Each edge of a regular tetrahedron is 6. Find the volumes of the tetrahedron and the cone inscribed in it.

| May 10, 2021 | education

The tetrahedron has all the side faces and the base of equilateral triangles.

Determine the area of ​​the base of the tetrahedron.

Sbn = AB * AC * Sin60 / 2 = 6 * 6 * Sin60 / 2 = 36 * √3 / 4 = 9 * √3 cm2.

Also Savs = BC * AH / 2.

АН = 2 * Saс / ВС = 2 * 9 * √3 / 6 = 3 * √3 cm.

Point O we divide the height AH in the ratio 2/1, then AO = 2 * √3 cm, OH = √3 cm.

In a right-angled triangle AOD, according to the Pythagorean theorem, OD ^ 2 = AD ^ 2 – AO ^ 2 = 36 – 12 = 24. OD = 2 * √6 cm.

Then the volume of the tetrahedron is: V = Sbas * OD / 3 = 9 * √3 * 2 * √6 / 3 = 6 * √18 = 18 * √2 cm3.

OH is the radius of the inscribed circle in the ABC triangle, then the volume of the inscribed cone is equal to: Vcon = π * r2 * OD / 3 = π * 3 * 2 * √6 / 3 = π * 2 * √6 cm3.

Answer: The volume of the tetrahedron is 18 * √2 cm3, the volume of the cone is π * 2 * √6 cm3.



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