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# A path $1m$ wide is built inside a square park of side $30m$ along its sides. Find the area of the path. Calculate the cost of constructing the path at the rate of Rs.$70$ per ${m^2}$.

Last updated date: 14th Jun 2024
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Hint: Here is the problem of mensuration we have. We have to find the area of a particular shape. We are approaching the solution by subtracting the area of small Square from the area of the bigger square (Fig.1). Also here we have to find the cost of constructing the path as the rate is given. To find the cost we will use the unitary method.

Suppose there are two squares ABCD and PQRS also ABCD is the boundary of the square park and the area between the two squares ABCD and PQRS is the mentioned path in the question. We need to find the area of the path and the cost of constructing the path.
Area of the path = Area of square ABCD - Area of the square PQRS … equation (1)
To find the area of the two squares we need to know the length of the side of both squares.
Length of one side of square ABCD $= 30m$ [Given]
Length of one side of square PQRS $= 30 - 1 = 29m$ [ the width of the path is 1m]
We know that the area of the square is $A = {a^2}$ where $A$ stands for the area of the square and $a$ stands for the side of the square. Therefore,
Area of Square ABCD $= 4 \times {(30)^2} = 4 \times 900 = 3600{m^2}$
Area of Square PQRS $= 4 \times {(29)^2} = 4 \times 841 = 3364{m^2}$
Now, from equation (1),
Area of the path = Area of square ABCD - Area of the square PQRS
Area of the path$= 3600 - 3364 = 236{m^2}$
We have to also find the cost of constructing the path at the rate of Rs.$70$per${m^2}$.
Cost of constructing $1{m^2}$path $= 70Rs.$
$\therefore$ Cost of constructing $236{m^2}$path $= 236 \times 70 = 16,520Rs.$
Hence the area of the path is $236{m^2}$ and the cost of constructing the path at the rate of Rs.$70$per${m^2}$ is $16520Rs.$

Note:
Second method-
We can also find the area of the path by dividing the shape into rectangles as,
Area of Shape of the path$= Area(rect.AWZD) + Area(rect.SZRY) + Area(rect.YCBX) + Area(rect.XQPW)$