Question

A line EF parallel to the base, divides the trapezoid into two trapezoids of equal perimeters. Calculate the ratio in which each of the non-parallel sides is divided.

Answer 1

Let $FA=x\Rightarrow FE=6-x$
And $CB=Y\Rightarrow DC=4-y$

Perimeter of trapezium ABCF
= 9 + y + FC + x
Perimeter of trapezium FCDE
= FC + 4 – y + 3 + 6 – x
= FC + 13 –x – y
$\Rightarrow$ 9 + y + FC + x = FC + 13 – x – y
$\Rightarrow 9+y+x=13-x-y$
$2(x+y)=13-9=4\Rightarrow x+y=2....\left(1\right)$
But $\frac{6-x}{x}=\frac{4-y}{y}\Rightarrow y=\frac{2}{3}x$
From eq (1) $\frac{2x}{3}+x=2\Rightarrow x=\frac{6}{5}$
$\therefore$ Required ratio = EF:FA [or DC:CB]
$=\frac{6-x}{x}=\frac{6-\frac{6}{5}}{\frac{6}{5}}=\frac{24}{6}=\frac{4}{1}=4:1$