Question

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

$Area(△AQD)=Area(△APD)+Area(△CPB)$

• A. True
• B. False

Answer 1

The correct option is A True

To prove
$Area\left(△AQD\right)=Area\left(△APD\right)+Area\left(△CPB\right)$
$Area\left(△APD\right)=Area\left(△APC\right)$ Area of triangle with same base AP and between same set of parallel lines are equal.
$Area\left(△APD\right)=Area\left(△CPB\right)$
$=Area\left(△APC\right)=Area\left(△CPB\right)$
$=Area\left(△ABC\right)$
$=\frac{1}{2}AreaofparallelogramABCD$( AC is diagonal and diagonals of parallelogram divides it into equal areas)
$Area\left(△AQD\right)=Area\left(△APD\right)+Area\left(△CPB\right)$ Proved