Question

Find the height of a trapezium in which parallel sides are 25 cm and 77 cm and non-parallel sides and 26 cm and 60 cm.Given the area of the trapezium as $1644c{m}^2$.

Answer 1

$CF=DF-CD$

$=(77-60)cm$

$=17cm$

For $△BCF$

Permiter of triangle $△BCF=BC+BF+CF$

$=(25+26+17)cm$

$=68cm$

Semiperimeter of the triangle $\left(s\right)=\frac{P}{2}=\frac{68}{2}=34cm$

By Heron's formula

Area of triangle $△BCF=\sqrt{S(s-BC)(s-BF)(s-CF)}$

$=\sqrt{34(34-25)(34-26)(34-17)}$

$=\sqrt{34\times 9\times 8\times 17}$

$=\sqrt{41616}=204c{m}^2$

let $hcm$ be the length of perpendicular to $CF$

Area of $BCF=\frac{1}{2}\times CF\times BE$

$\Rightarrow 204=\frac{1}{2}\times 17\times h$

$\Rightarrow h=\frac{204\times 2}{14}=24cm$

Area trapezium of $ABCD=\frac{1}{2}\left[\right(60+77)\times 24]=\frac{1}{2}(137\times 29)=\frac{1}{2}\times 3288=1644$