How do I find the surface area and the volume of a regular tetrahedron with two 2 in edges?
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Regular tetrahedron is a triangular pyramid with all four faces being equilateral triangles.
Here, the triangles have all equal sides = 2 in., and height of the pyramid, h = `2sqrt6/3` in.
Volume of the regular tetrahedron is given by
V = 1/3*area of the base *height
=1/3*(sqrt3/4) (edge)^2 *h
= `1/3(sqrt3)/4 *2^2*(2sqrt6)/3`
= `(2sqrt2)/3 ` cubic inches
And Total surface area
= area of the four equilateral triangles
= `4sqrt3` sq. inches.
Hence area of the tetrahedron is `4sqrt3` sq.in. and its volume is `(2sqrt2)/3` cubic inches.
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