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# In Fig. 11.11, 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:**

Given, ABC is an __equilateral triangle__ of side 10 cm.

D, E and F are the midpoints of BC, CA and AB.

We have to find the area of the shaded region.

From the figure,

EF, FD and ED are the sectors made at the vertices A, B and C.

Area of shaded region = 3(area of sector)

Here, radius, r = 10/2 = 5 cm

Since ABC is an equilateral triangle, the corresponding angle θ is 60°

__Area of sector__ = πr²θ/360°

= (3.14)(5)²(60°/360°)

= (3.14)(25)(1/6)

= 13.08 cm²

Area of shaded region = 3(13.08)

= 39.25 cm²

Therefore, the area of the shaded region is 39.25 cm²

**✦ Try This:** In the given figure, ΔPQR is an equilateral triangle of side 8 cm and D, E, F are centres of circular arcs, each of radius 4 cm. Find the area of shaded region ( Use π = 3.14 and √3 = 1.732)

**☛ Also Check: **NCERT Solutions for Class 10 Maths Chapter 12

**NCERT Exemplar Class 10 Maths Exercise 11.3 Problem 12**

## In Fig. 11.11, 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)

**Summary:**

In Fig. 11.11, 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 39.25 cm²

**☛ Related Questions:**

- In Fig. 11.12, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area . . . .
- A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area . . . .
- In Fig. 11.13, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilatera . . . .

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