Question

Find the area, in square metres, of the trapezium whose bases and altitudes are as under:
(i) bases = 12 dm and 20 dm, altitude = 10 dm
(ii) bases = 28 cm and 3 dm, altitude = 25 cm
(iii) bases = 8 m and 60 dm, altitude = 40 dm
(iv) bases = 150 cm and 30 dm, altitude = 9 dm.

Answer 1

$$\begin{gathered} \text{(i)} \\ \text{Given:} \\ \text{Bases:} \\ \text{12 dm =}\frac{12}{10}\text{m = 1.2 m} \\ \text{And, 20 dm=}\frac{20}{10}\text{m=2 m} \\ \text{Altitude = 10 dm =}\frac{10}{10}\text{m = 1 m} \\ \text{Area of trapezium =}\frac{1}{2}\text{×(Sum of the bases)×(Altitude)} \\ =\frac{1}{2}\times (1.2+2)\text{m}\times \left(1\right)\text{m} \\ =1.6\times \text{m×m} \\ {\text{=1.6 m}}^2 \\ \text{(ii)} \\ \text{Given:} \\ \text{Bases:} \\ \text{28 cm =}\frac{28}{100}\text{m = 0.28 m} \\ \text{And, 3 dm =}\frac{3}{10}\text{m = 0.3 m} \\ \text{Altitude = 25 cm =}\frac{25}{100}\text{m = 0.25 m} \\ \text{Area of trapezium =}\frac{1}{2}\text{×(Sum of the bases)×(Altitude)} \\ =\frac{1}{2}\times (0.28+0.3)\text{m}\times (0.25)\text{m} \\ {\text{= 0.0725 m}}^2 \\ \text{(iii)} \\ \text{Given:} \\ \text{Bases:} \\ \text{8 m} \\ \text{And, 60 dm =}\frac{60}{10}\text{m = 6 m} \\ \text{Altitude = 40 dm =}\frac{40}{10}\text{m = 4 m} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the bases)×(Altitude)} \\ =\frac{1}{2}\times (8+6)\text{m}\times \left(4\right)\text{m} \\ =28\times \text{m×m} \\ {\text{=28 m}}^2 \\ \text{(iv)} \\ \text{Given:} \\ \text{Bases:} \\ \text{150 cm =}\frac{150}{100}\text{m = 1.5 m} \\ \text{And, 30 dm =}\frac{30}{10}\text{m = 3 m} \\ \text{Altitude = 9 dm =}\frac{9}{10}\text{m = 0.9 m} \\ \text{Area of trapezium=}\frac{1}{2}\text{×(Sum of the bases)×(Altitude)} \\ =\frac{1}{2}\times (1.5+3)\text{m}\times (0.9)\text{m} \\ =2.025\times \text{m×m} \\ {\text{=2.025 m}}^2 \end{gathered}$$