Question

In the given figure, $LMN$ is a right triangle in which $\angle M={90}^{\circ }$, P and Q are the midpoints of LM and LN respectively. If $LM=9cm$, $MN=12cm$ and $LN=15cm$, find:
(i) the perimeter of trapezium MNQP,
(ii) the area of trapezium MNQP

Answer 1

Since $P$ and $Q$ are mid-points of $LM$ and $LN$.

$\frac{LP}{LM}=\frac{LQ}{LN}=\frac{1}{2}$

In $△LPQ$ and $△LMN$

$\frac{LP}{LM}=\frac{LQ}{LN}$

$\angle PLQ=\angle MLN\left(Common\right)=>△LPQ\sim △LMN(ByS.A.Ssimilaritycriterion)$

$\therefore \frac{PQ}{MN}=\frac{LP}{LM}=\frac{1}{2}\left(CPST\right)$

$\frac{PQ}{12}=\frac{1}{2}(\because MN=12cm)$

$PQ=\frac{12}{2}=6cm$

$PM=\frac{1}{2}LM=\frac{9}{2}=4.5cm$ [Since $P$ and $Q$ are mid-points of $LM$ and $LN$]

$QN=\frac{1}{2}LN=\frac{15}{2}=7.5cm$

$\left(i\right)$ Perimeter of trapezium $MNQP$

$=MN+NQ+PQ+PM=12+7.5+6+4.5=30cm$

$\left(ii\right)$ Area of Trapezium $MNQP$

$=Base\times Height=MN\times PM=12\times 4.5=54c{m}^2$