Chapter 5, Problem 16PT

To determine

To find: the width of the frame using the given picture.

Answer to Problem 16PT

The width of the frame is $2\text{ inches}$ .

Explanation of Solution

Given:

The length of the picture is $10\text{ inches}$ and breath is $8\text{ inches}$ .

Area of the picture and frame is $168\text{sq inches}$ .

  

Concept used:

  $\text{Area of rectangle}=L\times B.$

The quadratic equations are in the form:

  ${\text{x}}^{\text{2}}\text{-sx+p=0}$ .

Where $\text{s=sum of root=α+β}$ and $\text{p=product of the roots=α×β}$ .

Calculation:

According to the given the length of the picture is $10\text{ inches}$ and breath is $8\text{ inches}$ .

Area of the picture and frame is $168\text{sq inches}$ .

Since, the frames and picture both have same shape rectangular therefore the are of the rectangle is:

Are of the rectangle form by the picture only is:

  $\begin{array}{l}\text{Area of rectangle}=L\times B.\\ \Rightarrow 10\times 8=80\text{ sq inches}\text{.}\end{array}$

  $\begin{array}{l}\text{area of whole rectangle=area of rectangle picture+2}\times 10\times x+2\times 8\times x+4\times x^2.\\ \text{area of whole rectangle}=80+20x+16x+4x^2.\end{array}$

The value of $x$ or width of the frame can be calculated through quadratic equation as:

  $\begin{array}{l}4x^2+36x+80=168.\\ 4\left(x^2+9x+20\right)=168.\\ x^2+9x+20-42=0.\\ x^2+9x-22=0.\end{array}$

Now use the middle term factorization to solved the quadratic equation:

  $\begin{array}{l}{x}^2+9x-22=0.\\ x^2+11x-2x-22=0.\\ x\left(x+11\right)-2\left(x+11\right)=0.\\ x=2\text{ and }x=-11.\end{array}$

The distance cannot be negative therefore take $x=2$ .

Hence, the width of the frame is $2\text{ inches}$ .