Question

1. mr. mccall enclosed a rectangular dog run with 220 ft of fencing. if the length is 50 ft more than twice the width, what are the dimensions of the rectangle?

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) feet. Then the length \( l \) is \( 2w + 50 \) feet (since length is 50 ft more than twice the width).

Step2: Use perimeter formula

The perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \). We know the perimeter is 220 ft, so substitute the values of \( l \) and \( P \):

$$ 220 = 2((2w + 50) + w) $$

Step3: Simplify the equation

First, simplify inside the parentheses:

$$ 220 = 2(3w + 50) $$

Then, divide both sides by 2:

$$ 110 = 3w + 50 $$

Step4: Solve for \( w \)

Subtract 50 from both sides:

$$ 110 - 50 = 3w $$

$$ 60 = 3w $$

Divide both sides by 3:

$$ w = \frac{60}{3} = 20 $$

Step5: Find the length

Now that we have \( w = 20 \), substitute into the length formula \( l = 2w + 50 \):

$$ l = 2(20) + 50 = 40 + 50 = 90 $$

Answer:

The width of the rectangle is 20 feet and the length is 90 feet.