Question

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

Answer 1

ar(quad. ABCD) = ar(∆​ABD) + ar (∆DBC)
ar(∆ABD) = $\frac{1}{2}$⨯ base ⨯ height = $\frac{1}{2}$⨯ BD ⨯ AL ...(i)
ar(∆DBC) = $\frac{1}{2}$ ⨯ BD ⨯ CL ...(ii)
From (i) and (ii), we get:
​ar(quad ABCD) = $\frac{1}{2}$ ⨯BD ⨯​ AL + $\frac{1}{2}$ ⨯ BD ⨯​ CL
$\Rightarrow$​ar(quad ABCD) = $\frac{1}{2}$ ⨯ BD ⨯ ​(AL + CL)
Hence, proved.