Question

ABCD is a parallelogram whose area is 100 sq.cm. P is any point inside the parallelogram. Find the area of $△APB+△CPD$

Answer 1

REF.Image

Given ABCD is a parallelogram area is 100 sq. cm

P is a point inside the parallelogram

Construction : Let us drawn a line passing through P such that PQ||AB.

AB||CD _____ (1)

AB || PQ _____ (2)

from (1) & (2) CD || PQ

AB||PQ (By construction)

PA||QB ($\because$ PA and QB is a part of parallelogram DA and CB)

$\therefore$ ABQP is a parallelogram.

Similarly we can prove that PQCD is a parallelogram.

parallelogram ABQP and $\Delta APB$ lie on same base and same parallels $\therefore ar\left(PAB\right)=\frac{1}{2}ar\left(ABQP\right)$

or $2ar\left(\Delta PAB\right)=ar\left(ABQP\right)$

Parallelogram PQCD and $\Delta DPC$ lie on same base and same parallels $\therefore ar\left(\Delta DPC\right)=\frac{1}{2}ar\left(\right(11gmPQCD\left)\right)$

or $2.ar\left(\Delta DPC\right)=ar\left(11gmPQCD\right)$

ar (11 gm PQCD) = $100c{m}^2$ (given)

$\Rightarrow ar\left(11gmABPQ\right)+ar\left(11PQCD\right)=100c{m}^2$

$2.ar\left(\Delta PAB\right)+2ar\left(ADPC\right)=100c{m}^2$

$2.ar\left[\right(\Delta PAB)+(\Delta DPC\left)\right]=100c{m}^2$

$ar.\left(\Delta PAB\right)+ar\left(\Delta DPC\right)=\frac{100}{2}c{m}^2$

$ar\left(\Delta PAB\right)+ar\left(\Delta DPC\right)=50c{m}^2$

$\therefore ar\left(\Delta APB\right)+ar\left(\Delta CPD\right)=50c{m}^2$