Find the area of the trapezium ABCD as given in the figure in which ADCE is a rectangle?
Answer
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Hint: We start solving the problem by finding the length and breadth of the rectangle ADCE and recalling the definition of area of the rectangle with length ‘l’ units and breadth ‘b’ units is \[A=l\times bsq.units\]. We find the area of the rectangle using this formula. We then find the base length and height of the triangle ECB and use the definition of area of the triangle with base length ‘b’ units and height ‘h’ units is $A=\dfrac{1}{2}\times b\times hsq.units$. We then add the obtained area of the rectangle and triangle to get the area of trapezium ABCD.
Complete step by step answer:
According to the problem, we need to find the area of the given trapezium ABCD in which ADCE is rectangle.
Let us redraw the given figure.
From the figure, we can see that area of trapezium ABCD = area of rectangle ADCE + area of triangle EBC ---(1).
We can see that the rectangle ADCE has length 8 cm and breadth 3 cm. We know that the area of the rectangle with length ‘l’ units and breadth ‘b’ units is \[A=l\times bsq.units\].
Let us assume the area of the rectangle to be ${{A}_{1}}$. So, we get ${{A}_{1}}=8\times 3=24c{{m}^{2}}$ ---(2).
We can see that the triangle ECB has base length 8 cm and height 3 cm. We know that the area of the triangle with base length ‘b’ units and height ‘h’ units is $A=\dfrac{1}{2}\times b\times hsq.units$.
Let us assume the area of triangle ${{A}_{2}}=\dfrac{1}{2}\times 8\times 3=12c{{m}^{2}}$ ---(3).
From equation (1), we get the area of trapezium ABCD = ${{A}_{1}}+{{A}_{2}}$.
Area of trapezium ABCD = $\left( 24+12 \right)c{{m}^{2}}$.
Area of trapezium ABCD = $36c{{m}^{2}}$.
So, we have found the area of the trapezium ABCD as $36c{{m}^{2}}$.
∴ The area of the trapezium ABCD is $36c{{m}^{2}}$.
Note: Whenever we are asked to find the area of trapezium, we need to divide it into the two or figures which will be suitable for finding the area. We should not make calculation mistakes while solving this problem. Similarly, we can expect problems to find the perimeter of the trapezium ABCD, in which we need to add the lengths of the all sides of the trapezium with dividing it into two or more figures as shown.
Complete step by step answer:
According to the problem, we need to find the area of the given trapezium ABCD in which ADCE is rectangle.
Let us redraw the given figure.
From the figure, we can see that area of trapezium ABCD = area of rectangle ADCE + area of triangle EBC ---(1).
We can see that the rectangle ADCE has length 8 cm and breadth 3 cm. We know that the area of the rectangle with length ‘l’ units and breadth ‘b’ units is \[A=l\times bsq.units\].
Let us assume the area of the rectangle to be ${{A}_{1}}$. So, we get ${{A}_{1}}=8\times 3=24c{{m}^{2}}$ ---(2).
We can see that the triangle ECB has base length 8 cm and height 3 cm. We know that the area of the triangle with base length ‘b’ units and height ‘h’ units is $A=\dfrac{1}{2}\times b\times hsq.units$.
Let us assume the area of triangle ${{A}_{2}}=\dfrac{1}{2}\times 8\times 3=12c{{m}^{2}}$ ---(3).
From equation (1), we get the area of trapezium ABCD = ${{A}_{1}}+{{A}_{2}}$.
Area of trapezium ABCD = $\left( 24+12 \right)c{{m}^{2}}$.
Area of trapezium ABCD = $36c{{m}^{2}}$.
So, we have found the area of the trapezium ABCD as $36c{{m}^{2}}$.
∴ The area of the trapezium ABCD is $36c{{m}^{2}}$.
Note: Whenever we are asked to find the area of trapezium, we need to divide it into the two or figures which will be suitable for finding the area. We should not make calculation mistakes while solving this problem. Similarly, we can expect problems to find the perimeter of the trapezium ABCD, in which we need to add the lengths of the all sides of the trapezium with dividing it into two or more figures as shown.
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