Chapter 8.2, Problem 32WE

To determine

To find: the perimeter of the rectangle.

Answer to Problem 32WE

  $28\text{ cm}$

Explanation of Solution

Given:

The length of rectangle is 2 cm longer than its width.

Diagonal of rectangle is 10 cm.

Concept Used:

  $\text{Perimeter of rectangle}=2\left(Length+width\right)$

Draw a rectangle as given in the question:

Let the width of the rectangle be x cm, then the length of the rectangle is $x+2$ cm.

Consider the right triangle $△ABC$ .

Now, $\overline{AC}$ is the hypotenuse and has length 10 and the lengths of the legs are $AB=x+2$ and $BC=x$ .

So, by theorem 8-2,

  $\begin{array}{c}A{C}^2=B{C}^2+A{B}^2\\ {10}^2=x^2+{\left(x+2\right)}^2\\ 100=x^2+x^2+4x+4\\ 100=2x^2+4x+4\end{array}$

This gives that,

  $\begin{array}{c}2x^2+4x+4-100=0\\ 2x^2+4x-96=0\\ x^2+2x-48=0\end{array}$

  $\begin{array}{c}{x}^2+8x-6x-48=0\\ x\left(x+8\right)-6\left(x+8\right)=0\\ \left(x+8\right)\left(x-6\right)=0\end{array}$

Thus,

  $\begin{array}{c}x-6=0\\ x=6\end{array}$

Or

  $\begin{array}{c}x+8=0\\ x=-8\end{array}$

Since, length can’t be negative, so $x=6$ .

Hence, width of rectangle is 6 cm.

Length of rectangle = $6+2=8\text{ cm}$

  $\begin{array}{c}\text{Perimeter of rectangle}=2\left(Length+width\right)\\ =2\left(8+6\right)\\ =2\times 14\\ =28\text{ cm}\end{array}$