Question

ABCD is a parallelogram. P is the mid point of AB. BD and CP intersect at Q such that $CQ:QP=3:1.Ifar\left(△PBQ\right)=10c{m}^2$
Find the ara of parallelogram ABCD.

Answer 1

In ||gm ABCD, P is mid point on AB, PC and BD intersect each otehr at Q

CQ: QP = 3 :1
$ar\left(△PBQ\right)=10c{m}^2$
BD is its diagonal
$\therefore ar\left(△ABD\right)=ar\left(△BCD\right)=\frac{1}{2}ar||gmABCD\therefore ar\left(||gmABCD\right)=2ar\left(△ABD\right)In△PBCCQ:QP=3:1\therefore △PBQand△CQB$ have same vertice B
$\therefore 3\times area△PBQ=ar\left(△CBQ\right)\Rightarrow Area\left(△CBQ\right)=3\times 10=30c{m}^2\therefore ar\left(△PBC\right)=30+10=40c{m}^2$
Now $△ABDand△PBC$ are between the same parallel but base $PB=\frac{1}{2}AB$
$\therefore ar\left(△ABD\right)=2ar\left(△PBC\right)=2\times 40=80c{m}^{2}Butar\left(||gmABCD\right)=2ar\left(△ABD\right)=2\times 80=160c{m}^2$