To solve for the perimeter of the rectangle, we start by noting the relationships and given information:
1. The length of the rectangle ([tex]\( L \)[/tex]) is five times its width ([tex]\( W \)[/tex]). Therefore, we can express the length as:
[tex]\[ L = 5W \][/tex]
2. The area of the rectangle is given as [tex]\( 125 \, \text{in}^2 \)[/tex]. The area of a rectangle is calculated by multiplying its length by its width:
[tex]\[ \text{Area} = L \times W \][/tex]
Substituting the given area and the expression for [tex]\( L \)[/tex]:
[tex]\[ 125 = (5W) \times W \][/tex]
Simplifying this equation:
[tex]\[ 125 = 5W^2 \][/tex]
3. To find the width [tex]\( W \)[/tex], we solve the equation:
[tex]\[ W^2 = \frac{125}{5} \][/tex]
[tex]\[ W^2 = 25 \][/tex]
[tex]\[ W = \sqrt{25} \][/tex]
[tex]\[ W = 5 \, \text{in} \][/tex]
4. Now that we have the width, we can find the length:
[tex]\[ L = 5W \][/tex]
[tex]\[ L = 5 \times 5 \][/tex]
[tex]\[ L = 25 \, \text{in} \][/tex]
5. The perimeter [tex]\( P \)[/tex] of a rectangle is calculated using the formula:
[tex]\[ P = 2(L + W) \][/tex]
Substituting the values for [tex]\( L \)[/tex] and [tex]\( W \)[/tex]:
[tex]\[ P = 2(25 + 5) \][/tex]
[tex]\[ P = 2 \times 30 \][/tex]
[tex]\[ P = 60 \, \text{in} \][/tex]
Therefore, the perimeter of the rectangle is [tex]\( 60 \, \text{in} \)[/tex].
