## Book: RS Aggarwal - Mathematics

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

#### Subject: Maths - Class 10th

##### Q. No. 44 of Exercise 18B

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44
##### In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm, ∠AED = 90° and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the area of the shaded region.

Here in order to find the area of the shaded region we have to subtract the area of the semicircle and the triangle from the area of the rectangle.

Given AB = 80 cm, BC = 70 cm, DE = 42 cm, AED = 90°

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

H2 = P2 + B2 (pythagoras theorem)

702 = 422 + AE2 (putting the given values)

4900 = 1764 + AE2

4900 – 1764 = AE2

3136 = AE2

AE = √3136

AE = 56 cm

Area of ∆AED = 1/2×AE×DE

(Area of triangle = 1/2×base×height)

On putting values we get,

Area of ∆AED = 1/2×56×42

Area of ∆AED = 28×42

Area of ∆AED = 1176 cm2 eqn1

R = 35 cm

Area of semicircle = 11×175

Area of semicircle = 1925 cm2 eqn2

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

Area of rectangle = 80×70 = 5600 cm2 eqn3

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

Area of shaded region = 5600 -1925 – 1176 (fromeqn1, eqn2 and eqn3)

Area of shaded region = 2499 cm2

Area of the shaded region is 2499 cm2.

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