The altitude of a triangle is 3 less than twice its base.The area of the triangle is 10m^2.
Find the altitude and base of the triangle.
base ---- x
height ---2x-3
area of triangle = (1/2) base x heigh
(1/2)(x)(2x-3) = 10
carry on
X=4 and x=-5/2
To find the altitude and base of the triangle, we can use the formula for the area of a triangle:
Area = (1/2) * base * altitude
Given that the area of the triangle is 10 square meters, we have the equation:
10 = (1/2) * base * altitude
We are also given that the altitude is 3 less than twice the base. Let's represent the base as "b" and the altitude as "h".
So, h = 2b - 3
Substituting this expression for the altitude in the equation for the area, we have:
10 = (1/2) * b * (2b - 3)
Now, we can solve this equation to find the values of base (b) and altitude (h).
First, let's simplify the equation:
10 = (1/2) * b * (2b - 3)
10 = b * (2b - 3) / 2
20 = b * (2b - 3)
20 = 2b^2 - 3b
Rearranging the equation and setting it equal to zero:
2b^2 - 3b - 20 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Let's factor it:
(2b + 5)(b - 4) = 0
This factorization gives us two possible values for b:
1) 2b + 5 = 0 => b = -5/2
2) b - 4 = 0 => b = 4
Since the base of a triangle cannot be negative, we can discard the value -5/2 as extraneous.
So, the base of the triangle (b) is equal to 4 meters.
To find the altitude (h), we substitute this base value back into the expression we obtained earlier:
h = 2b - 3
h = 2(4) - 3
h = 8 - 3
h = 5
Therefore, the altitude of the triangle is 5 meters and the base is 4 meters.