Considering that the tetrahedron OABC is a regular triangular pyramid, its base ABC is a regular triangle, and the height of the crossing point O is a/2, because the height of the triangular prism ABC-a' b' c' is its side AA' = A, and considering that the two tetrahedrons OABC and OA' b' c' are symmetrical, the height of the tetrahedron OABC is OH = A/2, where H is the center of gravity of the regular triangle ABC.

Triangle OHA is a right triangle, OH = A/2, HA=a/√3 (using the knowledge of plane geometry), and hypotenuse OA = A (√ 2 1)/6 = R can be calculated by Pythagorean theorem.

The surface area of the sphere is s = 4 π r 2 = 7 π a 2/3.