Question

Find the area enclosed by each of the following figures [fig. (i) - (ii)] as the sum of the areas of a rectangle and a trapezium.

Answer 1

Figure (i)

Area of given figure = Area of trapezium + area of rectangle

Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth

Area of given figure = \(\frac{1}{2}\)x (18 + 7) x 8 + 18 x 18

Area of given figure = 100 +324 = 424

Therefore area of given figure is 424 cm²

Figure (ii)

Area of given figure = Area of trapezium + area of rectangle

Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth

Area of given figure = \(\frac{1}{2}\)x (15 + 6) x 8 + 15 x 20

Area of given figure = 84 + 300 = 384

Therefore area of given figure is 384 cm²

Figure (iii)

Using pythagorous theorem in the right angled triangle

5² = 4² + x²

x² = 25 – 16

x² = 9

x = 3 cm

Area of given figure = Area of trapezium + area of rectangle

Area of given figure = \(\frac{1}{2}\)x (sum of parallel sides) x altitude + length x breadth

Area of given figure = \(\frac{1}{2}\)x (14 + 6) x 3 + 4 x 6

Area of given figure = 30 + 24 = 54

Therefore area of given figure is 54 cm²