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Question Number 40686 by MrW3 last updated on 26/Jul/18

Commented by MrW3 last updated on 26/Jul/18

Find radius R of the circumsphere of pyramid. I got R=(((√3)b^2 )/(2(√(3b^2 −a^2 ))))

$${Find}\:{radius}\:{R}\:{of}\:{the}\:{circumsphere}\:{of} \\ $$$${pyramid}. \\ $$$$ \\ $$$${I}\:{got}\:{R}=\frac{\sqrt{\mathrm{3}}{b}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$

Commented by ajfour last updated on 26/Jul/18

Commented by ajfour last updated on 26/Jul/18

R^2 =x^2 +(a^2 /3) ...(i) b^2 =(a^2 /3)+(R+x)^2 ....(ii) from (ii) x=[((√(3b^2 −a^2 ))/(√3))−R] using this in (i) R^2 =[((√(3b^2 −a^2 ))/(√3))−R]^2 +(a^2 /3) ⇒ ((2R)/(√3))((√(3b^2 −a^2 )))= b^2 ⇒ R = ((b^2 (√3))/(2(√(3b^2 −a^2 )))) .

$${R}^{\mathrm{2}} ={x}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{\mathrm{3}}\:\:\:\:\:\:\:...\left({i}\right) \\ $$$${b}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{3}}+\left({R}+{x}\right)^{\mathrm{2}} \:\:\:\:\:\:\:....\left({ii}\right) \\ $$$${from}\:\left({ii}\right)\:\:\:{x}=\left[\frac{\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\sqrt{\mathrm{3}}}−{R}\right] \\ $$$${using}\:{this}\:{in}\:\left({i}\right) \\ $$$${R}^{\mathrm{2}} =\left[\frac{\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\sqrt{\mathrm{3}}}−{R}\right]^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}{R}}{\sqrt{\mathrm{3}}}\left(\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)=\:{b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:{R}\:=\:\frac{{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:. \\ $$

Commented by MrW3 last updated on 26/Jul/18

thanks sir! I tried in this way: h=(√(b^2 −((a/(√3)))^2 )) ((2R)/b)=(b/h) ⇒R=(b^2 /(2h))=(b^2 /(2(√(b^2 −(a^2 /3)))))=(((√3)b^2 )/(2(√(3b^2 −a^2 ))))

$${thanks}\:{sir}! \\ $$$${I}\:{tried}\:{in}\:{this}\:{way}: \\ $$$${h}=\sqrt{{b}^{\mathrm{2}} −\left(\frac{{a}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{2}{R}}{{b}}=\frac{{b}}{{h}} \\ $$$$\Rightarrow{R}=\frac{{b}^{\mathrm{2}} }{\mathrm{2}{h}}=\frac{{b}^{\mathrm{2}} }{\mathrm{2}\sqrt{{b}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{3}}}}=\frac{\sqrt{\mathrm{3}}{b}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$

Commented by ajfour last updated on 26/Jul/18

Its better Sir!

$${Its}\:{better}\:{Sir}! \\ $$

Commented by MrW3 last updated on 26/Jul/18

Thank you sir! Is there a way to calculate R for any pyramid with sides a,b,c (base) and p,q,r?

$${Thank}\:{you}\:{sir}! \\ $$$${Is}\:{there}\:{a}\:{way}\:{to}\:{calculate}\:{R}\:{for}\:{any} \\ $$$${pyramid}\:{with}\:{sides}\:{a},{b},{c}\:\left({base}\right)\:{and}\: \\ $$$${p},{q},{r}? \\ $$

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