## Book: RS Aggarwal - Mathematics

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

#### Subject: Maths - Class 10th

##### Q. No. 45 of Exercise 18B

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45
##### In the given figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. [Use π = 3.14.]

Here in order to find the area of the shaded region (region excluding the triangle) we have to subtract the area of the triangle from the area of the rectangle and then add the area of the semicircle.

Given AB = 20 cm, DE = 12 cm, AE = 9 cm and AED = 90°

Here we see that the triangle AED is right angle triangle, therefore, we can apply Pythagoras theorem i.e.

AD2 = 122 + 92 (putting given values)

Area of ∆AED = 1/2×AE×DE

(Area of triangle = 1/2×base×height)

On putting values we get,

Area of ∆AED = 9×6

Area of ∆AED = 54 cm2 eqn1

Here radius of semicircle = BC/2 = 15/2

R = 7.5 cm

Area of semicircle = 1.07×56.25

Area of semicircle = 88.3125 cm2 eqn2

Area of rectangle = ℓ×b

(ℓ = length of rectangle, b = breadth of rectangle)

Area of rectangle = 20×15

(putting the values of ℓ & b)

Area of rectangle = 300 cm2 eqn3

Area of shaded region = Area of rectangle + Area of semicircle – Area of ∆

Area of shaded region = 300 + 88.3125 – 53 (from eqn1, eqn2, eqn3)

Area of shaded region = 334.3125 cm2

Area of shaded region is 334.3125 cm2.

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