SOLUTION: The altitude of a right circular cylinder and the radius of its base are each 12 inches long. Two regular triangular prisms are inscribed in and circumscribed about the cylinder. F

Question 1170318: The altitude of a right circular cylinder and the radius of its base are each 12 inches long. Two regular triangular prisms are inscribed in and circumscribed about the cylinder. Find the total surface areas of both prism
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let $h$ be the altitude of the right circular cylinder and $r$ be the radius of its base. We are given $h = 12$ inches and $r = 12$ inches.
**Inscribed Triangular Prism**
For the inscribed triangular prism, the base is an equilateral triangle inscribed in a circle of radius $r$. Let $s$ be the side length of the equilateral triangle.
The relationship between $s$ and $r$ is $r = \frac{s}{\sqrt{3}}$, so $s = r\sqrt{3}$.
Since $r = 12$, $s = 12\sqrt{3}$.
The area of the equilateral triangle is $A_{inscribed} = \frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} (12\sqrt{3})^2 = \frac{\sqrt{3}}{4} (144 \cdot 3) = 108\sqrt{3}$.
The lateral surface area of the inscribed prism is $3 \cdot s \cdot h = 3 \cdot 12\sqrt{3} \cdot 12 = 432\sqrt{3}$.
The total surface area of the inscribed prism is $2 A_{inscribed} + 3sh = 2(108\sqrt{3}) + 432\sqrt{3} = 216\sqrt{3} + 432\sqrt{3} = 648\sqrt{3}$.
**Circumscribed Triangular Prism**
For the circumscribed triangular prism, the base is an equilateral triangle circumscribed about a circle of radius $r$.
The relationship between the side length $S$ of the equilateral triangle and the radius $r$ is $r = \frac{S}{2\sqrt{3}}$, so $S = 2\sqrt{3} r$.
Since $r = 12$, $S = 2\sqrt{3} \cdot 12 = 24\sqrt{3}$.
The area of the equilateral triangle is $A_{circumscribed} = \frac{\sqrt{3}}{4} S^2 = \frac{\sqrt{3}}{4} (24\sqrt{3})^2 = \frac{\sqrt{3}}{4} (576 \cdot 3) = 432\sqrt{3}$.
The lateral surface area of the circumscribed prism is $3 \cdot S \cdot h = 3 \cdot 24\sqrt{3} \cdot 12 = 864\sqrt{3}$.
The total surface area of the circumscribed prism is $2 A_{circumscribed} + 3Sh = 2(432\sqrt{3}) + 864\sqrt{3} = 864\sqrt{3} + 864\sqrt{3} = 1728\sqrt{3}$.
**Total Surface Areas**
The total surface area of the inscribed prism is $648\sqrt{3}$ square inches.
The total surface area of the circumscribed prism is $1728\sqrt{3}$ square inches.
Final Answer: The final answer is $\boxed{648\sqrt{3}, 1728\sqrt{3}}$

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