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.
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
But for a tetrahedron the bottom is not square it is a triangle. Does this formula still apply?
V = (1/3)bh. The area of the base is easy to calculate as it is an equilateral triangle.
To find the volume of a regular tetrahedron, you can use the formula:
Volume = (sqrt(2) / 12) * edge length^3
In this case, the edge length is given as 8 inches. So, let's substitute this value into the formula and calculate the volume.
Volume = (sqrt(2) / 12) * 8^3
First, let's calculate 8^3:
8^3 = 8 * 8 * 8 = 512
Now, substitute this value back into the formula:
Volume = (sqrt(2) / 12) * 512
To simplify it further, let's calculate sqrt(2) using a calculator:
sqrt(2) ≈ 1.414
Now, substitute this value back into the formula:
Volume ≈ (1.414 / 12) * 512
Divide 1.414 by 12:
Volume ≈ (0.1178) * 512
Multiply 0.1178 by 512:
Volume ≈ 60.2368
Rounding it to two decimal places, the volume of the tetrahedron is approximately 60.24 cubic inches.