# 016 Radius of the sphere circumscribing a regular triangular pyramid

**Example 016**

Find the area of the surface and the volume of the sphere circumscribed about a regular tetrahedron of edge 25 cm. See Figure 015.

**Solution 016**

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*

**Another Solution**

$h = \dfrac{a\sqrt{6}}{3}$

$h = \dfrac{25\sqrt{6}}{3}$

$h = 20.41 \, \text{ cm}$

The center of circumscribing and inscribed spheres is coincident with the centroid of regular tetrahedron. Thus,

$r = \frac{1}{4}h$ and

$R = \frac{3}{4}h$

$r = 5.10 \, \text{ cm}$ → radius of inscribed sphere (*okay*)

$R = 15.31 \, \text{ cm}$ → radius of circumscribing sphere (*okay!*)