Chapter 11, Problem 12MPS

To determine

To find: The perimeter of rectangle.

Answer to Problem 12MPS

The problem has no solution because real value of length and width of rectangle cannot be found.

Explanation of Solution

Given information:

The length of rectangle is $1\text{cm}$ shorter than diagonal.

The length is $2\text{cm}$ longer than twice the width.

Calculation:

Assume that the length of rectangle is $l$ , width of rectangle is $w$ and the diagonal is $d$ .

Use Pythagoras theorem to calculate the length of diagonal.

  $d=\sqrt{l^2+b^2}$

Since, $l=d-1$ above equation becomes,

  $\begin{array}{l}\left(l+1\right)=\sqrt{{\left(l\right)}^2+w^2}\\ l^2+2l+1=l^2+w^2\\ w^2+2l+1=0\end{array}$ ..... (1)

Since, $l=2+2w$ , equation (1) becomes,

  $\begin{array}{l}{w}^2+2\left(2+w\right)+1=0\\ w^2+4+2w+1=0\\ w^2+2w+5=0\end{array}$

Solve the quadratic equation.

  $w^2+2w+5=0$

The above equation will have complex roots that mean real root is not possible, so the problem has no solution.

Therefore, the problem has no solution because real value of length and width of rectangle cannot be found.