I am so annoyed over this one question: The rectangle and isosceles triangle on the right are equivalent. What is the perimeter of the isosceles triangle? Since you cannot see it, the length of the rectangle is x and width is x-4 while the height of the triangle is x-3 and the base is x+2.
I have tried many times. My teacher told me that x is equal to 6, however I cannot get to that answer. I would always get 4 or 12. I am really horrible in algebra, hopefully someone can explain me the steps and show me the work?
So, this is what I have been doing:
A1 = A2
L * W = 0.5 * B * H
(x)(x-4) = (0.5)(x+2)(x-3)
I have tried many times. My teacher told me that x is equal to 6, however I cannot get to that answer. I would always get 4 or 12. I am really horrible in algebra, hopefully someone can explain me the steps and show me the work?
So, this is what I have been doing:
A1 = A2
L * W = 0.5 * B * H
(x)(x-4) = (0.5)(x+2)(x-3)
-
(x)(x - 4) = (0.5)(x + 2)(x - 3)
x^2 - 4x = (0.5)(x^2 - x - 6)
2(x^2 - 4x) = (2)(0.5)(x^2 - x - 6)
2x^2 - 8x = x^2 - x - 6
2x^2 - x^2 - 8x + x + 6 = 0
x^2 - 7x + 6 = 0
x^2 - 6x - x + 6 = 0
(x^2 - 6x) - (x - 6) = 0
x(x - 6) - (1)(x - 6) = 0
(x - 1)(x - 6) = 0
x = 1 or x = 6
The width of the rectangle is x - 4
If x = 1, x - 4 = -3
The width cannot possible be negative, so x = 1 cannot be a solution.
Thus, we reject x = 1.
When x = 6, x - 4 = 2, which is positive.
Hence, x = 6 is the solution.
x^2 - 4x = (0.5)(x^2 - x - 6)
2(x^2 - 4x) = (2)(0.5)(x^2 - x - 6)
2x^2 - 8x = x^2 - x - 6
2x^2 - x^2 - 8x + x + 6 = 0
x^2 - 7x + 6 = 0
x^2 - 6x - x + 6 = 0
(x^2 - 6x) - (x - 6) = 0
x(x - 6) - (1)(x - 6) = 0
(x - 1)(x - 6) = 0
x = 1 or x = 6
The width of the rectangle is x - 4
If x = 1, x - 4 = -3
The width cannot possible be negative, so x = 1 cannot be a solution.
Thus, we reject x = 1.
When x = 6, x - 4 = 2, which is positive.
Hence, x = 6 is the solution.