Question

1. the top of the john hancock building is a rectangle whose length is 60 ft more than the width. the perimeter is 520 feet. find the width and the length of the rectangle.

\\(w = 100\text{ ft}, l = 40\text{ ft}\\)

\\(w = 40\text{ ft}, l = 100\text{ ft}\\)

\\(w = 160\text{ ft}, l = 100\text{ ft}\\)

\\(w = 100\text{ ft}, l = 160\text{ ft}\\)

Explanation:

Define variables and equations

Let \(W\) represent the width of the rectangle in feet.
Let \(L\) represent the length of the rectangle in feet.
We are given that the length is \(60\text{ ft}\) more than the width:

$$L = W + 60$$

The perimeter of a rectangle is given by the formula:

$$P = 2L + 2W$$

Given that the perimeter is \(520\text{ feet}\), we write:

$$2L + 2W = 520$$

Substitute and simplify

Using the Algebraic Simplification knowledge point, we substitute \(L = W + 60\) into the perimeter equation:

$$2(W + 60) + 2W = 520$$

Distribute the \(2\):

$$2W + 120 + 2W = 520$$

Combine like terms:

$$4W + 120 = 520$$

Solve for width

Subtract \(120\) from both sides of the equation:

$$4W = 400$$

Divide both sides by \(4\):

$$W = 100\text{ ft}$$

Solve for length

Substitute \(W = 100\) back into the expression for \(L\):

$$L = W + 60$$

$$L = 100 + 60 = 160\text{ ft}$$

Thus, the width is \(100\text{ ft}\) and the length is \(160\text{ ft}\).

Answer:

• (A) \(W = 100\text{ ft}, L = 40\text{ ft}\)
• (B) \(W = 40\text{ ft}, L = 100\text{ ft}\)
• (C) \(W = 160\text{ ft}, L = 100\text{ ft}\)
• (D) \(W = 100\text{ ft}, L = 160\text{ ft}\) (Correct answer)