Question

$P$ and $Q$ are any two points lying on the sides $DC$ and $AD$ respectively of a parallelogram $ABCD$. Show that $area\;\left( {\Delta APB} \right) = area\;\left( {\Delta BQC} \right)$.

Answer 1

Hint:In the solution we will use the concept of parallelogram which states that if the triangle and parallelogram have the same base and parallels then the area of triangle is half of the parallelogram.

Complete step-by-step solution
Given: $P$ and $Q$ are any two points lying on the sides $DC$ and $AD$ respectively of a parallelogram $ABCD$.

Since $\Delta APB$ and parallelogram $ABCD$ are on the same base $AB$ and between the same parallels.
Therefore, $ar\left( {\Delta APB} \right) = \dfrac{1}{2}ar\left( {{\rm{Parallelogram}}\;\;ABCD} \right)$ …..(1)
Similarly,
Since $\Delta BQC$ and parallelogram $ABCD$ are on the same base $AB$ and between the same parallels.
Therefore, $ar\left( {\Delta BQC} \right) = \dfrac{1}{2}ar\left( {{\rm{Parallelogram}}\;\;ABCD} \right)$ …..(2)
From equation (1) and equation (2) we can say that
$area\;\left( {\Delta APB} \right) = area\;\left( {\Delta BQC} \right)$

Note: Here we have make sure of using the concept of parallelogram instead of concept of quadrilateral.So here we can say that when parallelograms have same base as well as same parallels then the area of parallelogram is doubled of the area of triangle.