#### Question

Find the area of Fig. 20.35 as the sum of the areas of two trapezium and a rectangle.

#### Solution

In the given figure, we have a rectangle of length 50 cm and width 10 cm, and two similar trapeziums with parallel sides as 30 cm and 10 cm at both ends.

Suppose x is the perpendicular distance between the parallel sides in both the trapeziums.

We have:

Total length of the given figure = Length of the rectangle + 2 x Perpendicular distancebetween the parallel sides in both the trapeziums

\[70 = 50 + 2 \times x\]

\[2\times x=70-50=20\]

\[x=\frac{20}{2}=10 cm\]

Now, area of the complete figure = (area of the rectangle with sides 50 cm and 10 cm) + 2 x (area of the trapezium with parallel sides 30 cm and 10 cm, and height 10 cm

\[=(50\times10)+2\times[\frac{1}{2}\times(30+10)\times(10)]\]

\[ = 500 + 2 \times [200]\]

\[ = 900 {cm}^2\]