Question

In the given figure , $ABCD$ is a trapezium with AB || DC. If $AB=9$$cm$, $DC=18$ $cm$, $CF=13.5$ $cm$, $AP=6$ $cm$ and $BE=15$ $cm$. .What is the value of $EC$ in mm?

Answer 1

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)
Thus, $△AEB\sim △\angle FEC$ (AAA rule)
Hence, $\frac{AB}{CF}=\frac{BE}{EC}$
$\frac{9}{13.5}=\frac{15}{EC}$
$EC=\frac{15\times 13.5}{9}$
$EC=22.5$ cm=$225mm$