Answer :

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)


AD2 = DE2 + AE2


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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