Question

The area of a rectangle is 54 (cm2). Its length and breadth are reduced by 5cm and 2cm, respectively, so as to become a square. Find the length of a side of the square.

Answer 1

Step-by-step explanation:

Area of rectangle=54 cm^2

Area of square=(1-5)(w-2) because each is reduced as such. We know that (1-5) must be equal to (w-2) if it is a square.

Factors of 54 include 1,54,2,26,3,18,6,9

The pair 6 and 9 fits in the solution that 1*w=54 with 1=9 and w=6

1-5=9-5=4

w-2=6-2=4

Therefore the length of the side of square is 4 cm

Answer 2

Given,

The area of the rectangle$\[ = 54{\rm{c}}{{\rm{m}}^2}\]$

The reduction in the length of the rectangle$\[ = 5{\rm{cm}}\]$

The reduction in the breadth of the rectangle$\[ = 2{\rm{cm}}\]$

Solution,

The formula for the area of a rectangle is $\[{\rm{length}} \times {\rm{breadth}}{\rm{.}}\]$

The formula for the area of a square is $\[{\left( {{\rm{side}}} \right)^2}{\rm{.}}\]$

Assume that the length and breadth of the rectangle are l and b.

Therefore,

$\[\begin{array}{l} \Rightarrow l \times b = 54\\ \Rightarrow b = \frac{{54}}{l}\end{array}\]$

Know that the sides of the square are equal in length.

So,

$\[\begin{array}{l} \Rightarrow l - 5 = b - 2\\ \Rightarrow \frac{{54}}{b} - 5 = b - 2\\ \Rightarrow 54 - 5b = {b^2} - 2b\\ \Rightarrow {b^2} + 3b - 54 = 0\end{array}\]$

Solve further,

$\[\begin{array}{l} \Rightarrow {b^2} + \left( {9 - 6} \right)b - 54 = 0\\ \Rightarrow {b^2} + 9b - 6b - 54 = 0\\ \Rightarrow b\left( {b + 9} \right) - 6\left( {b + 9} \right) = 0\\ \Rightarrow \left( {b + 9} \right)\left( {b - 6} \right) = 0\end{array}\]$

Understand that only the positive value is taken into consideration.

$\[ \Rightarrow b = 6\]$

Therefore, the length of the side of the square is

$\[\begin{array}{l} \Rightarrow 6 - 2\\ \Rightarrow 4\end{array}\]$

Hence, the length of a side of the square is $\[4\]$cm.