**Question:**

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

**Solution:**

The given figure is:

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

Suppose $\mathrm{x}$ is the perpendicular distance between the parallel sides in both the trapeziums.

We have:

Total length of the given figure $=$ Length of the rectangle $+2 \times$ Perpendicular distance between the parallel sides in both the trapeziums

$70=50+2 \times \mathrm{x}$

$2 \times \mathrm{x}=70-50=20$

$\mathrm{x}=\frac{20}{2}=10 \mathrm{~cm}$

Now, area of the complete figure $=$ (area of the rectangle with sides $50 \mathrm{~cm}$ and $10 \mathrm{~cm}$ ) $+2 \times$ (area of the traperium with parallel sides $30 \mathrm{~cm}$ and $10 \mathrm{~cm}$, and height $10 \mathrm{~cm}$ ) $=(50 \times 10)+2 \times\left[\frac{1}{2} \times(30\right.$