Question

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

• A. $PE=5.8$ cm
• B. $PE=4.8$ cm
• C. $PE=2.8$ cm
• D. $PE=3.8$ cm

Answer 1

The correct option is B $PE=4.8$ cm

Given ABCD is trapezium, $AB\parallel DC$

$\Rightarrow AB=9cm,DC=18cm,CF=13.5cm,AP=6cm,BE=15cm$

Consider $△APD$ and $△BPE$

$\Rightarrow \angle APD=\angle BPE$ ( Vertically opposite angles )

$\Rightarrow \angle ADP=\angle EBP$ ( Alternate angles since $DC\parallel AB$ and $AD\parallel BC$ )

$\therefore △APD\cong △BPE$

$\therefore \frac{PA}{PB}=\frac{PD}{PE}=\frac{AD}{BE}$

$\Rightarrow \frac{6}{BP}=\frac{PD}{PE}=\frac{AD}{15}$

$\Rightarrow \frac{6}{6}=\frac{AD}{15}$

$\Rightarrow AD=15cm.$

$\therefore$ From the equations

$\Rightarrow PE=\frac{15\times 6}{18}$

$=4.8cm.$

Hence, the answer is $4.8cm.$