## Trapezium Split in Ratio By Vertical Line

Suppose we have a trapezium
$ABCD$
with top and bottom sides 6 and 10 respectively, and height 7, split via a vertical line into trapeziums
$APQD$
and
$PBCQ$
that have areas in the ratio
$2:1$

Let
$AP=x$
then
$DQ=x+2$
.

The area of
$APQD$
is two thirds the area of
$ABCD$
.
Hence
$\frac{1}{2}(x+x+2) \times 7 =\frac{2}{3} \times \frac{1}{2}(10+6) \times 7=\frac{112}{3}$

$7x+7=\frac{112}{3} \rightarrow x = \frac{112/3-7}{7}= \frac{13}{3}$