the height of a triangle is 4 inches greater than twice its base. the area of the triangle is 168 square inches. what is the base of the triangle?
Area=1/2 B*H= 1/2 B (2B+4)
solve for B
h = 4 + 2b
b = ?
A = (b*h)/2 = 168
168 = [b(4+2b)]/2
168 = (4b + 2b^2)/2
336 = 4b + 2b^2
2b^2 + 4b - 336 = 0
2(b^2 + 2b - 168) = 0
2(b-12)(b+14) = 0
b=12 and b=-14
Ignore the negative answer, negative distances have no real meaning.
To find the base of the triangle, we can use the formula for the area of a triangle, which is equal to half the base multiplied by the height.
Let's start by assigning variables:
b = base of the triangle
h = height of the triangle
Given that the height of the triangle is 4 inches greater than twice its base, we can express this relationship as:
h = 2b + 4
We also know that the area of the triangle is 168 square inches. Substituting the variables into the formula for the area, we have:
168 = (1/2) * b * h
Substituting the expression for h:
168 = (1/2) * b * (2b + 4)
Now, we can solve for the base (b):
Multiply both sides of the equation by 2 to get rid of the fraction:
336 = b * (2b + 4)
Expand the expression on the right side:
336 = 2b^2 + 4b
Rearrange the equation into standard quadratic form:
2b^2 + 4b - 336 = 0
We can divide the entire equation by 2 to simplify it:
b^2 + 2b - 168 = 0
We can factorize this quadratic equation:
(b - 12)(b + 14) = 0
From this, we have two possible values for b:
b = 12 and b = -14. However, since we are dealing with measurements, we can ignore the negative answer since a negative value for the base of a triangle doesn't make sense.
Thus, the base of the triangle is 12 inches.