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The length of the longer side of rectangle R is 10 percent more than the length of a side of square S. The length of the shorter side of rectangle R is 10 percent less than the length of a side of square S. What is the ratio of the area of rectangle R to the area of square S? Express your answer as a common fraction.

Firewolf  Jun 24, 2018
 #1
avatar+2268 
+1

 

Geometrists use diagrams in order to aid visualization. Although one could argue that this problem probably does not need the assistance of a diagram, I created one anyway. Here it is:

 

 

Our final goal is to express the relationship via a ratio. Calculating the area of a square is easier here.  

 

\(A_{\text{square}}=a^2\)

 

Let's consider what the area of a rectangle could be. The general formula for the area of a rectangle is \(A=lw\) .

 

\(\hspace{10mm}\textcolor{red}{l}=\textcolor{purple}{a}\textcolor{green}{+}\textcolor{blue}{10\%*a}=a+0.1a=1.1a\\ \textcolor{red}{\text{length}} \text{ is} \textcolor{blue}{\text{ 10 percent}} \textcolor{green}{\text{ more than}} \hspace{2mm}\textcolor{purple}{a} \)

 

\(\hspace{8mm}\textcolor{red}{w}=\textcolor{purple}{a}\textcolor{green}{-}\textcolor{blue}{10\%*a}=a-0.1a=0.9a\\ \textcolor{red}{\text{width}} \text{ is} \textcolor{blue}{\text{ 10 percent}} \textcolor{green}{\text{ less than}} \hspace{2mm}\textcolor{purple}{a}\)

 

Calculating the area of the rectangle is now possible:

 

\(A_{\text{rectangle}}=lw\\ A_{\text{rectangle}}=1.1a*0.9a\\ A_{\text{rectangle}}=0.99a^2\)

 

Let's now construct the ratio.

 

\(\frac{A_{\text{rectangle}}}{A_{\text{square}}}=\frac{0.99a^2}{a^2}=\frac{0.99}{1}=\frac{99}{100}\)

 

Since the fraction had to be a "common" fraction, the numerator and denominator had to be integral values. 

TheXSquaredFactor  Jun 24, 2018
edited by TheXSquaredFactor  Jun 24, 2018
 #2
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+1

Just a minor typo. It should read: 0.99a^2 =0.99/1 x 100/100 =99 / 100.

Guest Jun 24, 2018
 #3
avatar+2268 
+1

Thank you for catching that \(1.1*0.9\neq 9.9\)

TheXSquaredFactor  Jun 24, 2018

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