An isosceles triangle has its vertex at the origin and the ends of its base on the parabola y=9-x^2. Express the area of the triangle as a function of the length of its base. Assume its base lies above the x-axis.
a = 1/2 bh
Assuming the base of the triangle is parallel to the x-axis, then
a = 1/2 b (9 - b/2)^2
To find the area of the isosceles triangle, we need to determine the length of its base. The base of the triangle will lie on the parabola y=9-x^2 and will be perpendicular to the x-axis.
Let's find the coordinates of the points where the parabola intersects the x-axis. To do this, we set y=0 in the equation of the parabola and solve for x:
0 = 9 - x^2
Rearranging the equation, we get:
x^2 = 9
Taking the square root of both sides, we find:
x = ±√9
x = ±3
So, the parabola intersects the x-axis at x = -3 and x = 3.
Since the vertex of the triangle is at the origin (0,0), the base will have a length of 2 times the x-coordinate where it intersects the x-axis. Therefore, the length of the base is 2 * 3 = 6 units.
Now, let's find the height of the triangle, which is the distance between the vertex at the origin and the parabola. We can find the height by substituting x = 3 into the equation of the parabola y = 9 - x^2:
y = 9 - (3)^2
y = 9 - 9
y = 0
So, the height of the triangle is 0 units.
Now that we have both the length of the base and the height, we can calculate the area of the triangle using the formula for the area of a triangle:
Area = (base * height) / 2
In this case, since the height is 0, the area of the triangle will also be 0.
Therefore, the area of the isosceles triangle, expressed as a function of the length of its base (b), is:
Area(b) = 0