Question

A trapezium PQRSPQRS. Meassuremnt in centimeters are given as shown in the figure. Find the area of the trapezium.

Answer 1

From $R$ draw $RT$ perpendicular $PQ$

$\therefore ,SR=PT=20$cm

$TQ=PT-PQ=29-20=9$cm

In $△TQR,\angle T={90}^{\circ }$

By using Pythagoras theorem we have

${RT}^2={RQ}^2-{TQ}^2={15}^2-9^2=225-81=144$

$\Rightarrow RT=12$units

$SP=RT=12$cm

$\therefore ,PTRS$ is a rectangle

Area of trapezium $PQRS=$Area of rectangle $PTRS+$area of triangle$TQR$

$=PT\times RT+\frac{1}{2}\times TQ\times RT$

$=12\times 20+\frac{1}{2}\times 9\times 12$

$=240+54=294$sq.cms

$\therefore$ Area of trapezium $PQRS=294$sq.cms