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Regular triangular prism sphere volume example

Question:

In the figure shown ABC-A1B1C1 is a regular triangular prism. Each side of the triangle is square root of 3. The height of the prism is 2. If all vertices lie on a sphere, what is the volume of the sphere?

how to find the volume of a sphere if all vertices of a regular triangular prism lie in a sphere

Solution:

Because ABC-A1B1C1 is a regular triangular prism, so, triangle ABC and triangle A1B1C1 are equilateral triangle and the side of the equilateral triangle is square root of 3. Top triangle ABC and bottom triangle A1B1C1 are congruent and parallel. The triangle plane is perpendicular to edge AA1 and AA1 // BB1 // CC1. AA1 = BB1 = CC1 = 2.

Point O is the center of the triangle ABC. O1 is the center of the triangle A1B1C1. Connect OO1, then OO1 // AA1 and OO1 = 2.

If all vertices of the regular triangular prism lie on a sphere, then the center of the sphere is the midpoint of OO1.

Let O2 be the center of the sphere, then AO2 = R. R is the radius of the sphere.

the center of the sphere when all vertices of a regular triangular lie on a sphere.

Because plane ABC is perpendicular to OO1. AO is a line on plane ABC, so, AO is perpendicular to OO1. Because O2 lies on OO1, so AO is perpendicular to OO2. So, triangle AOO2 is a right triangle.

In right triangle AOO2,
AO22 = AO2 + OO22
AO2 = R

Because ABC is an equilateral triangle, so, the height of the equilateral triangle is:

AD = (square root of 3)/2 × AC
= (square root of 3)/2 × (square root of 3)
= 3/2

AO is the distance from a vertex of the equilateral triangle to the center of the equilateral triangle.

AO = (2/3) × AD
= (2/3) × (3/2)
= 1
R2 = AO2 + OO22
= 12 + 12
= 2
R = (square root of 2)

The volume of the sphere is:

V = (4/3) pi R3
= (4/3) pi (square root of 2)3
= (4/3) pi 2 (square root of 2)
= (8/3) (square root of 2) pi
= 11.84

Therefore, the volume of the sphere is 11.84. Watch the video for more details.