Length or Width of a Rectangle from Perimeter Worksheets - Free Printable

To solve the problems in the worksheet, we need to use the formula for the perimeter of a rectangle:

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]

We will rearrange this formula as needed to find either the length or the width based on the given information.

Problem 1:

Perimeter = 38 yd, Width = 7 yd

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
\[
38 = 2 \times (\text{Length} + 7)
\]
Divide both sides by 2:
\[
19 = \text{Length} + 7
\]
Subtract 7 from both sides:
\[
\text{Length} = 12 \text{ yd}
\]

Problem 2:

Perimeter = 44 in, Length = 12 in

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
\[
44 = 2 \times (12 + \text{Width})
\]
Divide both sides by 2:
\[
22 = 12 + \text{Width}
\]
Subtract 12 from both sides:
\[
\text{Width} = 10 \text{ in}
\]

Problem 3:

Perimeter = 72 ft, Width = 10 ft

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
\[
72 = 2 \times (\text{Length} + 10)
\]
Divide both sides by 2:
\[
36 = \text{Length} + 10
\]
Subtract 10 from both sides:
\[
\text{Length} = 26 \text{ ft}
\]

Problem 4:

Perimeter = 100 ft, Length = 30 ft

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
\[
100 = 2 \times (30 + \text{Width})
\]
Divide both sides by 2:
\[
50 = 30 + \text{Width}
\]
Subtract 30 from both sides:
\[
\text{Width} = 20 \text{ ft}
\]

Problem 5:

Perimeter = 86 yd, Width = 35 yd

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
\[
86 = 2 \times (\text{Length} + 35)
\]
Divide both sides by 2:
\[
43 = \text{Length} + 35
\]
Subtract 35 from both sides:
\[
\text{Length} = 8 \text{ yd}
\]

Problem 6:

Perimeter = 30 in, Length = 9 in

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
\[
30 = 2 \times (9 + \text{Width})
\]
Divide both sides by 2:
\[
15 = 9 + \text{Width}
\]
Subtract 9 from both sides:
\[
\text{Width} = 6 \text{ in}
\]

Problem 7:

Perimeter = 58 in, Width = 8 in

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
\[
58 = 2 \times (\text{Length} + 8)
\]
Divide both sides by 2:
\[
29 = \text{Length} + 8
\]
Subtract 8 from both sides:
\[
\text{Length} = 21 \text{ in}
\]

Problem 8:

Perimeter = 12 ft, Length = 4 ft

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
\[
12 = 2 \times (4 + \text{Width})
\]
Divide both sides by 2:
\[
6 = 4 + \text{Width}
\]
Subtract 4 from both sides:
\[
\text{Width} = 2 \text{ ft}
\]

Problem 9:

Perimeter = 64 yd, Width = 14 yd

\[
\text{Perimeter} = 2 \times (\text{Length} + \text{Width})
\]
\[
64 = 2 \times (\text{Length} + 14)
\]
Divide both sides by 2:
\[
32 = \text{Length} + 14
\]
Subtract 14 from both sides:
\[
\text{Length} = 18 \text{ yd}
\]

Final Answers:

\[
\boxed{
\begin{array}{ll}
1) & \text{Length} = 12 \text{ yd} \\
2) & \text{Width} = 10 \text{ in} \\
3) & \text{Length} = 26 \text{ ft} \\
4) & \text{Width} = 20 \text{ ft} \\
5) & \text{Length} = 8 \text{ yd} \\
6) & \text{Width} = 6 \text{ in} \\
7) & \text{Length} = 21 \text{ in} \\
8) & \text{Width} = 2 \text{ ft} \\
9) & \text{Length} = 18 \text{ yd} \\
\end{array}
}
\]