Question

A piece of ground measures \[19m{\text{ }}2.5{\text{ }}dm\] by \[12m{\text{ }}5dm\]. From the center of each side of path \[\;2{\text{ }}m\] wide goes across to the center of the opposite side. Find the cost of paving these paths \[\;Rs{\text{ }}12.32\] per ${m^2}$.

Answer 1

Hint: In this problem we have to find the area of the space. Then we can calculate the cost of paving these paths and multiply the area of the surface to the cost per ${m^2}$.
We know that \[\;\;1{\text{ }}dm{\text{ }} = {\text{ }}0.1{\text{ }}m\].

Formula used:
\[{\text{Area of rectangle = Length x Breadth}}\]

Complete step by step answer:
Here, it is given that dimensions of the ground is
Length \[ = AB = 19m{\text{ }}2.5{\text{ }}dm\]

Here, it is given that dimensions of the ground is
Length \[ = AB = 19m{\text{ }}2.5{\text{ }}dm\]
As we know that \[\;\;1{\text{ }}dm{\text{ }} = {\text{ }}0.1{\text{ }}m\] we get \[\;\;2.5dm{\text{ }} = {\text{ }}0.25m\]
Thus \[19m2.5dm = 19 + 0.25m\]
\[AB = 19.25m\]
Also it is given that
Breadth \[ = BC = 12m{\text{ }}5dm\]
As we know that \[\;\;1{\text{ }}dm{\text{ }} = {\text{ }}0.1{\text{ }}m\] we get \[\;\;5dm{\text{ }} = {\text{ }}0.5m\]
Thus \[12m5dm = 12.5m\]
\[BC = 12.5{\text{ }}m\]
Also given that\[\;2{\text{ }}m\] wide path goes across to the center of the opposite sides.
That is width of the path is \[\;w = 2m\]
From the figure we have to find the area of path \[EF{\text{ }} \times {\text{ }}w\]
\[{\text{Area of the path }}\left( {\text{A}} \right){\text{ = Area of EFGH + Area of PQRS - Area of Square KLMN}}\]
Here, Area of EFGH=
Area of \[PQRS = QR{\text{ }} \times {\text{ }}w\]
As they are in the shape of rectangles.
And, Area of Square \[KLMN = {w^2}\]
So, arrange all the areas and find area of the path
\[A{\text{ }} = {\text{ }}EF{\text{ }} \times {\text{ }}w{\text{ }} + {\text{ }}QR{\text{ }} \times {\text{ }}w{\text{ }} - {\text{ }}{w^2}\]… (1)
When we see the picture, \[EF = AB,QR = BC\]
Hence by replacing them in equation (1) we get,
\[\Rightarrow A = {\text{ }}AB{\text{ }} \times {\text{ }}w{\text{ }} + {\text{ }}BC{\text{ }} \times {\text{ }}w{\text{ }} - {\text{ }}{w^2}\]
Let us put the known values in the above formula we get,
\[\Rightarrow A = {\text{ }}19.25{\text{ }} \times {\text{ }}2{\text{ }} + {\text{ }}12.5{\text{ }} \times {\text{ }}2{\text{ }} - {\text{ }}{2^2}\]
Let us now solve the equation we get,
\[\Rightarrow A = {\text{ }}38.5{\text{ }} + {\text{ }}25{\text{ }} - {\text{ }}4\]
\[\Rightarrow A = {\text{ }}63.5{\text{ }} - {\text{ }}4\]
\[\Rightarrow A{\text{ }} = {\text{ }}59.5{\text{ }}{m^2}\]
Therefore, Area of the path \[A{\text{ }} = {\text{ }}59.5{\text{ }}{m^2}\]
Again we have to find the cost paving there path
The total cost paving the path is given by multiplying the cost given with the area found
That is the cost paving these paths \[ = 1.32 \times 59.5\]\[ = Rs.{\text{ }}78.54\]

Hence, the total cost paving these path is \[Rs.78.54\]

Additional information:
The area of a rectangle is calculated in units by multiplying the width (or breadth) by the Length of a rectangle. The area of the square will be equal to the square of side-length as the sides of the square are equal. We say that the area of the square is the square of the side length.

Note:
This problem is a real-life problem. So, we need to recreate the problem as a diagram it really helps the students to easily understand the problem and solution. We have to give the most preferred for the diagram which is clearly and representing all the parameters is the most important one.