Question

Find the area of a trapezium-shaped field, the lengths of whose parallel sides are $42$ metre and $30$ metre and other sides are $18$ metre and $18$ metre.

Answer 1

In the figure, ABCD is a trapezium-shaped field in which parallel sides $AB=42m$ and $CD=30m$ and non-parallel sides AD and BC equal to $18m$.
Draw EC || AD and CF ⊥ AB
Therefore
$EB=AB-AE=42m-30m$
$=12m$ and $EC=18m$
$\therefore$ Area of $\Delta BCE=\frac{b}{4}\sqrt{4a^2-b^2},$
where $a=18m$ and $b=12m$
$=\frac{12}{4}\sqrt{4(18{)}^2-(12{)}^2}$
$=3\sqrt{4\times 324-144}$
$=3\sqrt{1296-144}=3\sqrt{1152}$
$=3\times 33.94=101.82m^2$
Now, area of $\Delta BCE=\frac{1}{2}\times BE\times CF$
$\Rightarrow 101.82=\frac{1}{2}\times 12\times CF$
$\therefore CF=\frac{101.82}{6}=16.9m$
Area of parallelogram $AECD=AE\times CF$
$=30\times 16.9=507m^2$
$\therefore$ Area of trapezium ABCD = area of parallelogram AECD + area of ABCE$=507+101.82=610.92m^2$.