The cross section of a canal is in the shape of an isosceles trapezium which is 4 m wide at the bottom and 5 m wide at the top. If the depth of the canal is 2 m and it is 120 m long, what is the maximum capacity of this canal?

The canal is 4 m wide at the bottom and 5 m at the top, thus the canal is in the shape of a trapezium whose volume will be equal to the product of area of trapezium and the width.

Length of canal = 120 m

Volume of canal = $$(\frac{1}{2} \times $$(sum of parallel sides) $$\times$$ height $$) \times$$ length

=> Volume = $$(\frac{1}{2} \times (5+4) \times 2)\times 120$$

= $$9 \times 120 = 1080$$ $$m^3$$

=> Ans - (D)