Answer to Question #150235 in Geometry for solid mensuration

Question

The surface area of a sphere inscribed in a regular tetrahedron is 144 cm2
22. What is the altitude of the tetrahedron?.

a) 20 cm
b) 24 cm
c) 28 cm
d) 48 cm

Answer 1

Let $a=$ base of a regular tetrahedron, $l=$ slant height of a a regular tetrahedron, $r=$ radius of a sphere inscribed in a regular tetrahedron.

The surface area of a sphere

$$A=4\pi r^2$$

Given $A=144cm^2$

$$4\pi r^2=144$$ $$r^2=\dfrac{36}{\pi}$$

$$r=\dfrac{6\sqrt{\pi}}{\pi}\approx3.4(cm)$$ $$SD=l=a\dfrac{\sqrt{3}}{2}, ED=\dfrac{a}{2\sqrt{3}}$$

From the right triangle $SED$

$$SD^2=SE^2+ED^2$$

$$l^2=h^2+(\dfrac{a}{2\sqrt{3}})^2$$

$$\dfrac{3}{4}a^2=h^2+\dfrac{1}{12}a^2$$

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

$$\angle ESD=30\degree, \angle EDS=60\degree$$

Right triangle $SOF$

$$SO=2OF=2r$$

$$h=SE=SO+OE=2r+r=3r$$

$$h=3r=\dfrac{18\sqrt{\pi}}{\pi}(cm)\approx10.155(cm)$$

The altitude of the tetrahedron is $h=\dfrac{18\sqrt{\pi}}{\pi}cm\approx10.155cm$

No right answer.