Given Info: We are given a rectangular photograph of size 8X10 inches. The photograph is surrounded by a uniform width border whose area is 144 square inches.

Interpreting the Problem: In order to find the width of the border, we need to first assume the width of the border in terms of x, and then use the relation of the difference of the areas of 'Photograph+Border' and 'Photograph' to calculate the width of the border.

Solution: Let the width of the border be a.

Then the width of the larger rectangle formed as indicated in the figure will be 8+2a (Uniform width on both sides)

The length of the larger rectangle formed as indicated in the figure will be 10+2a (Unifrom width on both sides)

This is shown in the figure

Attachment:

6.png

Now calculating areas

Area of the larger rectangle = (8+2a)*(10+2a) = \(4a^2+36a+80\)

Area of the photograph = 8*10 = 80

Now the difference of the areas of larger rectangle and photograph is given to us as 144 square inches.

Forming an equation based on above information

Area of larger rectangle- Area of the photograph = \(4a^2+36a+80\)-\(80\)=144

\(4a^2+36a\)=\(144\)

\(4a^2+36a-144=0\)

\(a^2+9a-36=0\)

\(a^2+12a-3a-36=0\)

\((a-3)(a+12)=0\)

Only solution valid for this quadratic equation is a=3 (Because the length of the border of photograph cannot be negative)

So the width of the border will be equal to 3 inches.

Hence, option A is correct.

. If I look at the diagram, if the total area is supposed to be 224, shouldn't the length be 10+a and 8+a?