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10-30-2007

My attempt.

You know that the surface area of the tetrahedron is $a^2\sqrt{3}$ , making the side length $\frac{19}{\sqrt{3}}$ . This gives the volume: $\frac{6859\sqrt{3^3}\sqrt{2}}{324}$

You all may laugh at this failing attempt.

10-30-2007	#72
hunter34 Guest Posts: n/a	My attempt. You know that the surface area of the tetrahedron is $a^2\sqrt{3}$ , making the side length $\frac{19}{\sqrt{3}}$ . This gives the volume: $\frac{6859\sqrt{3^3}\sqrt{2}}{324}$ You all may laugh at this failing attempt.