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.
21. The surface area of a sphere
Given "A=144cm^2"
"4\\pi r^2=144""r^2=\\dfrac{36}{\\pi}""r=\\dfrac{6\\sqrt{\\pi}}{\\pi}\\approx3.4(cm)"
22.
From the right triangle "SED"
"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"
"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"
23. If a sphere is inscribed in a cube of side "a", then
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