Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn the figure,$ar(DRC) = ar(DPC)$ and $ar(BDP) = ar(ARC)$. Show that both the quadrilaterals $ABCD$ and $DCPR$ are trapeziums.
(N/A) We have $ar(\Delta DRC) = ar(\Delta DPC)$ [Given].
Since they are on the same base $DC$ and have equal areas,they must lie between the same parallels.
$\Rightarrow DC \parallel RP$.
Since one pair of opposite sides of quadrilateral $DCPR$ is parallel,$DCPR$ is a trapezium.
Again,we have $ar(\Delta BDP) = ar(\Delta ARC)$ $(1)$.
Also,$ar(\Delta DPC) = ar(\Delta DRC)$ $(2)$.
Subtracting $(2)$ from $(1)$,we get:
$[ar(\Delta BDP) - ar(\Delta DPC)] = [ar(\Delta ARC) - ar(\Delta DRC)]$
$\Rightarrow ar(\Delta BDC) = ar(\Delta ADC)$.
Since they are on the same base $DC$ and have equal areas,they must lie between the same parallels.
$\Rightarrow AB \parallel DC$.
Since one pair of opposite sides of quadrilateral $ABCD$ is parallel,$ABCD$ is a trapezium.
Since they are on the same base $DC$ and have equal areas,they must lie between the same parallels.
$\Rightarrow DC \parallel RP$.
Since one pair of opposite sides of quadrilateral $DCPR$ is parallel,$DCPR$ is a trapezium.
Again,we have $ar(\Delta BDP) = ar(\Delta ARC)$ $(1)$.
Also,$ar(\Delta DPC) = ar(\Delta DRC)$ $(2)$.
Subtracting $(2)$ from $(1)$,we get:
$[ar(\Delta BDP) - ar(\Delta DPC)] = [ar(\Delta ARC) - ar(\Delta DRC)]$
$\Rightarrow ar(\Delta BDC) = ar(\Delta ADC)$.
Since they are on the same base $DC$ and have equal areas,they must lie between the same parallels.
$\Rightarrow AB \parallel DC$.
Since one pair of opposite sides of quadrilateral $ABCD$ is parallel,$ABCD$ is a trapezium.
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