Question

In the given figure (not drawn to scale), ABCD is a rectangle. ED = 16 cm, FG = FC = 8 cm, BF = 20 cm and FG is perpendicular to BC. If CD is half of AD, then find the area of the blue region (in cm${}^2$).

• A. 225
• B. 220
• C. 320
• D. 325

Answer 1

The correct option is B 220

The white region is made up of two triangles and a square (∵ FG = FC and FG ⟂ BC).

Given, ED = 16 cm, FG = FC = 8 cm, and BF = 20 cm

AD = BC = BF + FC = 20 + 8 = 28 cm
AD = 28 cm = AE + ED = AE + 16 cm
⇒ AE = 12 cm

CD = $\frac{1}{2}$AD = $\frac{1}{2}$ × 28 = 14 cm
CD = DH + HC = DH + FG
⇒ 14 cm = DH + 8 cm
⇒ DH = 6 cm

Area of the triangle AEB = $\frac{1}{2}$ × 12 × 14 = 6 × 14 = 84 cm${}^2$
Area of the triangle DGH = $\frac{1}{2}$ × 6 × 8 = 3 × 8 = 24 cm${}^2$
Area of the square GHCF = 8 × 8 = 64 cm${}^2$
Area of the rectangles ABCD = 28 × 14 = 392 cm${}^2$

Area of the blue portion = Area of the rectangles ABCD – Area of the triangle AEB – Area of the triangle DGH – Area of the square GHCF Area of the blue portion
= 392 – 84 – 24 – 64 = 392 – 160 = 220 cm${}^2$