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Question-40581

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Question Number 40581 by ajfour last updated on 24/Jul/18
Commented by ajfour last updated on 24/Jul/18
Find radius of inscribed sphere which is tangent to all faces of the triangular pyramid.
$${Find}\:{radius}\:{of}\:{inscribed}\:{sphere} \\ $$$${which}\:{is}\:{tangent}\:{to}\:{all}\:{faces}\:{of} \\ $$$${the}\:{triangular}\:{pyramid}. \\ $$
Commented by ajfour last updated on 24/Jul/18
See Q.40594 for an alternate solution.
$${See}\:{Q}.\mathrm{40594}\:\:{for}\:{an}\:{alternate}\:{solution}. \\ $$
Answered by MrW3 last updated on 24/Jul/18
height of pyramid: h=(√(b^2 −((2/3)×(((√3)a)/2))^2 ))=(√(b^2 −(a^2 /3))) V=(1/3)×(a/2)×(((√3)a)/2)×(√(b^2 −(a^2 /3)))=((a^2 (√(3b^2 −a^2 )))/(12)) S=(a/2)×(((√3)a)/2)+3×(a/2)×(√(b^2 −(a^2 /4)))=((a((√3)a+3(√(4b^2 −a^2 ))))/4) ⇒r=((3V)/S)=((a(√(3b^2 −a^2 )))/( (√3)a+3(√(4b^2 −a^2 ))))
$${height}\:{of}\:{pyramid}: \\ $$$${h}=\sqrt{{b}^{\mathrm{2}} −\left(\frac{\mathrm{2}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{{b}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{3}}} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{{a}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}×\sqrt{{b}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{3}}}=\frac{{a}^{\mathrm{2}} \sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{12}} \\ $$$${S}=\frac{{a}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}+\mathrm{3}×\frac{{a}}{\mathrm{2}}×\sqrt{{b}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}=\frac{{a}\left(\sqrt{\mathrm{3}}{a}+\mathrm{3}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right)}{\mathrm{4}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}{V}}{{S}}=\frac{{a}\sqrt{\mathrm{3}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\:\sqrt{\mathrm{3}}{a}+\mathrm{3}\sqrt{\mathrm{4}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }} \\ $$
Commented by MrW3 last updated on 24/Jul/18
Let M be the center of the sphere. M has the same distance r to all faces of the pyramid. The big pyramid can be divided into 4 small ones: M_ABC with V_1 =(r/3)A_(ΔABC) M_ABD with V_2 =(r/3)A_(ΔABD) M_ACD with V_3 =(r/3)A_(ΔACD) M_BCD with V_4 =(r/3)A_(ΔBCD) V=V_1 +V_2 +V_3 +V_4 =(r/3)(A_(ΔABC) +...)=((rS)/3) with S=surface of pyramid ⇒r=((3V)/S)
$${Let}\:{M}\:{be}\:{the}\:{center}\:{of}\:{the}\:{sphere}.\:{M} \\ $$$${has}\:{the}\:{same}\:{distance}\:\:{r}\:{to}\:{all}\:{faces}\:{of} \\ $$$${the}\:{pyramid}.\:{The}\:{big}\:{pyramid}\:{can}\:{be} \\ $$$${divided}\:{into}\:\mathrm{4}\:{small}\:{ones}: \\ $$$${M\_ABC}\:{with}\:{V}_{\mathrm{1}} =\frac{{r}}{\mathrm{3}}{A}_{\Delta{ABC}} \\ $$$${M\_ABD}\:{with}\:{V}_{\mathrm{2}} =\frac{{r}}{\mathrm{3}}{A}_{\Delta{ABD}} \\ $$$${M\_ACD}\:{with}\:{V}_{\mathrm{3}} =\frac{{r}}{\mathrm{3}}{A}_{\Delta{ACD}} \\ $$$${M\_BCD}\:{with}\:{V}_{\mathrm{4}} =\frac{{r}}{\mathrm{3}}{A}_{\Delta{BCD}} \\ $$$${V}={V}_{\mathrm{1}} +{V}_{\mathrm{2}} +{V}_{\mathrm{3}} +{V}_{\mathrm{4}} =\frac{{r}}{\mathrm{3}}\left({A}_{\Delta{ABC}} +…\right)=\frac{{rS}}{\mathrm{3}} \\ $$$${with}\:{S}={surface}\:{of}\:{pyramid} \\ $$$$\Rightarrow{r}=\frac{\mathrm{3}{V}}{{S}} \\ $$
Commented by ajfour last updated on 24/Jul/18
what is S , sir and how r=((3V)/S) ?
$${what}\:{is}\:{S}\:,\:{sir}\:{and}\:{how}\:{r}=\frac{\mathrm{3}{V}}{{S}}\:? \\ $$
Commented by MrW3 last updated on 24/Jul/18
Commented by MrW3 last updated on 24/Jul/18
When we know the area and perimeter of a triangle then we can get the radius of its inscribed circle: ((rP)/2)=A⇒r=((2A)/P) I used this method for calculating the radius of inscribed sphere.
$${When}\:{we}\:{know}\:{the}\:{area}\:{and}\:{perimeter} \\ $$$${of}\:{a}\:{triangle}\:{then}\:{we}\:{can}\:{get}\:{the}\:{radius} \\ $$$${of}\:{its}\:{inscribed}\:{circle}: \\ $$$$\frac{{rP}}{\mathrm{2}}={A}\Rightarrow{r}=\frac{\mathrm{2}{A}}{{P}} \\ $$$${I}\:{used}\:{this}\:{method}\:{for}\:{calculating} \\ $$$${the}\:{radius}\:{of}\:{inscribed}\:{sphere}. \\ $$
Commented by ajfour last updated on 24/Jul/18
Understood Sir, thanks; its the best way to arrive at the answer.
$${Understood}\:{Sir},\:{thanks};\:{its}\:{the}\:{best} \\ $$$${way}\:{to}\:{arrive}\:{at}\:{the}\:{answer}. \\ $$
Commented by MrW3 last updated on 24/Jul/18
If we should determine the radius of the inscribed sphere of a triangular pyramid with sides a,b,c,p,q,r, then this method should be the easiest one.
$${If}\:{we}\:{should}\:{determine}\:{the}\:{radius}\:{of} \\ $$$${the}\:{inscribed}\:{sphere}\:{of}\:{a}\:{triangular} \\ $$$${pyramid}\:{with}\:{sides}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}},\boldsymbol{{p}},\boldsymbol{{q}},\boldsymbol{{r}},\:{then} \\ $$$${this}\:{method}\:{should}\:{be}\:{the}\:{easiest}\:{one}. \\ $$

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