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)
[tex] A = LW = \sqrt{50} \times \sqrt{32} = \sqrt{1600} = 40 [/tex]
[tex] P = 2L + 2W = 2\sqrt{50} + 2\sqrt{32} = [/tex]
[tex] P = 2 \times 5\sqrt{2} + 2 \times 4 \sqrt{2} = [/tex]
[tex] P = 10 \sqrt{2} + 8\sqrt{2} = 18\sqrt{2} [/tex]
Answer to part c):
The area is 40.
The perimeter is 18sqrt(2).
Part d)
[tex] P = S_1 + S_2 + S_3 [/tex]
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
