ABCD is a parallelogram. P is any point on CD. If ar( )=15cm^2 and ar( )=20cm^2 then ar( )=

ABCD is a parallelogram. P is any point on CD. If ar( )=15cm^2 an…

MCQ

$ABCD$ is a parallelogram. $P$ is any point on $CD$. If $\text{ar}(\triangle\text{DPA})=15\text{cm}^2$ and $\text{ar}(\triangle\text{APC})=20\text{cm}^2$ then $\text{ar}(\triangle\text{APB})=$

A
$15 \mathrm{~cm}^2$
B
$20 \mathrm{~cm}^2$
✓
$35 \mathrm{~cm}^2$
D
$30 \mathrm{~cm}^2$

✓

Answer

Correct option: C.

$35 \mathrm{~cm}^2$

Area of trapazium $ABCP$ = Area of $\triangle\text{APB}$ + ar of $\triangle\text{BPC}\ ...(1)$
$\triangle\text{APC}$ and $\triangle\text{BPC}$ have same base $PC$ and are batween same parallels.
$\Rightarrow $ Area of $\triangle\text{APC}$ = Area of $\triangle\text{BPC}$ = $20 \mathrm{~cm}^2$ $...(2)$
From figure, $\text{Ar}(\triangle\text{ADP})+\text{Ar}(\triangle\text{APC})=\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{ABCD})$
$\Rightarrow\text{Ar}(||^{\text{gm}}\text{ABCD})=2(20+15)=70\text{cm}^2$
Area of trapazium ABCP $=\text{Ar}(||^{\text{gm}}\text{ABCD})-\text{Ar}(\triangle\text{ADP)}=70-15=55\text{cm}^2$
$\Rightarrow $ Area of $\triangle\text{APB}$ Area of trapezium $ABCP$ - Area of $\triangle\text{BPC}$
$=(55-20)\text{cm}^2$ [From $(1)$]
$=35\text{cm}^2$
Hence, correct option is $(c)$.