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Question Number 19955 by ajfour last updated on 19/Aug/17
Commented by ajfour last updated on 19/Aug/17
Find the volume of the regular tetrahedron in terms of r .
$${Find}\:{the}\:{volume}\:{of}\:{the}\:{regular} \\ $$$${tetrahedron}\:{in}\:{terms}\:{of}\:\boldsymbol{{r}}\:. \\ $$
Commented by ajfour last updated on 19/Aug/17
Answered by dioph last updated on 19/Aug/17
(2r)^2 = l^2 + ((l/( (√2))))^2 4r^2 = l^2 + (l^2 /2) r^2 = ((3l^2 )/8) ∴ l = ((2r(√2))/( (√3))) h^2 + (((l(√3))/3))^2 = l^2 h^2 = l^2 − (l^2 /3) h = ((l(√2))/( (√3))) V = (1/3)(((l^2 (√3))/4) × h) = ((l^3 (√2))/(12)) ⇒ V = ((8r^3 ×2(√2))/(3(√3)))×((√2)/(12)) = ((32r^3 )/(36(√3))) = ((8r^3 )/(9(√3)))
$$\left(\mathrm{2}{r}\right)^{\mathrm{2}} \:=\:{l}^{\mathrm{2}} \:+\:\left(\frac{{l}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{r}^{\mathrm{2}} \:=\:{l}^{\mathrm{2}} \:+\:\frac{{l}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} \:=\:\frac{\mathrm{3}{l}^{\mathrm{2}} }{\mathrm{8}}\:\therefore\:{l}\:=\:\frac{\mathrm{2}{r}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}} \\ $$$${h}^{\mathrm{2}} \:+\:\left(\frac{{l}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} \:=\:{l}^{\mathrm{2}} \\ $$$${h}^{\mathrm{2}} \:=\:{l}^{\mathrm{2}} \:−\:\frac{{l}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${h}\:=\:\frac{{l}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}} \\ $$$${V}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{{l}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}}\:×\:{h}\right)\:=\:\frac{{l}^{\mathrm{3}} \sqrt{\mathrm{2}}}{\mathrm{12}} \\ $$$$\Rightarrow\:{V}\:=\:\frac{\mathrm{8}{r}^{\mathrm{3}} ×\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{\mathrm{3}}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{12}}\:=\:\frac{\mathrm{32}{r}^{\mathrm{3}} }{\mathrm{36}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{8}{r}^{\mathrm{3}} }{\mathrm{9}\sqrt{\mathrm{3}}} \\ $$
Commented by ajfour last updated on 19/Aug/17
Excellent Sir, thanks a lot. but the first line, (l/( (√2))) ; i did not understand..
$${Excellent}\:{Sir},\:{thanks}\:{a}\:{lot}. \\ $$$${but}\:{the}\:{first}\:{line},\:\frac{{l}}{\:\sqrt{\mathrm{2}}}\:;\:{i}\:{did}\:{not} \\ $$$${understand}.. \\ $$
Commented by dioph last updated on 19/Aug/17
Commented by dioph last updated on 19/Aug/17
My poor attempt on drawing a tetrahedron in a way that its edges are the diagonals of the faces of a cube. The main diagonal of the cube is 2r, hence (2r)^2 = l^2 + ((l/( (√2))))^2
$$\mathrm{My}\:\mathrm{poor}\:\mathrm{attempt}\:\mathrm{on}\:\mathrm{drawing}\:\mathrm{a} \\ $$$$\mathrm{tetrahedron}\:\mathrm{in}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{its}\:\mathrm{edges} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{diagonals}\:\mathrm{of}\:\mathrm{the}\:\mathrm{faces}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{cube}. \\ $$$$\mathrm{The}\:\mathrm{main}\:\mathrm{diagonal}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{is} \\ $$$$\mathrm{2}{r},\:\mathrm{hence}\:\left(\mathrm{2}{r}\right)^{\mathrm{2}} \:=\:{l}^{\mathrm{2}} \:+\:\left(\frac{{l}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$
Commented by ajfour last updated on 19/Aug/17
thanks sir, all the more.
$${thanks}\:{sir},\:{all}\:{the}\:{more}. \\ $$

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