Introductory Algebra for College Students (7th Edition)

The rectangle has a length of $150$ yards and a width of $50$ yards.
We know that the perimeter is given by the formula: $$P = 2l + 2w$$ where $P$ is the perimeter of the rectangle, $l$ is the measure of its length, and $w$ is the measure of its width. In this problem, we set $w$ as the width. The length is three times the width, so $l = 3w$. We plug in what we have for $w$ and $l$ into the formula for perimeter, where we have $P = 400$: $$400 = 2(3w) + 2w$$ We can now solve for $w$, the width. Let's multiply first: $$400 = 6w + 2w$$ Add the variable terms together: $$400 = 8w$$ Divide both sides by $8$ to solve for $w$: $$w = 50$$ Now that we have the measure for width, we can substitute $50$ for $w$ into the original equation: $$400 = 2l + 2(50)$$ Multiply: $$400 = 2l + 100$$ Subtract $100$ from both sides to isolate the variable on one side and the constant terms on the other: $$300 = 2l$$ Divide both sides by $2$ to solve for $l$: $$l = 150$$ The rectangle has a length of $150$ yards and a width of $50$ yards.