Radius of circular garden is \[{\text{9 m}}\]. A circular path of width of \[{\text{3}}{\text{.5m}}\] is laid around and outside the garden. Find it’s area.

Hint: Here we are given a circular garden of radius \[{\text{9 m}}\] and the difference between outer path and inner circular path is mentioned as \[{\text{3}}{\text{.5m}}\]. So to find the required area calculate the area of both the circles and subtract the area of an outer circle from the inner circle, in order to get the area of the green shaded portion.

Complete step by step solution:

So from here we can say that , the radius of ground i.e. r is \[{\text{9 m}}\] and the radius of outer ground, R is \[{\text{9 + 3}}{\text{.5 = 12}}{\text{.5m}}\] (as mention that the difference mong both is of \[{\text{3}}{\text{.5m}}\]).
Now, we calculate the area of both the field and subtract the area among them as to get the area of the path,
So, \[{\text{A =} \pi }{{\text{R}}^{\text{2}}}{ - \pi }{{\text{r}}^{\text{2}}}\]
On taking \[{ \pi }\] common, we get,
\[ \Rightarrow {\text{A =} \pi (}{{\text{R}}^{\text{2}}}{\text{ - }}{{\text{r}}^{\text{2}}})\]
On substituting the values we get,
\[ \Rightarrow {\text{A =} \pi }\left[ {{{\left( {\dfrac{{{\text{25}}}}{{\text{2}}}} \right)}^{\text{2}}}{\text{ - (9}}{{\text{)}}^{\text{2}}}} \right]\]
On further simplification we get,
\[ \Rightarrow {\text{A =} \pi }\left[ {\dfrac{{{\text{625}}}}{{\text{4}}}{\text{ - 81}}} \right]\]
On taking LCM and simplification we get,
\[ \Rightarrow {\text{A = }}\dfrac{{{\text{301}}}}{{\text{4}}}{ \pi }{{\text{m}}^{\text{2}}}\]
On substituting the value of \[ \pi = 3.14 \], we get,
\[ \Rightarrow {\text{A = 236}}{\text{.285}}{{\text{m}}^{\text{2}}}\]

Hence, the area of circular path is \[{\text{236}}{\text{.285}}{{\text{m}}^{\text{2}}}\]




Note: The area of a circle is $ \pi $ times the radius squared.
Some of the important properties of the circle are as follows:
1) The circles are said to be congruent if they have equal radii.
2) The diameter of a circle is the longest chord of a circle.
3) Equal chords and equal circles have equal circumference.
4) The radius drawn perpendicular to the chord bisects the chord.