Question
A triangular lot in a park has a base with a measure that is 4 more than twice the height of the lot. The area of the lot is 99 square feet. Since the area of a triangle is found by using the formula A= 1/2bh, the area for this lot can be represented by the equation 99=1/2(2h+4)h, where h represents the height of the triangle.
What is the height of the lot?
Once again, I am not sure what the steps are for figuring out this problem. Someone please give me the steps for solving this problem PLEASE!! thank you
What is the height of the lot?
Once again, I am not sure what the steps are for figuring out this problem. Someone please give me the steps for solving this problem PLEASE!! thank you
Answers
R_scott
solve the given equation for h
rearranging and simplifying ... 0 = h^2 + 2 h - 99
factor or use the quadratic formula
rearranging and simplifying ... 0 = h^2 + 2 h - 99
factor or use the quadratic formula
Henry2,
99 = 1/2(2h+4)h.
99 = (h+2)h,
99 = h^2 + 2h,
h^2 + 2h - 99 = 0. Solve for h:
h = (-B +- Sqrt(B^2-4AC))/2A.
h = (-2 +- Sqrt(4 + 396))/2,
h = (-2 +- 20)/2 = -1 +- 10 = 9, and -11 Ft.
Use the positive number: h = 9 Ft.
99 = (h+2)h,
99 = h^2 + 2h,
h^2 + 2h - 99 = 0. Solve for h:
h = (-B +- Sqrt(B^2-4AC))/2A.
h = (-2 +- Sqrt(4 + 396))/2,
h = (-2 +- 20)/2 = -1 +- 10 = 9, and -11 Ft.
Use the positive number: h = 9 Ft.