Question-47145

Question Number 47145 by ajfour last updated on 05/Nov/18

Commented by ajfour last updated on 05/Nov/18

Q.47113 Find volume of pyramid in terms of a,b,c,p,q,r.

$${Q}.\mathrm{47113}\:\:\:{Find}\:{volume}\:{of} \\ $$$${pyramid}\:{in}\:{terms}\:{of}\:{a},{b},{c},{p},{q},{r}. \\ $$

Answered by ajfour last updated on 05/Nov/18

let A be origin. B(ccos φ,−csin φ,0) C(bcos θ,bsin φ,0) H(x,0,h) x^2 +h^2 =p^2 ...(i) (x−bcos θ)^2 +b^2 sin^2 θ+z^2 = r^2 ..(ii) (x−ccos φ)^2 +c^2 sin^2 φ+z^2 = q^2 ..(iii) (ccos φ−bcos θ)^2 +(bsin θ+csin φ)^2 =a^2 ..(iv) (i)−(ii) ⇒ 2bxcos θ = p^2 +b^2 −r^2 = l ...(I) (iii) ⇒ 2cxcos φ = p^2 +c^2 −q^2 =m ...(II) (iv) ⇒ 2bc(cos θcos φ−sin θsin φ) = b^2 +c^2 −a^2 = n ...(III) ⇒ ((lm)/(2x^2 ))−2bc(√((1−(l^2 /(4b^2 x^2 )))(1−(m^2 /(4c^2 x^2 ))))) = n ⇒ (((lm)/(2x^2 ))−n)^2 =(1/4)(4b^2 −(l^2 /x^2 ))(4c^2 −(m^2 /x^2 )) (lm−2nx^2 )^2 =(4b^2 x^2 −l^2 )(4c^2 x^2 −m^2 ) ⇒ 4n^2 x^4 −4lmnx^2 = 16b^2 c^2 x^4 −4x^2 (c^2 l^2 +b^2 m^2 ) ⇒ x^2 = ((b^2 m^2 +c^2 l^2 −lmn)/(4b^2 c^2 −n^2 )) and h^2 = p^2 −x^2 Volume = (h/3)S_(ABC) (may be).

$${let}\:{A}\:{be}\:{origin}. \\ $$$${B}\left({c}\mathrm{cos}\:\phi,−{c}\mathrm{sin}\:\phi,\mathrm{0}\right) \\ $$$${C}\left({b}\mathrm{cos}\:\theta,{b}\mathrm{sin}\:\phi,\mathrm{0}\right) \\ $$$${H}\left({x},\mathrm{0},{h}\right) \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +{h}^{\mathrm{2}} ={p}^{\mathrm{2}} \:\:…\left({i}\right) \\ $$$$\left({x}−{b}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+{z}^{\mathrm{2}} \:=\:{r}^{\mathrm{2}} \:\:..\left({ii}\right) \\ $$$$\left({x}−{c}\mathrm{cos}\:\phi\right)^{\mathrm{2}} +{c}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \phi+{z}^{\mathrm{2}} =\:{q}^{\mathrm{2}} \:\:..\left({iii}\right) \\ $$$$\left({c}\mathrm{cos}\:\phi−{b}\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\left({b}\mathrm{sin}\:\theta+{c}\mathrm{sin}\:\phi\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:..\left({iv}\right) \\ $$$$\left({i}\right)−\left({ii}\right)\:\Rightarrow \\ $$$$\:\:\:\mathrm{2}{bx}\mathrm{cos}\:\theta\:=\:{p}^{\mathrm{2}} +{b}^{\mathrm{2}} −{r}^{\mathrm{2}} \:=\:{l}\:\:\:…\left({I}\right) \\ $$$$\left({iii}\right)\:\Rightarrow \\ $$$$\:\:\:\mathrm{2}{cx}\mathrm{cos}\:\phi\:=\:{p}^{\mathrm{2}} +{c}^{\mathrm{2}} −{q}^{\mathrm{2}} \:={m}\:\:…\left({II}\right) \\ $$$$\left({iv}\right)\:\Rightarrow \\ $$$$\:\:\mathrm{2}{bc}\left(\mathrm{cos}\:\theta\mathrm{cos}\:\phi−\mathrm{sin}\:\theta\mathrm{sin}\:\phi\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \:\:=\:{n}\:\:\:\:…\left({III}\right) \\ $$$$\Rightarrow\:\:\frac{{lm}}{\mathrm{2}{x}^{\mathrm{2}} }−\mathrm{2}{bc}\sqrt{\left(\mathrm{1}−\frac{{l}^{\mathrm{2}} }{\mathrm{4}{b}^{\mathrm{2}} {x}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{m}^{\mathrm{2}} }{\mathrm{4}{c}^{\mathrm{2}} {x}^{\mathrm{2}} }\right)}\:=\:{n} \\ $$$$\Rightarrow\:\left(\frac{{lm}}{\mathrm{2}{x}^{\mathrm{2}} }−{n}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{4}{b}^{\mathrm{2}} −\frac{{l}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)\left(\mathrm{4}{c}^{\mathrm{2}} −\frac{{m}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right) \\ $$$$\left({lm}−\mathrm{2}{nx}^{\mathrm{2}} \right)^{\mathrm{2}} =\left(\mathrm{4}{b}^{\mathrm{2}} {x}^{\mathrm{2}} −{l}^{\mathrm{2}} \right)\left(\mathrm{4}{c}^{\mathrm{2}} {x}^{\mathrm{2}} −{m}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\mathrm{4}{n}^{\mathrm{2}} {x}^{\mathrm{4}} −\mathrm{4}{lmnx}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{16}{b}^{\mathrm{2}} {c}^{\mathrm{2}} {x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{2}} \left({c}^{\mathrm{2}} {l}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:{x}^{\mathrm{2}} \:=\:\frac{{b}^{\mathrm{2}} {m}^{\mathrm{2}} +{c}^{\mathrm{2}} {l}^{\mathrm{2}} −{lmn}}{\mathrm{4}{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{n}^{\mathrm{2}} } \\ $$$$\:\:{and}\:\:\boldsymbol{{h}}^{\mathrm{2}} \:=\:\boldsymbol{{p}}^{\mathrm{2}} −\boldsymbol{{x}}^{\mathrm{2}} \\ $$$$\:\:{Volume}\:=\:\frac{\boldsymbol{{h}}}{\mathrm{3}}\boldsymbol{{S}}_{{ABC}} \:\:\:\left({may}\:{be}\right). \\ $$

Commented by MrW3 last updated on 05/Nov/18