Question

A contractor is building a new house and wants to determine the size of its foundation. The length of the

foundation is 6 feet more than 2 times its width. The value of the area of the foundation (in square feet)
is 1046 more than the value of the perimeter of the foundation (in feet). Which of the following values
are correct? Show work.

A. Area: 1196 sq ft
B. Worth: 23 ft
C. Perimeter: 150 ft
D. Length 29 ft

Answer 1

A, B, C

Explanation:

You want to find the dimensions, perimeter, and area of a foundation such that the length is 6 feet more than twice the width, and the area is 1046 more that the value of the perimeter.

Setup

We can let x and y represent the length and width, respectively. Then the given relations translate to the equation ...

A = LW . . . . . . . area formula

P = 2(L +W) . . . . perimeter formula

y = 2x +6 . . . . . relation between dimensions

xy = 2(x +y) +1046 . . . . . relation between perimeter and area

Solution

Using the first equation to substitute for y, we have ...

x(2x +6) = 2(x +(2x +6)) +1046

2x² +6x = 6x +1058 . . . . . . . . . . . . eliminate parentheses

x² = 529 . . . . . . . . . . . . . subtract 6x and divide by 2

x = √529 = 23 . . . . . . . . . width; matches B

y = 2(23) +6 = 52 . . . . length

area = (23)(52) = 1196 . . . area; matches A

perimeter = 2(23 +52) = 2(75) = 150 . . . . matches C

The area is 1196 square feet. (A)

The width is 23 feet. (B)

The perimeter is 150 feet. (C)