The area ____ enclosed by each of the following figures is the sum of the areas of a rectangle and a trapezium.

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Detailed Solution

Let us split the given closed figure, ABCFED in two parts –
Therefore area of ABCFED = Area of Trapezium ABCD + Area of rectangle EFCD .... (a)
Now, Area of the trapezium is AT =

\frac{a + b}{2} \times h

Where “a” and “b” are the sides of the trapezium and “h” is the height of the trapezium.
Given that –
a= 7cm, b= 18cm, h= 8cm
Place the values in the formula of area

AT = (\frac{7 + 18}{2}) \times 8

Simplify the above equation –
AT =

(\frac{25}{2}) \times 8

Dividing common factors from the numerator and denominator –
AT= 25×4
AT= 100 cm2 .... (b)
Now, similarly find the area of the rectangle EFCD –
Since, both the sides of the rectangle are equal, apply the formula for the area of the square.
Area of the square is the product of its two sides.
As = l×l
Where Length of the side is, l= 18cm
Place the value –
As = 18×18
Simplification –
⇒As = 324 cm2 ..... (c)
Place values of the equation (b) and (c) in the equation (a)-
Area of ABCFED = 100+324
⇒Area = 424 cm2
Hence, the area of enclosed figure is 424 cm2
So, the correct answer is “ 424 cm2 ”.

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