Question

The front walkway from the street to Pam's house has an area of 250ft^2. Its length is two less than four times its width. Find the length and width of the walkway. Round to the nearest tenth of a foot.

Answer 1

To find the length and width of Pam's walkway, we represent the width as 'w', set up an equation based on the given area and the relationship between length and width, and solve for 'w'. Once 'w' is found, we plug it back into the length expression to obtain the dimensions of the walkway.

Step-by-step explanation:

We are given the total area of the walkway and a relationship between its length and width. To find the length and width of the walkway, we can represent the unknown width as w and then express the length as 4w - 2, since the length is two less than four times the width.

The area of a rectangle (which we can assume the walkway to be) is calculated by multiplying the length by the width. Accordingly, the equation representing the area of the walkway is:

w × (4w - 2) = 250

Solving this equation:

1. Expand the equation: 4w^2 - 2w = 250
2. Move all terms to one side to set the equation to zero: 4w^2 - 2w - 250 = 0
3. Divide the entire equation by 2 to simplify: 2w^2 - w - 125 = 0
4. Solve this quadratic equation for w. The values found will be the possible widths, and we choose the one that makes sense in the context (positive and practical).
5. Once we have w, we can find the length by substituting back into the length expression: L = 4w - 2.

After carrying out the steps above, we will arrive at the width and length of the walkway, accurate to the nearest tenth of a foot.

Answer 2

Step-by-step explanation:

rectangle is 250ft2. Substituting into the formula for the area of a rectangle, A=length×width, we have

250250=(4w−2)(w)=4w2−2w

In the standard form aw2+bw+c=0, this is

4w2−2w−250=0

Substituting the coefficients a=4, b=−2, and c=−250 into the quadratic formula, we have

w=−b±b2−4ac‾‾‾‾‾‾‾‾√2a=−(−2)±(−2)2−4(4)(−250)‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√2(4)=2±4,004‾‾‾‾‾√8

There are two solutions for w, which we can evaluate on a calculator.

w=≈2+4,004‾‾‾‾‾√88.2andw=≈2−4,004‾‾‾‾‾√8−7.7

The width of the rectangle must be positive, so w=8.2. The length is then given by

4w−2=4(8.2)−2=30.8

Thus, Pam's front walkway has a width of 8.2ft and a length of 30.8ft.