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Question Number 218411 by Spillover last updated on 09/Apr/25

Answered by mr W last updated on 09/Apr/25

Commented by mr W last updated on 09/Apr/25

S_1 =πr_1 ^2 S_2 =πr_2 ^2 (S_1 /S_2 )=((r_1 /r_2 ))^2 =4 ⇒(r_1 /r_2 )=2 V_1 =((S_1 h_1 )/3)=((πr_1 ^2 h_1 )/3) V_2 =((S_2 h_2 )/3)=((πr_2 ^2 h_2 )/3) V_C =V_1 −V_2 =volume of cone V_S =((4πR^3 )/3)=volume of sphere h_1 =h_2 +2R (h_1 /h_2 )=(r_1 /r_2 )=2 ((h_2 +2R)/h_2 )=2 ⇒(R/h_2 )=(1/2) ⇒h_2 =2R (r_2 /R)=((√(h_2 ^2 +r_2 ^2 ))/(h_2 +R)) (r_2 /R)=((√(4R^2 +r_2 ^2 ))/(2R+R)) 2r_2 ^2 =R^2 ⇒(r_2 ^2 /R^2 )=(1/2) (V_C /V_S )=((r_1 ^2 h_1 −r_2 ^2 h_2 )/(4R^3 )) =((4r_2 ^2 ×2h_2 −r_2 ^2 h_2 )/(4R^3 )) =((7r_2 ^2 ×2R)/(4R^3 ))=((7r_2 ^2 )/(2R^2 ))=(7/4) ✓

$${S}_{\mathrm{1}} =\pi{r}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} =\pi{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\frac{{S}_{\mathrm{1}} }{{S}_{\mathrm{2}} }=\left(\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{4}\:\Rightarrow\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\mathrm{2} \\ $$$${V}_{\mathrm{1}} =\frac{{S}_{\mathrm{1}} {h}_{\mathrm{1}} }{\mathrm{3}}=\frac{\pi{r}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} }{\mathrm{3}} \\ $$$${V}_{\mathrm{2}} =\frac{{S}_{\mathrm{2}} {h}_{\mathrm{2}} }{\mathrm{3}}=\frac{\pi{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} }{\mathrm{3}} \\ $$$${V}_{{C}} ={V}_{\mathrm{1}} −{V}_{\mathrm{2}} ={volume}\:{of}\:{cone} \\ $$$${V}_{{S}} =\frac{\mathrm{4}\pi{R}^{\mathrm{3}} }{\mathrm{3}}={volume}\:{of}\:{sphere} \\ $$$${h}_{\mathrm{1}} ={h}_{\mathrm{2}} +\mathrm{2}{R} \\ $$$$\frac{{h}_{\mathrm{1}} }{{h}_{\mathrm{2}} }=\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\mathrm{2} \\ $$$$\frac{{h}_{\mathrm{2}} +\mathrm{2}{R}}{{h}_{\mathrm{2}} }=\mathrm{2}\:\Rightarrow\frac{{R}}{{h}_{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{h}_{\mathrm{2}} =\mathrm{2}{R} \\ $$$$\frac{{r}_{\mathrm{2}} }{{R}}=\frac{\sqrt{{h}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} }}{{h}_{\mathrm{2}} +{R}} \\ $$$$\frac{{r}_{\mathrm{2}} }{{R}}=\frac{\sqrt{\mathrm{4}{R}^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{2}{R}+{R}} \\ $$$$\mathrm{2}{r}_{\mathrm{2}} ^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{r}_{\mathrm{2}} ^{\mathrm{2}} }{{R}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{V}_{{C}} }{{V}_{{S}} }=\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} {h}_{\mathrm{1}} −{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:=\frac{\mathrm{4}{r}_{\mathrm{2}} ^{\mathrm{2}} ×\mathrm{2}{h}_{\mathrm{2}} −{r}_{\mathrm{2}} ^{\mathrm{2}} {h}_{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:=\frac{\mathrm{7}{r}_{\mathrm{2}} ^{\mathrm{2}} ×\mathrm{2}{R}}{\mathrm{4}{R}^{\mathrm{3}} }=\frac{\mathrm{7}{r}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}{R}^{\mathrm{2}} }=\frac{\mathrm{7}}{\mathrm{4}}\:\checkmark \\ $$

Commented by Spillover last updated on 09/Apr/25

thank you

$${thank}\:{you} \\ $$

Answered by Nicholas666 last updated on 09/Apr/25

((volume(cone))/(volume(sphere))=((((14(√2))/3)πr^3 )/(((8(√2))/3)πr^3 ))=((14(√2))/(8(√2)))=((14)/8)=7/4

$$\:\frac{{volume}\left({cone}\right)}{{volume}\left({sphere}\right.}=\frac{\frac{\mathrm{14}\sqrt{\mathrm{2}}}{\mathrm{3}}\pi{r}^{\mathrm{3}} }{\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}}\pi{r}^{\mathrm{3}} }=\frac{\mathrm{14}\sqrt{\mathrm{2}}}{\mathrm{8}\sqrt{\mathrm{2}}}=\frac{\mathrm{14}}{\mathrm{8}}=\mathrm{7}/\mathrm{4} \\ $$$$ \\ $$

Commented by mr W last updated on 11/Apr/25

it makes little sense, when you just post a result without showing how to get it. people want to know the path, not only the final answer.

$${it}\:{makes}\:{little}\:{sense},\:{when}\:{you}\:{just} \\ $$$${post}\:{a}\:{result}\:{without}\:{showing}\:{how} \\ $$$${to}\:{get}\:{it}.\:{people}\:{want}\:{to}\:{know}\:{the} \\ $$$${path},\:{not}\:{only}\:{the}\:{final}\:{answer}. \\ $$

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