Question

find the dimensions of a rectangle whose width is 5 miles less than its length and whose area is 66 square miles

Answer 1

To solve for the dimensions of the rectangle, we assign the letter L to represent the length and W for the width (W = L - 5). The equation for area, A = L×W, becomes 66 = L×(L - 5), which is a quadratic equation that can be factored to reveal L = 11 miles. Hence, the dimensions are 11 miles by 6 miles.

Step-by-step explanation:

To find the dimensions of a rectangle whose width is 5 miles less than its length and whose area is 66 square miles, let's denote the length of the rectangle as L miles and the width as W miles. The problem states that W = L - 5. To find the dimensions, we need to express the area, which is the product of length and width, as an equation: Area = L × W or 66 = L × (L - 5).

Let's solve this quadratic equation by expanding and rearranging it into standard form:

• L² - 5L = 66
• L² - 5L - 66 = 0

Solving for L using the quadratic formula or factoring, we find that the possible values are:

• L = 11 miles
• L = -6 miles (but this is not physically possible for the dimensions of a rectangle)

Therefore, the length of the rectangle is 11 miles and the width is L - 5 = 11 - 5 = 6 miles.

So, the dimensions of the rectangle are 11 miles by 6 miles.

Answer 2

A = l * w
66 = l * (l - 5)
66 =
`l^(2)`-5l
0 =
`l^(2)`-5l - 66

l = 11 or 6 check which works.

A = 11 * 6, which fits.

length = 11 miles
width = 5 miles