A plane cuts a sphere of radius
𝑎
a at a distance
𝑎
2
2
a
from its center
𝑂
O.
Find the radius of the circle of intersection.
Now, imagine a solid cone is inscribed inside this sphere such that its base coincides with this circle of intersection and its vertex is at the center of the sphere. Find the volume of the cone in terms of
𝑎
a.
Compare this cone’s volume with the volume of the sphere.
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We have a sphere of radius a centered at point O. A plane cuts through this sphere at a distance of a over 2 from the center. Our goal is to find the radius of the circular intersection and then determine the volume of a cone inscribed in the sphere with its base on this circle.
To find the radius of the circular intersection, we use the Pythagorean theorem. We have a right triangle where one leg is the distance from the center to the plane, which is a over 2, another leg is the radius of the intersection circle r, and the hypotenuse is the sphere's radius a. This gives us r squared plus a over 2 squared equals a squared. Solving step by step, we get r equals a times square root of 3 over 2.
Now we construct the inscribed cone. The cone has its vertex at the sphere's center O and its base coincides with the circular intersection we found. The cone's properties are: vertex at point O, base radius equals a times square root of 3 over 2, and height equals a over 2, which is the distance from the center to the cutting plane.
To find the cone's volume, we use the formula V equals one third pi r squared h. Substituting our known values: r equals a times square root of 3 over 2, and h equals a over 2. This gives us V equals one third pi times 3 a squared over 4 times a over 2. Simplifying step by step, we get V equals pi a cubed over 8.
Now let's compare the volumes. The sphere's volume is four thirds pi a cubed, while our cone's volume is pi a cubed over 8. The ratio of cone volume to sphere volume equals 3 over 32, which is approximately 0.09375 or 9.375 percent. This means the inscribed cone occupies less than one tenth of the sphere's total volume, demonstrating how much space remains unused in this geometric configuration.