The following figure shows the cross-section ABCD of a swimming pool which is a trapezium in shape.

If the width DC of the swimming pool is 6.4 m, depth (AD) at the shallow end is 80 cm and depth (BC) at the deepest end is 2.4 m, find its area of cross-section.

Given:

DC = 6.4 m

AD = 80 cm = $\dfrac{80}{100}$ m = 0.8 m

BC = 2.4 m

Area of the cross-section = Area of trapezium ABCD

= $\dfrac{1}{2}$ x (sum of parallel sides) x height

= $\dfrac{1}{2}$ x (0.8 + 2.4) x 6.4

= $\dfrac{1}{2}$ x 3.2 x 6.4

= $\dfrac{1}{2}$ x 20.48

= 10.24 m²

Hence, the area of cross-section is 10.24 m².