Question

In the given figure, if the area of triangle $ADE$ is $60\text{ c}{{\text{m}}^{2}}$ , Now find the area of following figures: A. $\parallel gm\text{ }ABED$ B. Rectangle $ABCF$ C. $\Delta ABE$

Answer 1

Hint: For this question, we will first draw a diagonal $AE$ and then apply the properties of a parallelogram to find the areas of different figures given in the question. After applying the properties of a parallelogram write down the area in square units and get the answers ultimately.

Complete step by step answer:

Let’s consider the figure:

Now, we will start with the first part, now we have to find the area of $\parallel gm\text{ }ABED$ :
Now we see that $\Delta ADE$ and $\parallel gm\text{ }ABED$ are on the same base $AB$ and between the same parallels $DE\parallel AB$ therefore the area of triangle $\Delta ADE$ is half the area of $\parallel gm\text{ }ABED$.
$\text{Area of }ABED=2\left( \text{Area of }ADE \right)=2\times 60=120\text{ c}{{\text{m}}^{2}}$
Now we will see the second part, we have to find the area of rectangle $ABCF$ : we know that according to the theorem the area of a parallelogram is equal to the area of a rectangle on the same base and of the same altitude that is between the same parallels.
Therefore, $\text{Area of }ABCF=\text{Area of }ABED=120\,\text{c}{{\text{m}}^{\text{2}}}$
Now, we will go for the final part what we have to find here in the area of $ADE$ , now we know that in a parallelogram, a diagonal bisects a parallelogram into two equal triangles. So, we have $\parallel gm\text{ }ABED$ and the diagonal $AE$ bisects it into $\Delta ADE$ and $\Delta ABE$, therefore they both have the same area. Therefore, $\text{Area of }ABE=\text{ Area of }ADE=60\text{ c}{{\text{m}}^{2}}$.
Hence, the answer of all parts is: A. $\parallel gm\text{ }ABED=120\text{ c}{{\text{m}}^{2}}$ B. Rectangle $ABCF=120\text{ c}{{\text{m}}^{2}}$ C. $\Delta ABE=60\text{ c}{{\text{m}}^{2}}$
Note:
  For the last part that is an area of $\Delta ABE$ can be found out by another approach that is we know that the area of triangles on the same base and between the same parallel lines are equal that is: $\text{Area of }ABE=\text{Area of }ADE=60\text{ c}{{\text{m}}^{2}}$ . Always mention the units while writing the area of the figure.