Find the area,in square metres,of the trapezium whose bases and altitude area 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

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Solution

(i)

$I n a t r a p e z i u m . B a s e (b_{1}) = 12 d m, B a s e (b_{2}) = 20 d m, A l t i t u d e (h) = 10 d m ∴ A r e a = \frac{1}{2} (b_{1} + b_{2}) \times h = \frac{1}{2} (12 + 20) \times 10 d m^{2} = \frac{1}{2} \times 32 \times 10 = 160 d m^{2} = \frac{160}{10 \times 10} m^{2} = 1.6 m^{2}$

(ii)

$I n a t r a p e z i u m . B a s e (b_{1}) = 28 c m, B a s e (b_{2}) = 3 d m = 30 c m, A l t i t u d e (h) = 25 c m ∴ A r e a = \frac{1}{2} (b_{1} + b_{2}) \times h = \frac{1}{2} (28 + 30) \times 25 c m^{2} = \frac{1}{2} \times 58 \times 25 = 725 c m^{2} = \frac{725}{1000} m^{2} = 0.725 m^{2}$

(iii)

$I n a t r a p e z i u m . B a s e (b_{1}) = 8 m = 80 d m, B a s e (b_{2}) = 60 d m, A l t i t u d e (h) = 40 d m ∴ A r e a = \frac{1}{2} (b_{1} + b_{2}) \times h = \frac{1}{2} (80 + 60) \times 40 d m^{2} = \frac{1}{2} \times 140 \times 40 d m^{2} = 2800 d m^{2} = \frac{2800}{100} d m^{2} = 28 m^{2}$

(iv)

$I n a t r a p e z i u m . B a s e (b_{1}) = 150 c m = \frac{150}{100} = 1.5 m B a s e (b_{2}) = 30 d m = \frac{30}{10} = 3 m A l t i t u d e (h) = 9 d m = \frac{9}{10} = 0.9 m ∴ A r e a = \frac{1}{2} (b_{1} + b_{2}) \times h = \frac{1}{2} (1.5 + 3.0) \times 0.9 m^{2} = \frac{1}{2} \times 4.5 \times 0.9 m^{2} = 2.025 m^{2}$

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