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:
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