Question

The parallel sides of a trapezium are $23cm$ and $13cm$. Its non-parallel sides are $10cm$ each. Find the area of trapezium.

Answer 1

Given $AB=13cm,CD=23cm,AC=BD=10cm$ each

construction Draw $AP\perp CD$ & $BQ\perp CD$

Formulae:(i)Area of a trapezium $=\frac{1}{2}\times 15cm$ of parallel sides)$\times$(perpendicular distance between then)

(ii) Pythagoras theorem

In quad $ABPQ$

$AB\parallel PQ$

$AP\parallel BQ$ & $AP=BQ${parallel lines} - $\left(1\right)$

In $△APC$ & $△BQD$

$\angle APC=\angle BQD={90}^o$ {construction}

$AC=BD=10cm$= given

$AP=BQ$ {from $\left(1\right)$}

$\therefore$ By $RHS$ rule of congruence, $△APC\cong △BQD$

Now let $PC=QD=x$

$\Rightarrow 2x+PQ=CD$

$2x=23-13=10cm\Rightarrow x=5cm$

In $△APC$ & $△BQD$

$AP$ & $BQ=\sqrt{(10{)}^2-(5{)}^2}=5\sqrt{3}cm$ {Pythagoras Theorem}

Now area of Trapezium $ABCD$

$=\frac{1}{2}\times (AB+CD)\times AP$

$=\frac{1}{2}\times (13+23)\times 5\sqrt{3}c{m}^2$

$=\frac{1}{2}\times 36\times 5\sqrt{3}c{m}^2$

$=90\sqrt{3}c{m}^2$