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)
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.