Chapter 1.4, Problem 66PS

Solution

To find: the new dimension of mat on the top and bottom as well as left and right of painting.

The width of mat at top and bottom of painting is $4.25\text{inches}$

The length of mat at left and right of painting is $2\times 4.25=8.5\text{ inches}$ .

Given:

A painting of rectangular shape measures 25 inches by 21 inches and area of mat is 714 square inches.

Concept Used:

Factoring method to find the roots of the quadratic equation.

Area of rectangle $=\text{length }\times \text{ width}$ .

Calculation:

From the figure:

Length of mat and painting $=2x+2x+25\text{ inches}$ .

Width of mat and painting $=x+x+21\text{ inches}$ .

Given $\text{Area of mat}=714{\text{inches}}^2$ .

  $\begin{array}{l}\text{Area of painting}=25\times 21\\ =525{\text{inches}}^2\end{array}$

Here, Applying the area of rectangle formula,

  $\therefore \left(2x+2x+25\right)\left(x+x+21\right)=714+525$ .

Solving this equation:

  $\begin{array}{c}\left(2x+2x+25\right)\left(x+x+21\right)=714+525\\ \left(4x+25\right)\left(2x+21\right)=1239\\ 4x\left(2x+21\right)+25\left(2x+21\right)=1239\\ 8x^2+84x+50x+525=1239\\ 8x^2+134x+525-1239=0\\ 8x^2+134x-714=0\end{array}$

Applying the factoring method of quadratic equation:

  $\begin{array}{c}8x^2+134x-714=0\\ 8x^2+168x-34x-714=0\\ 8x\left(x+21\right)-34\left(x+21\right)=0\\ \left(x+21\right)\left(8x-34\right)=0\end{array}$

Using Zero Product property:

  $\begin{array}{c}x+21=0\text{or}8x-34=0\\ x=-21\text{or}x=\frac{34}{8}\end{array}$

So, $x=4.25\text{ inches}$ .

Here, Neglecting the value of $x=-21$ as distance cannot be negative.

The width of mat at top and bottom of painting is $4.25\text{inches}$ .

The length of mat at left and right of painting is $2\times 4.25=8.5\text{ inches}$ .

Conclusion:

The width of mat at top and bottom of painting is $4.25\text{inches}$ .

The length of mat at left and right of painting is $2\times 4.25=8.5\text{ inches}$ .