Question

The adjacent sides of a parallelogram ABCD measures 34 cm and 20 cm and the diagonal AC measures 42 cm. Find the area of the parallelogram.

Answer 1

Hint: We have been given the adjacent sides of a parallelogram ABCD. We know that the opposite sides in a parallelogram are the same, so we will get all the four sides of the parallelogram from this relation. We also have the length of a diagonal, so we will join the diagonal to form two triangles ABC and ADC. And to find the area of the parallelogram, we will find the area of the triangles ABC and ADC by using the Heron’s formula, which is given as, $\sqrt{S\left( S-a \right)\left( S-b \right)\left( S-c \right)}$, where a, b and c are the lengths of the sides of the triangle and S is the semi-perimeter, that is given by, $S=\dfrac{a+b+c}{2}$. We will then add the area of both the triangles to get the area of the parallelogram.

Complete step-by-step answer:
We have been given the measures of the adjacent sides and diagonal of a parallelogram as, 34 cm, 20 cm and 42 cm respectively and have to find the area of the parallelogram. We will first represent the parallelogram as shown below.

In the above figure, we have joined the diagonal AC and that divides the parallelogram into two triangles ABC and ADC. So, we will find the areas of these triangles and add them to find the area of the parallelogram. We will use Heron’s formula to find the area of the triangle ABC and ADC, which is given as, $\sqrt{S\left( S-a \right)\left( S-b \right)\left( S-c \right)}$, where a, b and c are the lengths of the sides of the triangle and S is the semi-perimeter, that is given by, $S=\dfrac{a+b+c}{2}$.
So, we will first find the area of triangle ABC. Here we have a = 20 cm, b = 34 cm and c = 42 cm. So, we get the value of S as,
 $\begin{align}
  & S=\dfrac{a+b+c}{2} \\
 & \Rightarrow S=\dfrac{20+34+42}{2} \\
 & \Rightarrow S=\dfrac{96}{2} \\
 & \Rightarrow S=48 \\
\end{align}$
Now, we will use Heron’s formula to find the area of triangle ABC. So, we get,
$\begin{align}
  & Area\text{ }of\text{ }\Delta ABC=\sqrt{48\left( 48-20 \right)\left( 48-34 \right)\left( 48-42 \right)} \\
 & \Rightarrow Area\text{ }of\text{ }\Delta ABC=\sqrt{48\times 28\times 14\times 6} \\
 & \Rightarrow Area\text{ }of\text{ }\Delta ABC=\sqrt{112896} \\
 & \Rightarrow Area\text{ }of\text{ }\Delta ABC=336c{{m}^{2}}\ldots \ldots \ldots \left( i \right) \\
\end{align}$
Now, we will find the area of triangle ADC that has the same sides as triangle ABC. So, since it has the same sides, the area of triangle ADC will be equal to area of triangle ABC, so we get,
$Area\text{ }of\text{ }\Delta ADC=336c{{m}^{2}}\ldots \ldots \ldots \left( ii \right)$
So, now we can find the area of the parallelogram by taking the sum of both the areas of the triangles ABC and ADC. So, from equation (i) and (ii), we get,
Area of parallelogram ABCD $=336+336=672c{{m}^{2}}$.
Hence, the area of the parallelogram ABCD is $672c{{m}^{2}}$.

Note: The students may try to find the area of the parallelogram by using the formula for the area of the parallelogram, that is, $area=base\times height$. But that is not appropriate here. We should use Heron's formula as all the sides are available and it is easier to find the area of the two triangles formed by the diagonal and adding them to get the area of the parallelogram.