Question

BD is one of the diagonals of a quad.ABCD.If $AL\perp BD$ and $CM\perp BD$,show that ar(quad.ABCD)=$\frac{1}{2}\times BD\times (Al+CM)$.

Answer 1

ar(quad. ABCD) = ar$△ABD+ar△DBC$
ar$△ABD=\frac{1}{2}\times base\times height=\frac{1}{2}\times BD\times AL$
ar$△\left(DBC\right)=\frac{1}{2}\times BD\times CL$......(ii)
From (i) and (ii), we get:
ar(quad ABCD) = $\frac{1}{2}\times BD\times AL+\frac{1}{2}\times BD\times CL$
$\to ar\left(quadABCD\right)=\frac{1}{2}\times BD(AL+CL)$
Hence, proved.