# 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

= 4*sqrt3/4*(edge)^2

=`sqrt3*2^2`

= `4sqrt3` sq. inches.

Hence area of the tetrahedron is `4sqrt3` sq.in. and its volume is `(2sqrt2)/3` cubic inches.

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