Correct answer: D. (pi * R^3 / 12) * (2 + sqrt(2))
The solution requires a multi-step geometric and algebraic approach.
1. · *Relate the cone's base radius (r) to the sphere's radius (R). · *
The area of the cone's base is `π · r²`. A great circle of the sphere has an area of `π · R²`. The problem states the base area is half the great circle area: `π · r² = (1/2) · π * R²`. This simplifies to `r² = R² / 2`.
2. · *Establish a geometric relationship between r, h, and R. · *
Consider a 2D cross-section of the sphere and the inscribed cone that passes through the sphere's center O and the cone's vertex. This cross-section shows an isosceles triangle (the cone) inscribed in a circle (the sphere). The vertices of a key right-angled triangle are the center of the sphere (O), the center of the cone's base, and any point on the circumference of the cone's base. The hypotenuse of this triangle is the radius of the sphere, `R`. One leg is the radius of the cone's base, `r`. The other leg is the distance from the sphere's center to the cone's base. Since the cone's height `h` is greater than `R`, the base is on the opposite side of the center from the vertex, and this distance is `h - R`. The Pythagorean theorem gives: `r² + (h - R)² = R²`.
3. · *Solve for the height (h). · *
Substitute `r² = R² / 2` into the Pythagorean equation: `(R² / 2) + (h - R)² = R²`.
Expand the squared term: `R² / 2 + h² - 2hR + R² = R²`.
Simplify to get a quadratic equation in `h`: `h² - 2Rh + R² / 2 = 0`.
Use the quadratic formula, `h = [-b ± √(b² - 4ac)] / (2a)`, with `a=1`, `b=-2R`, `c=R^2/2`:
`h = [2R ± √((-2R)² - 4(1)(R^2/2))] / 2`
`h = [2R ± √(4R² - 2R²)] / 2`
`h = [2R ± √(2R²)] / 2`
`h = [2R ± R · √(2)] / 2 = R · (1 ± √(2)/2)`.
4. · *Select the correct value for h. · *
The problem states that `h > R`. Since `√(2)/2` is approximately 0.707, the two possible values for `h` are approx. `R · (1+0.707) = 1.707R` and `R · (1-0.707) = 0.293R`. We must choose the larger value that is greater than `R`.
So, `h = R · (1 + √(2)/2)`.
5. · *Calculate the volume of the cone. · *
The volume of a cone is `V = (1/3) · π * r² · h`.
Substitute the expressions for `r²` and `h`:
`V = (1/3) · π * (R² / 2) · [R · (1 + √(2)/2)]`
`V = (π · R³ / 6) · (1 + √(2)/2)`
To simplify, find a common denominator inside the parenthesis: `1 + √(2)/2 = (2 + √(2)) / 2`.
`V = (π · R³ / 6) · [(2 + √(2)) / 2]`
`V = (π · R³ · (2 + √(2))) / 12`.
This matches choice D. · *Distractor Analysis: · *
- Choice A results from correctly setting up the problem but choosing the smaller root for the cone's height, `h = R · (1 - √(2)/2)`, which violates the given condition that `h > R`.
- Choice B results from an incorrect application of the Pythagorean theorem, such as `r² + h² = R²`, which misunderstands the geometry of the inscribed cone.
- Choice C results from correctly performing all the geometric and algebraic steps but using the incorrect formula for the volume of a cone, `V = π · r² · h`, omitting the `1/3` factor. The result is three times the actual volume.