# Ex.11.3 Q5 Perimeter and Area - NCERT Maths Class 7

## Question

From a circular sheet of radius \(4 \rm\,cm\), a circle of radius \(3\rm\, cm\) is removed. Find the area of the remaining sheet. \(\begin{align}(\text{Take}\, π = 3.14)\end{align}\)

## Text Solution

**What is known?**

Radius of circular sheet from which a circle of radius \(3\rm\, cm\) is removed.

**What is the unknown?**

The area of the remaining sheet.

**Reasoning:**

To find the area of the remaining sheet, first find the area of both the circular sheets and then subtract the area of circle sheet of radius \(3\rm\, cm\) from the area of circular sheet of radius \(4 \rm\,cm\).

**Steps:**

Radius of circular sheet (r_{1}) \(=\) \(4 \rm\,cm\)

Radius of removed circle (r_{2}) \(=\) \(3\rm\, cm\)

Area of remaining sheet \(=\) Area of circular sheet \(-\) Area of removed circle

\[\begin{align} &= \pi {r_1}^2 - \pi {r_2}^2\\ &= \pi \left( {{r_1}^2 - {r_2}^2} \right)\\ &= 3.14{\rm{ }}\left[ {\left( {{4^2}} \right) - \left( {{3^2}} \right)} \right]\\ &= 3.14\left( {16 - 9} \right)\\ &= 3.14 \times 7\\ &= 21.98{\rm{ }}\rm\,c{m^2}\end{align}\]

Thus, the area of remaining sheet is \(21.98 \rm\,cm^2.\)

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