Question

The radius of a field in the form of a sector is \[21m\]. The cost of constructing a wall around the field is \[{\rm{Rs}}.1500\] at the rate of \[{\rm{Rs}}.20\] per meter.
What will be the cost of tilling the whole field at the costs of \[{\rm{Rs}}.15\] per square meter.

Answer 1

Hint:
Here we will find the wall constructed at the given cost using the unitary method. Then we will find the length of the minor arc. Then we will use the formula of the sector to find the area of the required sector. We will then multiply the rate of tilling to the area of the sector to get the tilling the whole field.

Complete step by step solution:
Here we need to find the cost of tilling the whole field at the cost of tilling the field.
It is given that the radius \[\left( r \right)\] of the sector is equal to \[21m\].
Let the length of the minor arc of the sector be \[l\].
It is given that the length of the wall constructed at the cost of \[{\rm{Rs}}.20\]\[ = 1m\]
Using the unitary method, we get
Length of the wall constructed at the cost of \[{\rm{Rs}}.1500\] \[ = \dfrac{1}{{20}} \times 1500 = 75m\]
Therefore, we have got the total length of the given sector and we know that the total length of the sector is equal to \[l + 2r\].
So we can write it as
\[ \Rightarrow l + 2r = 75\]
Now, we will substitute the value of radius, 21m of the sector in the above equation. Therefore, we get
\[ \Rightarrow l + 2 \times 21 = 75\]
On multiplying the numbers, we get
\[ \Rightarrow l + 42 = 75\]
Now, subtracting the number 42 from both sides, we get
\[ \Rightarrow l + 42 - 42 = 75 - 42\]
\[ \Rightarrow l = 33m\]
Now, we will find the area of the sector using the formula and the formula of the area of the sector is given by \[{\rm{Area}} = \dfrac{\theta }{{360}} \times \pi {r^2}\].
Now, we will use this formula here.
Area of the sector \[ = \dfrac{\theta }{{360}} \times \pi {r^2}\]
Now, we will rearrange the terms, we get
\[ \Rightarrow \] Area of the sector \[ = \dfrac{r}{2} \times \dfrac{{\pi r\theta }}{{180}}\]
We know that \[l = \dfrac{{\pi r\theta }}{{180}}\]
Now, we will substitute the value of the length of the minor arc here.
\[ \Rightarrow \] Area of the sector \[ = \dfrac{r}{2} \times l\]
Substituting \[r = 21\] and \[l = 33\] in the above equation, we get
\[ \Rightarrow \] Area of the sector \[ = \dfrac{{21}}{2} \times 33 = 346.5{m^2}\]
As it is given that the cost of tilling \[1{m^2}\] region \[ = {\rm{Rs}}.15\]
Using the unitary method here, we get
Cost of tilling \[346.5{m^2}\] region \[ = 346.5 \times {\rm{Rs}}.15 = {\rm{Rs}}.5197.5\]

Hence, the cost of tilling the whole field is \[{\rm{Rs}}.5197.5\].

Note:
Here the field was in the form of a sector. A sector is defined as the part of a circle enclosed by two radii of a circle and their intercepted arc. We have found out the cost using a unitary method. The unitary method is a method where first, the per-unit quantity is calculated, and then the number of units are multiplied. Here, we can multiply the cost of tilling by the Length of the wall constructed to find the cost of tilling the whole field. This will give us the wrong answer.