Chapter 11.5, Problem 42SR

(a)

To determine

Find the maximum dimensions of the reduced photograph.

Answer to Problem 42SR

3.6 inches by 4.5 inches.

Explanation of Solution

Given:

Tal resized a photograph that was 8 inches by 10 inches so that it would fit in a 4-inch by 4-inch area on a yearbook page.

Calculation:

The area of the reduced photograph = $A_{reduced}=4\times 4=16\text{ in}{\text{.}}^2$

The area of the original photograph = $A_{original}=8\times 10=80\text{ in}{\text{.}}^2$

Now, the ratio of areas :

  $\frac{A_{original}}{A_{reduced}}=\frac{80}{16}=\frac{5}{1}$

So, the ratio of the sides is $\frac{\sqrt{5}}{1}$

Since the original dimensions of the photographs was 8 inches by 10 inches.

So,let the reduced dimensions are x and y .

Thus,

  $\begin{array}{l}\frac{\sqrt{5}}{1}=\frac{8}{x}\text{ and }\frac{\sqrt{5}}{1}=\frac{10}{y}\\ \Rightarrow x=\frac{8}{\sqrt{5}}\text{ and }y=\frac{10}{\sqrt{5}}\\ \Rightarrow x\approx 3.6\text{ and }y\approx 4.5\end{array}$

So, the reduced dimensions are 3.6 inches by 4.5 inches.

(b)

To determine

Find the percent of reduction of the length .

Answer to Problem 42SR

55%

Explanation of Solution

Given:

Tal resized a photograph that was 8 inches by 10 inches so that it would fit in a 4-inch by 4-inch area on a yearbook page.

Calculation:

From part (a) , the reduced dimensions are 3.6 inches by 4.5 inches which were 8 inches by 10 inches originally.

Now , we find the reduction in any one of the lengths :

10 − 4.5 = 5.5 inches

Now, we find the reduction percentage:

  $\frac{5.5}{10}\times 100\%=55\%$

So, the length was reduced by 55%.