Triangles ABC and BCD are equilateral triangles. See
the regular tetrahedron for more information. Triangle AGD is isosceles triangle. DE and AF are altitudes of AGD and intersect at O, thus O is the center of the circumscribing sphere. E and F are the centroids of ABC and BCD, respectively.
From right triangle BGA
$AG^2 + BG^2 = AB^2$
$AG^2 + 12.5^2 = 25^2$
$AG = 21.65 \, \text{ cm} = DG$
$AE = \frac{2}{3}AG = \frac{2}{3}(21.65)$
$AE = 14.43 \, \text{ cm} = DF$
$EG = \frac{1}{3}AG = \frac{2}{3}(21.65)$
$EG = 7.22 \, \text{ cm} = FG$
From right triangle AED
$ED^2 + AE^2 = AD^2$
$ED^2 + 14.43^2 = 25^2$
$ED = 20.42 \, \text{ cm} = FA$
From isosceles triangle AGD
$\sin \frac{1}{2}(\angle AGD) = \dfrac{\frac{1}{2}AD}{AG}$
$\sin \frac{1}{2}(\angle AGD) = \dfrac{\frac{1}{2}(25)}{21.65}$
$\frac{1}{2}(\angle AGD) = 35.26^\circ$
$\angle AGD = 70.52^\circ$
From right triangle GEO
$\tan \frac{1}{2}(\angle AGD) = \dfrac{OE}{EG}$
$\tan 35.26^\circ = \dfrac{OE}{7.22}$
$OE = 5.10 \, \text{ cm} = OF$ → radius of the inscribed sphere, r
$OD = ED - OE = 20.42 - 5.10$
$OD = 15.32 \, \text{ cm} = OA$ → radius of the circumscribing sphere, R
Area of circumscribing sphere:
$A = 4\pi R^2 = 4\pi (15.32^2)$
$A = 2\,949.36 \, \text{ cm}^2$ answer
Volume of circumscribing sphere:
$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (15.32^3)$
$V = 15\,061.38 \, \text{ cm}^3$ answer
