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Question Number 47195 by MrW3 last updated on 06/Nov/18

Find the volume of the pyramid which is folded from a trangular paper with sides a, b and c.

$${Find}\:{the}\:{volume}\:{of}\:{the}\:{pyramid}\:{which} \\ $$$${is}\:{folded}\:{from}\:{a}\:{trangular}\:{paper}\:{with} \\ $$$${sides}\:\boldsymbol{{a}},\:\boldsymbol{{b}}\:{and}\:\boldsymbol{{c}}. \\ $$

Answered by MJS last updated on 06/Nov/18

in this case it′s easy to use Euler′s formula because the sides of the bottom triangle are (a/2), (b/2), (c/2) and the skew sides have the same values we get V=((√2)/(96))(√(a^4 (b^2 +c^2 )+b^4 (a^2 +c^2 )+c^4 (a^2 +b^2 )−(a^6 +b^6 +c^6 )))= =((√2)/(96))(√((a^2 +b^2 −c^2 )(a^2 −b^2 +c^2 )(−a^2 +b^2 +c^2 )))

$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{formula} \\ $$$$\mathrm{because}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bottom}\:\mathrm{triangle} \\ $$$$\mathrm{are}\:\frac{{a}}{\mathrm{2}},\:\frac{{b}}{\mathrm{2}},\:\frac{{c}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{the}\:\mathrm{skew}\:\mathrm{sides}\:\mathrm{have}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{values} \\ $$$$\mathrm{we}\:\mathrm{get}\:{V}=\frac{\sqrt{\mathrm{2}}}{\mathrm{96}}\sqrt{{a}^{\mathrm{4}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{b}^{\mathrm{4}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{c}^{\mathrm{4}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} \right)}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{96}}\sqrt{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left(−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)} \\ $$

Commented by MrW3 last updated on 06/Nov/18

thanks alot sir!

$${thanks}\:{alot}\:{sir}! \\ $$

Commented by ajfour last updated on 06/Nov/18

Sir please elaborate..

$${Sir}\:{please}\:{elaborate}.. \\ $$

Commented by MJS last updated on 07/Nov/18

Commented by MJS last updated on 07/Nov/18

a, b, c are the basic edges p is skew to a q is skew to b r is skew to c Euler′s formula gives the volume of a general tetrahedron we know that V=((Bh)/3) with B being the area of the basic triangle ⇒ h=((3V)/B)=(ε/(4B)) with ε being the root in Euler′s formula B=(Δ/4) with Δ=(√((a+b+c)(a+b−c)(a−b+c)(−a+b+c))) ⇒ h=(ε/Δ)

$${a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{basic}\:\mathrm{edges} \\ $$$${p}\:\mathrm{is}\:\mathrm{skew}\:\mathrm{to}\:{a} \\ $$$${q}\:\mathrm{is}\:\mathrm{skew}\:\mathrm{to}\:{b} \\ $$$${r}\:\mathrm{is}\:\mathrm{skew}\:\mathrm{to}\:{c} \\ $$$$\mathrm{Euler}'\mathrm{s}\:\mathrm{formula}\:\mathrm{gives}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{a}\:\mathrm{general} \\ $$$$\mathrm{tetrahedron} \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:{V}=\frac{{Bh}}{\mathrm{3}}\:\mathrm{with}\:{B}\:\mathrm{being}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{basic}\:\mathrm{triangle}\:\Rightarrow\:{h}=\frac{\mathrm{3}{V}}{{B}}=\frac{\epsilon}{\mathrm{4}{B}}\:\mathrm{with}\:\epsilon\:\mathrm{being} \\ $$$$\mathrm{the}\:\mathrm{root}\:\mathrm{in}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{formula} \\ $$$${B}=\frac{\Delta}{\mathrm{4}}\:\mathrm{with}\:\Delta=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)} \\ $$$$\Rightarrow\:{h}=\frac{\epsilon}{\Delta} \\ $$

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