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.

H^{2} = P^{2} + B^{2} (pythagoras theorem)

AD^{2} = DE^{2} + AE^{2}

⇒ 70^{2} = 42^{2} + AE^{2} (putting the given values)

⇒ 4900 = 1764 + AE^{2}

⇒ 4900 – 1764 = AE^{2}

⇒ 3136 = AE^{2}

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 cm^{2}→ eqn1

R = 35 cm

⇒ Area of semicircle = 11×175

∴ Area of semicircle = 1925 cm^{2}→ eqn2

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

⇒ Area of rectangle = 80×70 = 5600 cm^{2}→ 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 cm^{2}

Area of the shaded region is 2499 cm^{2}.

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