Question

In a parallelogram $ABCD$, point $P$ lies in $DC$ such that $DP:PC=3:2$. If area of $\Delta DPB=30sq.cm$ , find the area of the parallelogram $ABCD$.

• A. $100sq.cm$
• B. $110sq.cm$
• C. $120sq.cm$
• D. $130sq.cm$

Answer 1

The correct option is A $100sq.cm$

Area of a parallelogram $ABCD=$ area of congruent triangles $ABD$ and $BDC$.
We know congruent triangles have equal area.
Given area of triangle $DPB=30$ sq cm
Hence, area of triangle $PBC=20$ sq cm .... $[DP:PC=3:2]$
Hence, area of triangle $BDC=50$ sq cm.
Area of triangle $ABD=$ area of triangle $BDC=50$ sq cm.
Therefore, area of parallelogram $ABCD=50+50=100$ sq.cm.