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**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 $\vartriangle ABC$.

**Complete step-by-step answer:**

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.

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