Question

1. the length of a rectangle is one more than four times its width. if the perimeter of the rectangle is 62 meters, find the dimensions of the rectangle.

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) meters. Then the length \( l \) is \( 4w + 1 \) meters (since length is one more than four times width).

Step2: Recall perimeter formula

The perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \). We know \( P = 62 \) meters. Substitute \( l = 4w + 1 \) into the formula:
\[
62 = 2((4w + 1) + w)
\]

Step3: Simplify the equation

First, simplify the expression inside the parentheses:
\[
62 = 2(5w + 1)
\]
Then distribute the 2:
\[
62 = 10w + 2
\]

Step4: Solve for \( w \)

Subtract 2 from both sides:
\[
62 - 2 = 10w
\]
\[
60 = 10w
\]
Divide both sides by 10:
\[
w = \frac{60}{10} = 6
\]

Step5: Find the length

Now that we have \( w = 6 \), substitute into the length formula \( l = 4w + 1 \):
\[
l = 4(6) + 1 = 24 + 1 = 25
\]

Answer:

The width of the rectangle is 6 meters and the length is 25 meters.