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Find the area of the given figure?
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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.

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