Question

Two parallel sides of a trapezium are 60cm and 77 cm and other sides are 25 cm and 26 cm. Find the area of the trapezium.

Answer 1

In trapezium ABCD, AB||DC

AB= 77 cm, BC= 26 cm, CD= 60 cm
DA= 25 cm
Through, C, draw CE||DA meeting AB at E
$\therefore$ AE=CD = 60 cm
and EB= 77-60= 17 cm,
CE=DA = 25 cm
Now area of $\Delta BCE$, with sides 17 cm, 26 cm, 25 cm
$s=\frac{a+b+c}{2}=\frac{17+26+25}{2}=\frac{68}{2}=34$
$\therefore$ Area $\Delta EBC=\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{34(34-17)(34-26)(34-25)}=\sqrt{34\times 17\times 8\times 9}=\sqrt{2\times 17\times 17\times 2\times 2\times 2\times 3\times 3}=17\times 2\times 2\times 3=204c{m}^2$
Draw $CL\perp EB$,
$\therefore CL=\frac{\text{area of}\Delta \text{EBC}\times \text{2}}{base}=\frac{204\times 2}{17}$
=24 cm
Now area of trap. ABCD = $\frac{1}{2}$ (Sum of parallel sides$\times$ height)
=$\frac{1}{2}(77+60)\times 24$
= $\frac{1}{2}\times 137\times 24=137\times 12$
=1644 $c{m}^2$