# Ex.12.3 Q11 Areas Related to Circles Solution - NCERT Maths Class 10

## Question

On a square handkerchief, nine circular designs each of radius \(7\, \rm{cm}\) are made (see Figure). Find the area of the remaining portion of the handkerchief.

## Text Solution

**What is known?**

On a square handkerchief, \(9\) circular designs each of radius \({(r)} = \rm7cm\) are made.

**What is unknown?**

Area of the remaining portion of the handkerchief.

**Reasoning:**

Radius of circular design \(\text{= 7 cm}\)

\(\therefore \;\)Diameter each circular design \(=\,7 \times 2 = 14{\text{cm}}\)

Visually and logically since all the \(3\) circular design are touching each other and cover the entire length of the square.

\(\therefore \;\)Side of the square(s)= three times diameter of circular design

\[\begin{align}= 3 \times 14cm = 42cm\end{align}\]

Also, visually from the figure it is clear that:

Area of the remaining portion of handkerchief \(= \) Area of square \( - \,\,9 \times \) (Area of each circular design)

\[ = {s^2} - 9\left( {\pi {r^2}} \right)\]

Which can be easily solved since side \((s)\) of square and radius \({(r)}\) are known.

**Steps:**

Radius of each circular design \(\rm{}r = 7\; \rm{cm}\)

\(\because\;\) Diameter of each circular design \(\rm{}= 2r =2\times 7\,cm=\text{ 14 cm}\)

From the figure, it is observed that

Side of the square \(\begin{align}s = 3 \times 14cm = 42cm\end{align}\)

\[\begin{align}&= 14 \,\rm{cm}\, + 14 cm + 14 \,\rm{cm}\\ &= 42\,\rm{cm}\end{align}\]

Area of the remaining portion of the handkerchief = Area of square \( - \,9\; \times \) (Area of each circular design)

\[\begin{align} &={{s^2} - 9\left( {\pi {r^2}} \right)}\\ &= {{{(42)}^2} - 9 \times \frac{{22}}{7} \times {{(7)}^2}}\\&= {1764 - 9 \times 22 \times 7}\\ &= {1764 - 1386}\\&= {378\,\,{\text{c}}{{\text{m}}^2}}\end{align}\]