Question

Find the area of Fig. 34 in the following ways:
(i) Sum of the areas of three triangles
(ii) Area of a rectangle − sum of the areas of five triangles

Answer 1

We have,
(i) P is the midpoint of AD.
Thus AP = PD = 25 cm and AB = CD = 20 cm
From the figure, we observed that,
Area of Δ APB = Area of Δ PDC
Area of Δ APB = $\frac{1}{2}$x AB x AP
= $\frac{1}{2}$x 20 cm x 25 cm = 250 cm²
Area of Δ PDC = Area of Δ APB = 250 cm²

Area of Δ RPQ = $\frac{1}{2}$x Base x Height
= $\frac{1}{2}$x 25 cm x 10 cm = 125 cm²
Hence,
Sum of the three triangles = (250 + 250 + 125) cm²
= 625 cm²

(ii) Area of the rectangle ABCD = 50 cm x 20 cm = 1000 cm²
Thus,
Area of the rectangle − Sum of the areas of three triangles (There is a mistake in the question; it should be area of three triangles)
= (1000 − 625 ) cm² = 375 cm²