Asked by Anna
Find the volume of a regular tetrahedron in which each edge is 8 in. long.
I am so confused on this one and I really need some help.
I am so confused on this one and I really need some help.
Answers
Answered by
drwls
The base (square) area is 8x8 = 64 square inches. Multiply that by 1/3 of the height.
Getting the height is the hard part. Consider the triangle formed by the apex, one corner, and a point that is the center of the base. Use the Pythagorean theorem on that triangle to get the pyramid height, H.
(4sqrt2)^2 + H^2 = 8^2
H^2 = 64 - 32 = 32
H = 4 sqrt 2 inches
Volume = (1/3)(64 in^2)*4 sqrt 2 in
= (4/3)*64*sqrt2 = 120.7 in^3
Getting the height is the hard part. Consider the triangle formed by the apex, one corner, and a point that is the center of the base. Use the Pythagorean theorem on that triangle to get the pyramid height, H.
(4sqrt2)^2 + H^2 = 8^2
H^2 = 64 - 32 = 32
H = 4 sqrt 2 inches
Volume = (1/3)(64 in^2)*4 sqrt 2 in
= (4/3)*64*sqrt2 = 120.7 in^3
Answered by
Anna
But for a tetrahedron the bottom is not square it is a triangle. Does this formula still apply?
Answered by
Marth
V = (1/3)bh. The area of the base is easy to calculate as it is an equilateral triangle.
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