How to Solve Rectangle Problems: Finding the Width When Length is Double the Width

Understanding basic geometry principles helps solve many problems involving rectangles—especially when dimensions follow simple ratios. One common question is: If a rectangle has a length that is twice its width and a perimeter of 36 units, what is the width? This article breaks down the math behind the problem with clear steps, making it easy for students and math enthusiasts alike.

Understanding the Context

Understanding the Rectangle’s Dimensions

A rectangle has two consistent measurements: length and width. According to the problem, the length is twice the width. We can express this relationship using algebra:

Let the width = w
Then the length = 2w

A rectangle has a length that is twice its width. If the perimeter of the rectangle is 36 units, what is the width of the rectangle?

Key Insights

Using the Perimeter Formula

The perimeter P of a rectangle is calculated with the formula:
P = 2 × (length + width)

Substituting the values we have:
36 = 2 × (2w + w)

Simplify inside the parentheses:
36 = 2 × (3w)
36 = 6w

Continue Reading

\text{LCM} = 2^3 \times 3^2 = 8 \times 9 = 72
The least common multiple is \boxed{72}.
Question: What is the sum of all values of $ q $ for which $ \sqrt{(q + 3)^2} = 5 $, symbolizing possible deviations in philosophical models?

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Final Thoughts

Solving for the Width

To find w, divide both sides of the equation by 6:
w = 36 ÷ 6
w = 6

Verifying the Answer

We now know the width is 6 units. Since length = 2 × width, the length is:
2 × 6 = 12 units

Calculate the perimeter to confirm:
P = 2 × (12 + 6) = 2 × 18 = 36 units, which matches the given value.

Final Answer

The width of the rectangle is 6 units.