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 6.4 cm, depth (AD) at the shallow end is 80 cm and depth (BC) at deepest end is 2.4 m, find Its area of the cross-section.

Answer 1

Area of the cross-section =Area of trapezium ABCD

$=\frac{1}{2}$ (Sum of parallel sides) $\times$ height

$=\frac{1}{2}$ (Sum of parallel sides) $\times$ height

$=\frac{1}{2}(80+240)\times 6.4$

=(320)(3.2)=(32)(32)

$=1024c{m}^2$ or $\frac{1024}{100}$ sq m

=10.24 sq.m.

$(\because$ AD=80 cm and BC =2.4 m=240 cm)