Question

In the following figure , $ABCD$ is a trapezium with $AB\parallel DC.$ If $AB=9$ $cm$, $DC=18$ $cm$, $CF=13.5$ $cm$, $AP=6$ $cm$ and $BE=15$ $cm$. Find the value of $PE.$

Answer 1

Given: Trapezium ABCD, $AB\parallel CD$,

Now, In $△APB$ and $△DPF$,
$\angle APB=\angle DPF$ [Vertically Opposite angles]
$\angle ABP=\angle PDF$ [Alternate angles]
$\angle PAB=\angle PFD$ [Alternate angles]

So, by $AAA$ criteria of similarity,
$△APB\sim △FPD$

Hence,

$\frac{AP}{PF}=\frac{AB}{DF}$

$\Rightarrow \frac{AP}{PF}=\frac{AB}{DC+CF}$

$\Rightarrow \frac{6}{PF}=\frac{9}{18+13.5}$

$\Rightarrow PF=\frac{6\times 31.5}{9}$

$\Rightarrow PF=21cm$

Now,

$AF=AP+PF$
$\Rightarrow AF=6+21$
$\Rightarrow AF=27cm$
In $△AEB$ and $△ECF$,
$\angle AEB=\angle CEF$ [Vertically opposite angles]
$\angle ABE=\angle ECF$ [Alternate angles of parallel lines $AB$ and $CF$]
$\angle BAE=\angle EFC$ [Alternate angles of parallel lines $AB$ and $CF$]

So, by $AAA$ criteria of similarity,
$△AEB\sim △\angle FEC$

Hence,

$\frac{AB}{CF}=\frac{AE}{EF}$
$\Rightarrow \frac{9}{13.5}=\frac{15}{EC}$
$\Rightarrow \frac{AE}{EF}=\frac{2}{3}$

Then, let $AE=2x$ and $EF=3x$
$AF=AE+EF$
$\Rightarrow 21=2x+3x$
$\Rightarrow x=4.2$

Thus, $AE=8.4cm$ and $EF=12.6cm.$
And,
$AE=AP+PE$
$\Rightarrow 8.4=6+PE$
$\Rightarrow PE=2.4cm$