Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
Solution:
From the figure, it is clear that the shaded region has an overlap region of area of a sector of angle 60° [Since ΔAOB is an equilateral triangle, each angle will be of 60° measure.]
∴ Area of shaded region = Area of a circle with radius 6 cm + Area of ΔOAB - Area of a sector of angle 60°
We know that, Area of an equilateral Δ = √3/4 (side)2
Area of the sector of angle θ = θ/360° × πr2 , where r is the radius of the circle.
Radius of circle (r) = 6 cm
Side of equilateral ΔOAB, (s) = 12 cm
We know each interior angle of equilateral Δ = 60°
Since they overlap, part of area of a sector OCD is in area of the circle and triangle.
∴ Area of shaded region = Area of circle + Area of ΔOAB - Area of sector OCDE
= πr2 + √3/4 (side)2 - θ/360° × πr2
= π(6 cm)2 + √3/4 (12)2 - 60°/360° × π (6 cm)2
= 36π cm² + 36√3 cm² - 6π cm²
= (30π + 36√3) cm²
= (30 × 22/7 + 36√3) cm²
= (36√3 + 660/7) cm²
☛ Check: NCERT Solutions Class 10 Maths Chapter 12
Video Solution:
Find the area of the shaded region in Fig. 12.22, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre
NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 4
Summary:
The area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre is 36√3 + 660/7 cm2.
☛ Related Questions:
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