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
$15cm^2$
B
$20cm^2$
✓
$35cm^2$
D
$30cm^2$

✓

Answer

Correct option: C.

$35cm^2$

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.

⇒ Area of $\triangle\text{APC}$ = Area of $\triangle\text{BPC}$ = $20cm^2$ ...(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 $=\text{Ar}(||^{\text{gm}}\text{ABCD})-\text{Ar}(\triangle\text{ADP)}=70-15=55\text{cm}^2$

⇒ 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).