Question

in fig., we have $AB||CD||EF$. if $AB=6cm$,$CD=xcm$, $EF=10cm$, $BD=4cm$ and $DE=ycm$, calculate the values of $x$ and $y$.

Answer 1

Consider $△ADB$ and $△EDF$,

$\angle ADB=\angle EDF\left(\text{Vertical angles}\right)$

$\angle ABD=\angle FED\left(\text{Alternate angles}\right)$

$\angle BAD=\angle EFD\left(\text{Alternate angles}\right)$

$\therefore △ADB\sim △EDF\left(\text{ASA criteria}\right)$

$\because$ Corresponding sides of similar triangles are proportional.

$\therefore \frac{BD}{DE}=\frac{AB}{FE}$

$\Rightarrow \frac{4}{y}=\frac{6}{10}$

$\Rightarrow 6y=40$

$\Rightarrow y=\frac{20}{3}$

Hence, $DE=\frac{20}{3}cm$

$\therefore BE=4+y=4+\frac{20}{3}=\frac{32}{3}$

Similarly, by AA criteria,

$△ABE\sim △CDE$

$\therefore \frac{DE}{BE}=\frac{DC}{BA}$

$\Rightarrow \frac{\left(\frac{20}{3}\right)}{\left(\frac{32}{3}\right)}=\frac{x}{6}$

$\Rightarrow x=\frac{5}{8}\times 6=\frac{15}{4}$

Hence, the value of x and y are $\frac{15}{4}$ and $\frac{32}{3}$ respectively.