Question

The following figure shows the cross-section ABCD of a swimming pool which is trapezium in shape. If the width DC of the swimming pool is $64$ m, depth (AD) at the shallow end is $80$ m and depth (BC) at the deepest end is $24$ m, find its area of the cross-section($m^2$).

Answer 1

Area of a trapezium $=\frac{1}{2}\times \left(\text{sum of the parallel sides}\right)\times height$

Area $=\frac{1}{2}\times (80+24)\times 64$

$=\frac{1}{2}\times 104\times 64$

$=3328m^2$