A triangular-shaped wall has a base of 2x+4 and a height of x+3. The area of the triangle is 56 in^2. What is the value of "x"?
A = 1/2bh
52 = 1/2(2x+4)(x +3)
multiply both sides of equation by 2
104 = (2x +4)(x +3)
2+2
-2,-3
To find the value of "x," we can set up an equation using the formula for the area of a triangle:
Area = (1/2) * base * height
Given that the area is 56 in^2 and the base is 2x + 4, and the height is x + 3, we can write:
56 = (1/2) * (2x + 4) * (x + 3)
To solve for "x," we can follow these steps:
1. Distribute the (1/2) to the terms inside the parentheses:
56 = (1/2) * (2x^2 + 6x + 4x + 12)
2. Simplify:
56 = x^2 + 5x + 6
3. Move all the terms to one side of the equation to set it equal to zero:
x^2 + 5x + 6 - 56 = 0
4. Combine like terms:
x^2 + 5x - 50 = 0
5. We can now factor the quadratic equation:
(x + 10)(x - 5) = 0
Setting each factor to zero, we have:
x + 10 = 0 or x - 5 = 0
6. Solve for "x":
x = -10 or x = 5
Therefore, the value of "x" can be either -10 or 5.