Question

The area of a square and a rectangle are equal. If side of the square is $40cm$ and breadth of the rectangle is $25cm$, find the length of the rectangle
(a) $81cm$
(b) $65cm$
(c) $64cm$

Answer 1

Hint: The areas of the square and the rectangle are given to be equal. So we have to equate the areas of the square and the rectangle. For this we need to use the formula for the area of the square which is given by ${{A}_{S}}={{a}^{2}}$ where $a$ is the side of the square. Also , we need to use the formula for the area of the rectangle which is given as ${{A}_{R}}=lb$ where $l$ and $b$ are respectively the length and the breadth of the rectangle. On equating them we will get a relation between $a$, $l$ and $b$. The values of $a$ and $b$ are given respectively as $40cm$ and $25cm$. Substituting them in the relation, we will get the required value of $l$.

Complete step by step solution:
According to the given information, the square and the rectangle can be drawn as

We know that the area of a square equal to the square of its side, that is,
$\Rightarrow {{A}_{S}}={{a}^{2}}$
Also, the area of a rectangle is equal to the product of its length and breadth, that is
$\Rightarrow {{A}_{R}}=lb$
According to the question, the areas of the square and the rectangle are equal. So we can equate the above two equations to get
$\Rightarrow {{a}^{2}}=lb$
The side of the square is given to be equal to $40cm$ and the breadth of the rectangle is given to be equal to $25cm$. Therefore we substitute $a=40cm$ and $b=25cm$ in the above equation to get
$\Rightarrow {{40}^{2}}=25l$
Dividing both the sides by $25$ we get
$\begin{align}
  & \Rightarrow l=\dfrac{{{40}^{2}}}{25} \\
 & \Rightarrow l=\dfrac{1600}{25} \\
 & \Rightarrow l=64cm \\
\end{align}$
Thus, the length of the rectangle is equal to $64cm$.
Hence, the correct answer is option (c).

Note: Ensure that the units of all the dimensions given in the question are the same. In the case of the above question, all the dimensions of the square and the rectangle are given in centimeters, so we need not to worry about the units. But the ignorance of the units is common in the solution of these types of problems.