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In the previous lesson, we have calculated the areas of different figures separately. Let us now try to calculate the areas of some combinations of plane figures.

We come across these types of figures in our daily life and also in the form of various exciting designs. Flower beds, drain covers, window designs, and table covers are some examples.

We come across these types of figures in our daily life and also in the form of various exciting designs. Flower beds, drain covers, window designs, and table covers are some examples.

This chapter's learning objective is that when an object forms two or more shapes like a circle, semi-circle, triangle and rectangle, we determine to find the unknown or shaded area using the known values.

We illustrate the process of calculating areas of these figures through some examples.

We illustrate the process of calculating areas of these figures through some examples.

Example:

Consider a below rectangular flower bed with semi-circular ends. Find the area of the flower bed, if \(A = 17\) \(cm\), \(B = 4\) \(cm\).

Let's find the radius of the semi-circle using the diameter.

The radius of the semi-circle \(r =\) \(\frac{4}{2}\) \(= 2\) \(cm\).

We know that area of the rectangle \(= l × b\) square units.

The area of the rectangle \(= 17 × 4\).

And the area of two semi-circles $=\frac{\mathrm{\pi}{r}^{2}}{2}+\frac{\mathrm{\pi}{r}^{2}}{2}=2\times \frac{\mathrm{\pi}{r}^{2}}{2}$ square units.

\(\text{The area of the flower bed \(=\) Area of rectangle \(+\) Area of two semi-circles}\).

The area of the flower bed $=l\times b+\overline{)2}\times \frac{\mathrm{\pi}{r}^{2}}{\overline{)2}}=l\times b+\mathrm{\pi}{r}^{2}$.

Let's substitute the known values.

$\begin{array}{l}=17\times 4+(\frac{22}{7}\times 2\times 2)\\ =68+(\frac{22}{7}\times 4)\\ =68+(\frac{88}{7})\\ =68+12.6\\ =80.6\phantom{\rule{0.147em}{0ex}}{\mathit{cm}}^{2}\end{array}$

**Therefore**,

**the area of the flower bed**\(=\) 80.6 \(cm^2\).