# Ex.11.3 Q13 Perimeter and Area - NCERT Maths Class 7

## Question

A circular flower bed is surrounded by a path \(4 \rm\,m\) wide. The diameter of the flower bed is \(66 \rm\,m\). What is the area of this path? \((π = 3.14)\)

## Text Solution

**What is known?**

A circular flower bed of diameter \(66 \rm\,m\) which is surrounded by a path \(4 \rm\,m\) wide.

**What is unknown?**

The area of the path.

**Reasoning:**

Since the diameter of the circular flower bed is \(66\rm\,m,\) so, the radius of the circular bed will be \(33\rm\,m\). The radius of the outer circle, which includes radius of the circular path as well width of the path is \(33\rm\,m + 4\rm\,m = 37\rm\,m\). Now the areas of the inner circle of radius \(33\rm\,m\) and outer circle of radius \(37 \rm\,m\) can be calculated and by subtracting area of the inner circle from the area of the outer circle you will get the area of this path.

**Steps:**

Diameter of the circular flower bed \(=\) \(66 \rm\,m\)

\(\therefore\) Radius of circular flower bed (r) \(=\) \(\begin{align}\frac{{66}}{2}\end{align}\) \(= 33 \rm\,m\)

\(\therefore\) Radius of circular flower bed with \(4 \rm\,m\) wide path \((R) = 33 + 4 = 37 \rm\,m\)

Area of path = Area of bigger circle-Area of smaller circle

\(\begin{align}&=\pi {{\rm{R}}^2} - \pi {r^2}\\&= \pi \left( {{{\rm{R}}^2} - {r^2}} \right)\\ &= 3.14\left[ {{{\left( {37} \right)}^2} - {{\left( {33} \right)}^2}} \right]\\ &= 3.14\left [{\left({37 + 33} \right)\left( {37-33} \right)} \right] \end{align}\)

Using this formula,

\(\begin{align} (a^2 -b^2)= [( {a + b} ) ( {a - b} )]\end{align}\)

\(\begin{align} & = 3.14 \times 70 \times 4\\ &= 879.20\,\rm {m^2}\end{align}\)

Therefore, the area of the path is \(879.20 \rm\,m^2.\)