Question

A rectangle is $25mby9m$. Find the side of the square whose area equals the area of the rectangle. By how much does the perimeter of the rectangle exceed the perimeter of the square?

Answer 1

Step 1 :- Find the area of the rectangle.

$$\begin{gathered} Areaofrec\tan gle=l\times b \\ Areaofrec\tan gle=25\times 9=225m^2 \end{gathered}$$

Step 2 :- Find the side of the square

Now, the area of the rectangle is equal to the area of the square.

$$\begin{gathered} Areaofsquare=225m^2 \\ {\left(side\right)}^2=225 \\ sideofsquare=\sqrt{225} \\ sideofsquare=15m \end{gathered}$$

Step 3 :- Find the perimeter of the rectangle and square

Now, finding the perimeter of the rectangle and square,

$$\begin{gathered} Perimeterofrec\tan gle=2\left(l+b\right) \\ Perimeterofrec\tan gle=2\left(25+9\right) \\ Perimeterofrec\tan gle=68m \end{gathered}$$

$$\begin{gathered} Perimeterofsquare=4\times side \\ Perimeterofsquare=4\times 15=60m \end{gathered}$$

The difference between the perimeter of a rectangle and a square is $68-60=8m$.

Hence, the perimeter of the rectangle exceeds the perimeter of the rectangle by $8m$.