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# Find the area of the given figure?A. $28c{{m}^{2}}$B. $24c{{m}^{2}}$C. $32c{{m}^{2}}$D. $36c{{m}^{2}}$

Last updated date: 19th Jul 2024
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Hint: We draw the perpendicular line $EB\bot AC$. We break into rectangle and right-angled triangles and find their individual areas. We use the formulas of the area to find them and add to find the final solution.

We first need to break the given image of the trapezium into two parts: a rectangle and triangle according to the following image.

We draw $EB\bot AC$. Therefore, due to the parallel sides of the quadrilateral EBCD, it becomes rectangular as it has angles equal to ${{90}^{\circ }}$. We also have a right-angle triangle $\Delta ABE$.
We find individual areas and add them to find the area of the given figure in the problem.
For the rectangle EBCD the opposite sides are equal to each other.
Therefore, $ED=BC=7$ and also $EB=DC=4$.
The formula of area of rectangle is multiplication of unequal sides.
So, the area of EBCD is $4\times 7=28c{{m}^{2}}$.
The side AB can be found from the length of the sides AC and BC.
We have $AB=AC-BC=9-7=2$.
The formula of the area of the right-angle triangle is half of the multiplication of sides holding the right angle.
So, the area of $\Delta ABE$ is $\dfrac{1}{2}\times 4\times 2=4c{{m}^{2}}$.
Therefore, the area of EACD is the sum of the area of EBCD with the area of $\Delta ABE$.
This gives $ar\left( EACD \right)=28+4=32c{{m}^{2}}$.

So, the correct answer is “Option C”.

Note: The quadrilateral cannot be square as it doesn’t have equal sides although it has angles of ${{90}^{\circ }}$. The area for regular trapezium comes from this where we have half of the multiplication of sum of the opposite non-oblique sides and the height of the figure.