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# How to find the height of a regular tetrahedron?

Question:

In the figure shown, P-ABC is a regular tetrahedron with side length a. Prove: the height of the regular tetrahedron is (square root of 6) a/3 in which a is the side of the regular tetrahedron.

Proof:

Because P-ABC is a regular tetrahedron, so, each side is equal to a. PA = PB = PC = AB = BC = CA = a. Each triangle face is an equilateral triangle with side length a.

Draw a line from the vertex P perpendicular to the plane ABC, this line intersects the plane ABC at point O. Because P-ABC is a regular tetrahedron, so, the point O lies on the center of the equilateral triangle ABC.

From the property of the equilateral triangle, OA is the distance from the center of the equilateral triangle to a vertex. From our previous lesson, OA = (square root of 3) a/3 in which a is the length of the side of the equilateral triangle.

Because line segment PO perpendicular to the plane ABC, and because OA lies on the plane ABC, so, the line segment PO perpendicular to the line segment OA. So, triangle POA is a right triangle.

In right triangle POA,
PO2 = PA2 - OA2
= a2 - [(square root of 3) a/3]2
= a2 - (3/9) a2
= a2 (1 - 3/9)
= a2 (9/9 - 3/9)
= a2 (6/9)
apply square root to both side of the equation
PO = (square root of 6) a/3

Therefore, the height of a regular tetrahedron is (square root of 6) a/3 in which a is the side of the regular tetrahedron. Please watch the video for more details.