Chapter 5.5, Problem 28PPS

To determine

To find: The dimensions of the rectangle.

Answer to Problem 28PPS

The required dimensions of the rectangle are 19cm and 6cm.

Explanation of Solution

Given information:

The perimeter of the rectangle is 50 centimeters.

The length of the rectangle is one more than 3 times the width of the rectangle.

Calculation:

Let the length of the rectangle be $x$

Let the width of the rectangle be $y$

As the length is one more than 3 times the width

So, the length, $x=3y+1$

We know that perimeter of a rectangle $=2\left(length+width\right)$

As the perimeter of the rectangle is 50 centimeters

  $\Rightarrow 2\left(x+y\right)=50$

  $\begin{array}{l}\Rightarrow 2\left(3y+1+y\right)=50\\ \Rightarrow 4y+1=\frac{50}{2}\\ \Rightarrow 4y+1=25\\ \Rightarrow 4y=25-1\\ \Rightarrow 4y=24\\ \Rightarrow y=\frac{24}{4}\\ \Rightarrow y=6\end{array}$

Now,

  $\begin{array}{l}\Rightarrow x=3y+1\\ \Rightarrow x=3·6+1\\ \end{array}$

  $\begin{array}{l}\Rightarrow x=18+1\\ \Rightarrow x=19\end{array}$

Therefore the length and width of the rectangle are 19 and 6 respectively.

So the required dimensions are 19 and 6.