Base x Height / 2 = Area of triangle.
8:5 x:7
8/5=x/7 Solve for x
x=11 1/5 or 11.2 liters
The height of the object at the time of launch is 100 meters.
To find this, we simply have to put 0 in for x, as this is when there has been no time (at launch)
h(x)=-5(x-4)^2+180
h(0)=-5(0-4)^2+180
h(0)=-5(-4)^2+180
h(0)=-5(16)+180
h(0)=-80+180
h(0) = 100
If Q is tangent to circle P, that means that it is perpendicular to the radius at the point of tangency. This means that ΔPRQ would be a right triangle. We can use the Pythagorean theorem to solve this.
9² + 12² = (x+9)²
We use x+9 because the unknown segment x would combine with the radius, 9, to give that side of the triangle (the hypotenuse).
81 + 144 = (x+9)(x+9) ---- remember that squared means multiplied by itself
225 = (x+9)(x+9)
Multiply the right hand side:
225 = x² + 9x + 9x + 81
225 = x² + 18x + 81
Subtract 225 from both sides:
225 - 225 = x² + 18x + 81 - 225
0 = x² + 18x - 144
Use the Quadratic Formula to solve:

This gives us

Since -18 - 30 = -48, and a negative number doesn't make sense in this situation, we have
x = (-18+30)/2 = 12/2 = 6.