Question

A rectangle's width is 6 inches less than its length and its perimeter is 72 inches.

Answer 1

Let the length of the rectangle be \( L \) inches. According to the problem, the width \( W \) is 6 inches less than the length, so we can express the width as:

\[
W = L - 6
\]

The formula for the perimeter \( P \) of a rectangle is given by:

\[
P = 2L + 2W
\]

We are told that the perimeter is 72 inches. Substituting the expression for the width into the perimeter formula, we have:

\[
72 = 2L + 2(L - 6)
\]

Now, distribute the 2 in the second term:

\[
72 = 2L + 2L - 12
\]

Combine like terms:

\[
72 = 4L - 12
\]

To isolate \( 4L \), add 12 to both sides:

\[
72 + 12 = 4L
\]
\[
84 = 4L
\]

Next, divide both sides by 4 to solve for \( L \):

\[
L = \frac{84}{4} = 21
\]

Now that we have the length, we can find the width:

\[
W = L - 6 = 21 - 6 = 15
\]

Thus, the dimensions of the rectangle are:
- Length: \( 21 \) inches
- Width: \( 15 \) inches

To verify, we can check the perimeter:

\[
P = 2L + 2W = 2(21) + 2(15) = 42 + 30 = 72 \text{ inches}
\]

This confirms that our values are correct. Therefore, the length of the rectangle is 21 inches and the width is 15 inches.