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
Given "A=144cm^2"
"4\\pi r^2=144""r^2=\\dfrac{36}{\\pi}""r=\\dfrac{6\\sqrt{\\pi}}{\\pi}\\approx3.4(cm)"
No correct answer.
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