We just expanded the second floor in installed some furniture. the second floor is represented by the bigger box, while the rectangle and triangle represent tables. if the length of the second floor is represented by 3x-4 and the perimeter is represented by 12x^2-25x-7.

Part a)

The area of a rectangle is

A = LW

The perimeter of a rectangle is the sum of the lengths of the sides and is

P = 2L + 2W

We are told L = 3x - 4.
We are also told P = 12x^2 - 25x - 7

We plug in the length and perimeter into the perimeter formula and solve for the width, W.

P = 2L + 2W

12x^2 - 25x - 7 = 2(3x - 4) + 2W

12x^2 - 25x - 7 = 6x - 8 + 2W

Subtract 6x from both sides, and add 8 to both sides.

12x^2 - 31x + 1 = 2W

Switch sides.

2W = 12x^2 - 31x + 1

Divide both sides by 2.

W = 6x^2 - 15.5x + 0.5

Answer to part a): The width is 6x^2 - 15.5x + 0.5

Part b)

A = LW = (3x - 4)(6x^2 - 15.5x + 0.5) =

= 18x^3 - 46.5x^2 + 1.5x - 24x^2 + 62x - 2

= 18x^3 - 70.5x^2 + 63.5x - 2

Answer to part b): The area is 18x^3 - 70.5x^2 + 63.5x - 2

Part c)









Answer to part c):
The area is 40.
The perimeter is 18sqrt(2).

Part d)



P = 9x^2 - 6x + 1 + x - 5 + 3x^3 - 25x - 44x + 10

P = 3x^3 + 9x^2 - 74x + 6

Answer to part d)
The perimeter is 3x^3 + 9x^2 - 74x + 6

Answer to part c) The area of the