# 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 each angle of an equilateral Δ is of measure 60° area of a circle with radius 6 cm and area of triangle OAB with side 12 cm.

∴ Area of shaded region = Area of a circle with radius 6 cm + Area of ΔOAB - Area of a sector of angle 60°

= πr^{2} + √3/4 (side)^{2 }- θ/360° × πr^{2}

Using the formula, Area of an equilateral Δ = √3/4 (side)^{2}

Area of the sector of angle θ = θ/360° x πr^{2}

Here, 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

= 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

**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 - Chapter 12 Exercise 12.3 Question 4

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

The area of the shaded region 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 cm^{2}