Question

Let $ABCD$ be parallelogram whose equations for the diagonals $AC$ and $BD$ are $x+2y=3$ and $2x+y=3$, respectively. If length of diagonal $AC=4$ units and area of parallologram $ABCD=8$ sq. units then the length of other diagonal $BD$ is

• A. $\frac{10}{3}$
• B. $2$
• C. $\frac{20}{3}$
• D. None of these

Answer 1

The correct option is C $\frac{20}{3}$

The above lines can be written as
$y=\frac{-x}{2}+\frac{3}{2}$
$y=-2x+3$
Hence angle between the lines will be
$tan\alpha =|\frac{-2+\frac{1}{2}}{1+1}|$
$=\frac{3}{4}$ (taking the positive value)
Let P be the point of intersection of the diagonals.
Therefore area of triangle CPB is given by
$Area=pc.\left(\frac{pbsin\alpha }{2}\right)$
$=\frac{8}{4}$
$2.\left(\frac{pbsin\alpha }{2}\right)=\frac{8}{4}$
$pbsin\alpha =2$
$pb\frac{3}{5}=2$
$PB=\frac{10}{3}$
Hence BD will be twice of PB
$=\frac{20}{3}$