Answer :
Answer:
A, B, C
Step-by-step 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)