Question

The area of a circular plot is 9856 sq.m. Calculate the cost of fencing the plot at the rate of Rs. 6 per$mete{r^2}$.

Answer 1

Hint: First of all using the formula of area of a circle ${\text{A}} = {\text{$\pi$ }}{{\text{r}}^2}$ we find the radius of the circular plot then using the radius find the circumference of a circular plot. To calculate the cost of fencing multiplies the given rate of per sq.m. into circumference in order to get the required answer which is cost.

Complete step by step answer:

Given: Area of a circular plot is 9856 sq.m.
Area of a circle $ = {\text{$\pi$ }}{{\text{r}}^2}$
On substituting the value of Area of circle we get,
$ \Rightarrow {\text{$\pi$ }}{{\text{r}}^2} = 9856$
On substituting the value of \[\pi = \dfrac{{22}}{7}\], we get,
$ \Rightarrow {{\text{r}}^2} = \dfrac{{9856}}{{\dfrac{{22}}{7}}}$
On further simplification we get,
$ \Rightarrow {{\text{r}}^2} = 3136$
\[ \Rightarrow \]${\text{r}} = \sqrt {3136} $
\[ \Rightarrow \]r = 56m.
Circumference of a circle $ = 2{\text{$\pi$ r}}$
$ = 2 \times \dfrac{{22}}{7} \times 56$\[ \ldots \]\[\left[ {\pi = \dfrac{{22}}{7}} \right]\]
$ = 2 \times 22 \times 8$
$ = 352{\text{m}}$
Cost of fencing of a circular plot per 1 meter $ = {\text{Rs}}.6$
∴ Cost of fencing of a circular plot of 325m $ = ?$
\[ \Rightarrow \]$\dfrac{{Rs.6 \times 325m}}{{1m}} = Rs.2112$
Hence, the cost of the fencing of a circular plot is Rs. 2112.

Note: In order to find the cost of fencing of the circular plot we have to calculate its circumference. The area of a circular plot is given because of using that we can find the radius of a given plot.
We can also understand this with unitary method as cost of \[{\text{x(metre)}}\]is \[{\text{Rs}}{\text{.y}}\]
So the cost for \[{\text{z(metre)}}\]is \[?\]
So, it will be given as \[{\text{ = }}\dfrac{{{\text{yz}}}}{{\text{x}}}\].
Many times the same question occurs with different values, in general, we have given the area of a circular plot is n sq.m. Calculate the cost of fencing the plot at the rate of Rs. x per$meter{r^2}$then we have to use the above method.