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# An example of regular triangular prism and sphere

Question:

ABC-A1B1C1 is a regular triangle prism. Each side of the triangle is a. The edge of the regular triangular prism is 2a. If each vertex of the regular triangular prism lies on a sphere, what is the radius of the sphere? Solution:

Because this is a regular triangular prism, so, each side of the triangle is a (a is a number.) Each edge is parallel to each other and equal to 2a. A1A is perpendicular to the base plane ABC. Point O is the center of the equilateral triangle ABC. Point O1 is the center of the equilateral triangle A1B1C1. Point O2 is the midpoint of OO1. The point O2 is the center of the sphere because the distance from O2 to each vertex of the regular triangular prism is equal.

Connect AO2 and AO. Because OO1 parallel to AA1 and AA1 is perpendicular to the plane ABC, so, OO1 is perpendicular to the plane ABC. Because AO lies on the plane ABC, so, OO1 is perpendicular to AO. The angle O1OA is 90 degrees. Because O2 lies on OO1, so, the angle O2OA is 90 degrees. The triangle O2OA is a right triangle. AO is the projection of AO2 on the plane ABC.

In right triangle AOO2,
AO2 = r (r is the radius of the sphere)
O2O = OO1/2 = 2a/2 = a

AO is the distance from a vertex of an equilateral triangle to its center, so AO = square root of 3 times a divide by 3 (in which, a is the length of the side of the equilateral triangle.)

r2 = a2 + [square root of 3) a/3]2
= a2 + (3/9)a2
= a2 + (1/3)a2
= (4/3) a2
r = [2/(square root of 3)] a
= (2/3)(square root of 3) a

Therefore, the radius of the sphere is two-third of square root of 3 times the side length of the equilateral triangle. Please watch the video for more details.