a conical vessel 6 meters across the top and 4 meters deep is filled with water. A portion was spilled out by slanting the cone such a way that its slant height is in vertical position. determine the cross section area of the water surface that remains on the tank.
To determine the cross-sectional area of the water surface that remains in the tank, we need to find the radius of the water surface and then calculate the area.
1. First, let's calculate the slant height of the cone. The slant height is the distance from the tip of the cone to any point on the circumference of the top base.
Using the Pythagorean theorem, we can calculate the slant height (l) as follows:
l = √(r^2 + h^2)
Where r is the radius of the top base and h is the height of the cone.
Given that the diameter (or width) of the top base is 6 meters, we can find the radius (r) by dividing the diameter by 2:
r = 6 / 2 = 3 meters.
The depth (or height) of the cone is given as 4 meters.
Substituting the values into the formula, we have:
l = √(3^2 + 4^2)
l = √(9 + 16)
l = √25
l = 5 meters
2. Now that we have the slant height, let's find the radius of the remaining water surface.
In the vertical position, the slant height becomes the height of the water surface. So, the height of the remaining water surface is 5 meters.
Using the formula for the slant height of the cone, we can find the radius (r) of the remaining water surface:
r = √(l^2 - h^2)
r = √(5^2 - 4^2)
r = √(25 - 16)
r = √9
r = 3 meters
3. Finally, let's calculate the cross-sectional area (A) of the water surface using the formula for the area of a circle:
A = π * r^2
Substituting the value of the radius, we have:
A = π * 3^2
A = π * 9
A ≈ 28.27 square meters
Therefore, the cross-sectional area of the water surface that remains in the tank is approximately 28.27 square meters.