Question

In the above figure, $ABCD$ is a parallelogram and $P$ is mid-point of $AB$. If $\text{Area}\left(APCD\right)=36c{m}^2$, then $\text{Area}\left(△ABC\right)=$?

• A. $36c{m}^2$
• B. $48c{m}^2$
• C. $24c{m}^2$
• D. None of these

Answer 1

The correct option is C $24c{m}^2$

Given, ABCD is a parallelogram having P as mid point of AB.
Join the diagonal AC and PC.
We know that the diagonal of a parallelogram divides it into two triangles of equal area.
Therefore, Ar (ABC) = Ar (ADC) = x ${cm}^2$(let)
Also, let area of APC = BPC = y ${cm}^2$(let) [Again, the median of a triangle divides a triangle into two triangles of equal area]
Given, Ar(APCD) = 36 ${cm}^2$
$⟹Ar\left(ACD\right)+Ar\left(APC\right)=36⟹x+y=36...\left(1\right)$
Again, Ar (ABC) = Ar (ADC)
or, Ar (ABC) = Ar (APC) + Ar (BPC)
or, x = y + y
or, x = 2y .... (2)
Putting value of x in (1), we get
$2y+y=36⟹y=12{cm}^2$
Therefore, x = 24 ${cm}^2$
Hence, Ar(ABC) = 24 ${cm}^2$