Question

A circular flower bed is surrounded by a path 4m hide the diameter of the flowe bed is 66.What is the area of the path.

Answer 1

$\underline{\underline{\huge{\textbf{Solution:}}}}$

$\begin{center}\setlength{\unitlength}{1.3 pt}\begin{picture}(100,100)(0,0)\put(0,0){\circle{50}} \put(0,0){\circle{22}} \put(0,0){\circle*{1.5}}\put(-16,0){\line(1,0){31.5}}\put(10,1.5){ 4 }\put(-8,1.5){ 66 }\put(-15,1.5){ 4 }\end{picture}\setlength{\unitlength}{1 pt}\end{center}$

Given,

Diameter of the Circular Flower bed, r= 66 m

Radius of the circular bed = $\mathbf{\frac{66\: m}{2}}$ = 33 m

Width of the Circular Path = 4 m

Radius of the Circular Flower bed with 4 m wide path, R = 33 + 4 = 37 m

$\textbf{Area\: of\: the\: 4m\: wide\: path}$ = $\textbf{Area\: of\: Outer\: Circle}$ - $\textbf{Area\: of\: Inner\: Circle}$

$=\pi{r}^{2} - \pi {R}^{2}$

$\implies \pi \left( r^2 - R^2 \right)$

$\implies \pi \left( 37^2 - 33^2 \right)$

Using the Identity

$\LARGE{\boxed{\quad\bigstar \; \; \mathbf{(a-b)^{2} = (a+b)(a-b)}}}$

$\implies \pi \: (37+33)(37-33)$

$\implies 3.14 \times 70 \times 4$

$\implies 879.2\: m^{2}$

$\therefore$

$\blacksquare\;\: \textsf{Area \: of \: the \: circular \: path} \: =\: \LARGE{\underline{\Large{\mathbf{879.2 \: m^{2}}}}}$