Question

practice #5 perimeter of a rectangle the perimeter of the rectangle is 56 cm. the length is 2 cm more than the width. label the rectangle’s length. create an equation for the perimeter of the rectangle. solve the equation and identify the width and length.

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) (in cm). Then the length \( l \) is \( w + 2 \) (since length is 2 cm more than width).

Step2: Recall perimeter formula

The perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \). We know \( P = 56 \) cm. Substitute \( l = w + 2 \) into the formula:
\[
56 = 2((w + 2) + w)
\]

Step3: Simplify the equation

First, simplify inside the parentheses:
\[
56 = 2(2w + 2)
\]
Then distribute the 2:
\[
56 = 4w + 4
\]

Step4: Solve for \( w \)

Subtract 4 from both sides:
\[
56 - 4 = 4w
\]
\[
52 = 4w
\]
Divide both sides by 4:
\[
w = \frac{52}{4} = 13
\]

Step5: Find the length

Now that \( w = 13 \) cm, the length \( l = w + 2 = 13 + 2 = 15 \) cm.

Answer:

Width \( w = 13 \) cm, Length \( l = 15 \) cm