Question

ABCD is a parallelogram. P is any point on CD. If $ar\left(△DPA\right)=15c{m}^2$ and $ar\left(△APC\right)=20c{m}^2,$ then $ar\left(△APB\right)$ =

• A. $15c{m}^2$
  
• B. $20c{m}^2$
  
• C. $35c{m}^2$
• D. $30c{m}^2$

Answer 1

The correct option is C

$35c{m}^2$

In ||gm ABCD, P is any pomt on CD
AP, AC and PB are joined
$ar\left(△DPA\right)=15c{m}^{2}ar\left(△APC\right)=20c{m}^2$
Adding, $ar\left(△ADC\right)=15+20=35c{m}^2$

$\because AC$ divides it into two triangle equal in area
$\therefore ar\left(△ACB\right)=ar\left(△ADC\right)=35c{m}^2$
$\because △APB$ and $△ACB$ are on the same base and between the same parallels
$\therefore ar\left(△APB\right)=ar\left(△ACB\right)=35c{m}^2$