Question

The perimeter of the rectangular field is \[2.5\] km. What is the greatest possible area it may contain?

Answer 1

Hint:
We are required to find the greatest possible area from the rectangle, whose perimeter is given to us. To do that, we will first find the sides, based upon the perimeter. And then based upon the sides, we will calculate the greatest area.
Formula Used: We will use the following formulas in this question –
\[{\text{Perimeter of a square}} = 4a\] unit, where, \[a\] is the side of the square.
\[{\text{Area of a square}} = {a^2}\] unit sq., where, \[a\] is the side of the square.

Complete step by step solution:
We are given that the perimeter of the rectangular field is \[2.5\] km. We are required to find the greatest possible area. To do that we will have to find the sides first.
We know that out of all the rectangles with the same perimeter, the square has the greatest area. We will use the formula for the perimeter of the square to find the sides.
Since we are given the perimeter as \[2.5\] km, so we will substitute \[2.5\] for the perimeter of the square in the equation \[{\text{Perimeter of a square}} = 4a\].
\[\begin{array}{l}2.5 = 4a\\ \Rightarrow a = \dfrac{{2.5}}{4}\end{array}\]
Dividing the terms, we get
\[a = 0.625\]
Thus the side of the square is \[0.625\] km.
Now, we will find the Area of the square.
We will substitute \[a = 0.625\] in the equation \[{\text{Area of a square}} = {a^2}\].
\[{\text{Area of a square}} = {\left[ {0.625} \right]^2}\]
Apply the exponent on the term, we get
\[{\text{Area of a square}} = 0.390625{\text{ k}}{{\text{m}}^2}\]
Thus, the area of the greatest rectangle with perimeter \[2.5\] km is \[0.390625k{m^2}\].

Note:
Square is a special case of a rectangle. It is a special rectangle which has all of its sides equal. Of all the rectangles that have the same perimeter, a square always has the maximum area. This can be proved as shown below –
Let there be a rectangle with sides \[a\] and \[b\], where \[a > b\].
Let there be a square with side \[c\].
If the perimeter of the square is equal to the perimeter of the rectangle then,
\[a + b = 2c\]
Since \[a > b\] this implies that \[a > c\] and \[b < c\]. Thus, for any non-negative number \[y\],
\[\begin{array}{l}a = c + y\\b = c - y\end{array}\]
Now, find the area of the rectangle,
\[\begin{array}{l}{\text{Area of rectangle}} = a \times b\\ = \left[ {c + y} \right]\left[ {c - y} \right]\\ = {c^2} - {y^2}\end{array}\]
Hence, we can see that the area of the rectangle is lesser than that of the square by the square of the difference of the sides. Thus, the square has the greatest length amongst all the rectangles with the same perimeter.