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# 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° × πr^{2} , 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

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

= 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 cm^{2}.

**☛ Related Questions:**

- From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. 12.23. Find the area of the remaining portion of the square.
- In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.
- In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
- Fig. 12.26 depicts a racing track whose left and right ends are semicircular. Fig. 12.26 The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :(i) the distance around the track along its inner edge(ii) the area of the track.

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