The area of a trapezium whose parallel sides are of lengths $40metres$  and

$24metres$  and whose non parallel sides are equal, each being $10metres$  is

Draw $CE\parallel DA$ , intersects $AB$  at $E$

In quadrilateral $AECD,AE\parallel DC$  and $CE\parallel DA$

$\therefore$ $AECD$  is a parallelogram

$AE=DC=24cm$

$AD=CE=10cm$

$\therefore$ $EB=AB-AC$

$=40-24=16$

Area of $\Delta BEC$

$=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

Where $S=\frac{16+10+10}{2}=\frac{36}{2}=18$

$=\sqrt{18\times 2\times 8\times 8}$

$=\sqrt{9\times \left(2\times 2\right)\times \left(8\times 8\right)}$

$=\sqrt{\left(3\times 3\right)\times \left(2\times 2\right)\times \left(8\times 8\right)}$

$=3\times 2\times 8$

$=48c{m}^2$

But area of $\Delta BEC=\frac{1}{2}\left(base\right)\times \left(height\right)$

$=\frac{1}{2}\times 16\times h=48$

$=8\times h=48$

$h=\frac{48}{8}=6$

$h=6cm$

$\therefore$  distance between parallel sides $=6cm$

$\therefore$  Area of trapezium

$=\frac{1}{2}\left(\text{sum}\text{of}\text{parallel}\text{sides}\right)\text{×}\left(\text{Distance}\text{between}\text{parallel}\text{sides}\right)$

$=\frac{1}{2}\left(40+24\right)\times 6$

$=\frac{1}{2}\left(64\right)\times 6$

$=192c{m}^2$