Find the area, in square metres, of the trapezium whose bases and altitudes are as under:
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.

Solution:

(i) Given:

Bases:

$12 \mathrm{dm}=\frac{12}{10} \mathrm{~m}=1.2 \mathrm{~m}$

And, $20 \mathrm{dm}=\frac{20}{10} \mathrm{~m}=2 \mathrm{~m}$

Altitude $=10 \mathrm{dm}=\frac{10}{10} \mathrm{~m}=1 \mathrm{~m}$

Area of trapezium $=\frac{1}{2} \times($ Sum of the bases $) \times($ Altitude $)$

$=\frac{1}{2} \times(1.2+2) \mathrm{m} \times(1) \mathrm{m}$

$=1.6 \times \mathrm{m} \times \mathrm{m}$

$=1.6 \mathrm{~m}^{2}$

(ii) Given:

Bases:

$28 \mathrm{~cm}=\frac{28}{100} \mathrm{~m}=0.28 \mathrm{~m}$

And, $3 \mathrm{dm}=\frac{3}{10} \mathrm{~m}=0.3 \mathrm{~m}$

Altitude $=25 \mathrm{~cm}=\frac{25}{100} \mathrm{~m}=0.25 \mathrm{~m}$

Area of trapezium $=\frac{1}{2} \times($ Sum of the bases $) \times($ Altitude $)$

$=\frac{1}{2} \times(0.28+0.3) \mathrm{m} \times(0.25) \mathrm{m}$

$=0.0725 \mathrm{~m}^{2}$

(iii) Given:

Bases:

$8 \mathrm{~m}$

And, $60 \mathrm{dm}=\frac{60}{10} \mathrm{~m}=6 \mathrm{~m}$

Altitude $=40 \mathrm{dm}=\frac{40}{10} \mathrm{~m}=4 \mathrm{~m}$

Area of trapezium $=\frac{1}{2} \times($ Sum of the bases $) \times($ Altitude $)$

$=\frac{1}{2} \times(8+6) \mathrm{m} \times(4) \mathrm{m}$

$=28 \times \mathrm{m} \times \mathrm{m}$

$=28 \mathrm{~m}^{2}$

(iv) Given:

Bases:

$150 \mathrm{~cm}=\frac{150}{100} \mathrm{~m}=1.5 \mathrm{~m}$

And, $30 \mathrm{dm}=\frac{30}{10} \mathrm{~m}=3 \mathrm{~m}$

Altitude $=9 \mathrm{dm}=\frac{9}{10} \mathrm{~m}=0.9 \mathrm{~m}$

Area of trapezium $=\frac{1}{2} \times($ Sum of the bases $) \times($ Altitude $)$

$=\frac{1}{2} \times(1.5+3) \mathrm{m} \times(0.9) \mathrm{m}$

$=2.025 \times \mathrm{m} \times \mathrm{m}$

$=2.025 \mathrm{~m}^{2}$