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?

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.

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.