Question

In the figure, $\text{AB = 25 cm}$, $\text{DN = 20 cm}$. Segment $\text{DM}$ and $\text{DN}$ are the altitudes. If the area of the parallelogram $\text{ABCD = 300 sq cm}$, then the area of $\Delta \text{ABD}$ is?

Answer 1

Hint: We are given the length of the base of parallelogram $\text{ABCD}$. We will use the formula for the area of a parallelogram to find the length of the height. We will use the formula for the area of the triangle to find the area of $\Delta \text{ABD}$. The height of this triangle is the same as the height of the parallelogram.

Complete step-by-step solution:
The formula for the area of parallelogram is given by,
$\text{Area of parallelogram = base }\times \text{ height}$
In parallelogram $\text{ABCD}$, the base is segment $\text{AB}$ and the height is segment $\text{DM}$. We know that the area of the parallelogram $\text{ABCD = 300 sq cm}$, and the length of the base is $\text{AB = 25 cm}$.
Substituting these values in the formula for the area of parallelogram, we get
$300=25\times \text{DM}$
Solving the above equation, we get $\text{DM = }\dfrac{300}{25}=12\text{ cm}$.
The formula to find the area of a triangle is given by
$\text{Area of a triangle = }\dfrac{1}{2}\times \text{base}\times \text{height}$
Now, in $\Delta \text{ABD}$, the base is segment $\text{AB}$ and the height is segment $\text{DM}$. We know that the length of the base is $\text{AB = 25 cm}$ and the length of the height is $\text{DM = 12 cm}$. Substituting these values in the above formula, we get
$\begin{align}
  & \text{Area of }\Delta \text{ABD = }\dfrac{1}{2}\times 25\times 12 \\
 & =25\times 6 \\
 & =150\text{ sq cm}
\end{align}$
Therefore, the area of $\Delta \text{ABD}$ is $150\text{ sq cm}$.

Note: It is essential that we know the formulae for the areas of various standard objects. Looking at the diagram, we should be able to comprehend the relations between the given information regarding lengths with the length or area we are supposed to calculate. The parallelogram can also be seen as two triangles that are formed due to the diagonal. So, the area of one triangle out of the two will be half of the area of the parallelogram.