Question

In a parallelogram ABCD, P is a point in interior of parallelogram ABCD. If $ar\left(|{|}^{gm}ABCD\right)=18c{m}^2$, then $\left[ar\right(△APD)+ar(△CPB\left)\right]$ is:

• A. $9c{m}^2$
• B. $12c{m}^2$
• C. $18c{m}^2$
• D. $15c{m}^2$

Answer 1

The correct option is A $9c{m}^2$

$EF\parallel AB⟶\left(1\right)$ [By construction]

$\because AD\parallel BC$

$\because$ opposite sides of a $\parallel$gm are $\parallel$al.

$\therefore$ $AE\parallel BF⟶\left(2\right)$

From equation (1) and (2), quadrilateral ABEF is a $\parallel$gm. A quadrilateral is a $\parallel$gm if its opposite sites are parallel. Similarly, quadrilateral $CDEF$ is a $\parallel$gm.

$\because △APB$ and $\parallel$gm $ABFE$ are on the same base $AB$ and between the same $\parallel$als $AB$ and $EF$.

$\therefore ar\left(△APB\right)=\frac{1}{2}ar(\parallel gmABFE)⟶\left(3\right)$

$\because △PCD$ and $\parallel$gm $CDEF$ are on the same base $DC$ nd between the same $\parallel$als $DC$ and $EF$.

$\therefore ar\left(△PCD\right)=\frac{1}{2}ar(\parallel gmCDEF)⟶\left(4\right)$

Adding equations (3) and (4), we get

$ar\left(△APB\right)+ar\left(△PCD\right)=\frac{1}{2}[ar\parallel gmABEF+\parallel gmCDEF]$

$=\frac{1}{2}ar\parallel gmABCD$

$=\frac{1}{2}\times 18$

$=9{cm}^2$