Question

In parallelogram $ABCD$ . $P$ is a point on side $AB$ and $Q$ is a point on side $BC$, then

$△CPD$ and $△AQD$ are equal in area.

• A. True
• B. False

Answer 1

The correct option is A True

Given,
ABCD is a parallelogram P is a point on side AB and Q is a point on side BC
Construction:Join diagonals AC and BD.
Proof:
$CPD=BCD$ [triangle with same base and between same set of parallel lines have equal area] (i)
Area $BCD=\frac{1}{2}$area of parallelogram ABCD.[daigonal of parallelogram bisects area in two equal parts] (ii)
From (i) and (ii)
Area$CPD=\frac{1}{2}$area ABCD (III)
Similarly,
Area $AQD=AreaABD$
=$\frac{1}{2}$area ABCD (IV)
From (iii) and (iv)
$AreaCPD=AreaAQD$