## Book: RS Aggarwal - Mathematics

### Chapter: 18. Area of Circle, Sector and Segment

#### Subject: Maths - Class 10th

##### Q. No. 43 of Exercise 18B

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43
##### Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex of an equilateral triangle of side 12 cm as centre and a sector of circle of radius 6 cm with centre B is made. [Use √3 = 1.73 and π = 3.14.]

Area of shaded region can be calculated by subtracting the area of minor sector at vertex B from the sum of areas of the major sector at O and area of equilateral triangle.

Given Radius of circle at O = r = 6 cm

Side of equilateral triangle = a = 12 cm

Central angle at O = 360 -60 = 300°

Central angle at B = 60°

Area of the equilateral triangle =

where a = side of equilateral triangle

Area of the equilateral triangle = (putting the value of a)

Area of the equilateral triangle = (144×√3)/4

Area of the equilateral triangle = 36√3 cm2 eqn1

Area of sector = θ/360×πR2 where r = radius of the sector

Area of minor sector at B = 60/360×π×(62) (given)

Area of minor sector at B = 6π cm2 eqn2

Similarly,

Area of major sector at O = 30π cm2 eqn3

Area of shaded region = eqn1 + eqn3 – eqn2

On putting values

Area of shaded region = 36√3 + 30π-6π

Area of shaded region = 36√3 + 24π

(put π = 3.14 and √3 = 1.73

Area of shaded region = (36×1.73) + (24×3.14)

Area of shaded region = 62.28 + 75.36

Area of shaded region = 137.64 cm2

Area of the shaded region is 137.64 cm2.

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