Question

A field is in the shape of a trapezium whose parallel sides are $\text{25 m and 10 m}$. If its non-parallel sides are $\text{14 m and 13 m}$. Find its area ___ (in $c{m}^2$).

Answer 1

Let ABCD be the trapezium given to us with $\text{AB = 10 cm, CD = 25 cm, AD = 13 cm and BC = 14 cm}$.
If we drop perpendiculars from points A and B on the line CD it intersects the line CD on the points E and F respectively, which means that the length of the line segment EF is also 10cm. Let DE = x cm, so that,
FC = (25 - 10 - x)cm = (15 - x)cm ( as CD = 25cm = EF + DE + CF).
Let the height of the trapezium be h cm, and $\text{AE = BF = h cm}$.

In $△$ADE, using Pythagoras theorem,
$A{D}^2=D{E}^2+A{E}^2$
${13}^2=x^2+h^2$ ......................(i)
In $△$BFC, using Pythagoras theorem,
$B{C}^2=F{C}^2+B{F}^2$
${14}^2=(15–x{)}^2+h^2$ ............(ii)
Substituting the value of $h^2$ from (i) in (ii), we get,
${14}^2=(15-x{)}^2+{13}^2-x^2$
$196-169=225+x^2-30x-x^2$
$27=225-30x$
$x=\frac{198}{30}=6.6$
Substituting the value of x as 6.6 in (i), we get,
${13}^2={6.6}^2+h^2$
$169-43.56=h^2$
$h^2=125.44$
h = 11.2 cm
Area of a trapezium:-

$\frac{1}{2}\times \text{(Sum of parallel sides of a trapezium)}\times \text{(height of the trapezium)}$
= $\frac{1}{2}$$\times$(25+10)$\times$(11.2)
= $35\times 5.6$= $196c{m}^2$