Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# In the below figure, $ABCD$ is a parallelogram and $P$ is a mid-point of $AB$. If the area of a quadrilateral $APCD$ is $36c{m^2}$ , then what is the area of $\vartriangle ABC$?A.$36c{m^2}$B.$48c{m^2}$C.$24c{m^2}$D.None of these

Last updated date: 12th Aug 2024
Total views: 429.9k
Views today: 5.29k
Verified
429.9k+ views
Hint:Diagonal will bisect the area of the parallelogram and the line $PC$ will bisect the area of $\vartriangle ABC$.And in the quadrilateral $APCD$, area of $\vartriangle APC$ is one third of it. In this way you can find out the area of $\vartriangle ABC$.

Let us note down the given data initially,

$ABCD$ is a parallelogram and $P$ is a mid-point of $AB$.
area of quadrilateral $APCD$ is $36c{m^2}$.
We are asked to find out the area of $\vartriangle ABC$.
We know that in a parallelogram, Diagonal will bisect the area of the parallelogram.
So, area of $\vartriangle ADC$ $=$ area of $\vartriangle ABC$.
In the $\vartriangle ABC$, the side $PC$ is the median.
We all know that median also bisects the area of a triangle.
Hence area of $\vartriangle APC$ $=$ area of $\vartriangle PBC$.
From the above equations we can conclude that,
Area of quadrilateral $APCD$ $=$ $3 \times$ Area pf $\vartriangle APC$.
Then, Area of $\vartriangle APC$ $= \dfrac{1}{3} \times 36= 12c{m^2}$
Now,
Area of $\vartriangle ADC$ $=$ Area of quadrilateral $APCD$ $-$ Area of $\vartriangle APC$.
$= 36 - 12 \\ = 24c{m^2} \\$
But we already seen that the area of $\vartriangle ADC$ is equal to the area of $\vartriangle ABC$.
Hence, area of $\vartriangle ABC$ $=$ area of $\vartriangle ADC = 24c{m^2}$

So, the correct answer is “Option C”.

Note:We can do this problem in a little bit different manner. In the above solution we found the area of $\vartriangle ADC$ by removing the area of $\vartriangle APC$ as it is one third of that. Then as we know that diagonal bisects the area of parallelogram, the area of $\vartriangle ADC$ is equal to the area of $\vartriangle ABC$. We can notice one thing that the area of quadrilateral $APCD$ is $\dfrac{2}{3}$ of the required area. By solving this proportionality also, we will get the same solution.