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the length of a rectangle is 5 ft longer than its width. if the perimet…

Question

the length of a rectangle is 5 ft longer than its width. if the perimeter of the rectangle is 54 ft, find its length and width. length: ft width: ft

Explanation:

Step1: Set up variables

Let the width of the rectangle be \( w \) ft. Then the length \( l=(w + 5) \) ft.
The formula for the perimeter of a rectangle is \( P = 2(l + w) \). Given \( P=54 \) ft.
Substitute \( l=(w + 5) \) into the perimeter formula: \( 54=2((w + 5)+w) \).

Step2: Simplify the equation

First, simplify the right - hand side of the equation:
\( 54=2(2w + 5) \).
Using the distributive property \( a(b + c)=ab+ac \), we get \( 54 = 4w+10 \).
Subtract 10 from both sides: \( 54-10=4w+10 - 10 \), so \( 44 = 4w \).

Step3: Solve for \( w \)

Divide both sides by 4: \( w=\frac{44}{4}=11 \) ft.

Step4: Solve for \( l \)

Since \( l=w + 5 \), substitute \( w = 11 \) into the equation: \( l=11 + 5=16 \) ft.

Answer:

length: \( 16 \) ft, width: \( 11 \) ft