Question

In the given figure, diagonals PR and QS of the parallelogram PQRS intersect at point O and LM is parallel to PS.

Show that : Area ($\Delta$ POS) + Area ($\Delta$ QOR) = Area ($\Delta$ POQ) + Area ($\Delta$ SOR).

Answer 1

In given figure parallelogram PQRS , PR and QS diagonal intersect at O.

We know that the diagonal intersect at right angle each other

The line LM is parallel to PS

Then diagonal of parallelogram divided in equal area of triangles

Then area $\Delta$ POQ= area of $\Delta$ POS=area of $\Delta$QOR=area of $\Delta$ SOR

OR Area of $\Delta$ POS +area of $\Delta$ QOR=$\frac{1}{2}$ = area of parallelogram

Then area of $\Delta$ POS +area of $\Delta$QOR=area of $\Delta$POQ +area $\Delta$SOR