Question Number 220257 by abbb last updated on 10/May/25
$${proof}\:{that}\:{volume}\:{of}\:{frustum}\:{of} \\ $$$$\:{circular}\:{cone}\:{is}\:\frac{\mathrm{1}}{\mathrm{3}}{h}\left[{A}\mathrm{1}+{A}\mathrm{2}+\sqrt{{A}\mathrm{1}{A}\mathrm{2}}\right. \\ $$$${A}_{\mathrm{1}} {and}\:{A}_{\mathrm{2}} \:{are}\:\:{areas}\:{of}\:{base} \\ $$
Answered by MrGaster last updated on 10/May/25
$${V}=\frac{\mathrm{1}}{\mathrm{3}}{h}\left({A}_{\mathrm{1}} +{A}_{\mathrm{2}} +\sqrt{{A}_{\mathrm{1}} {A}_{\mathrm{2}} }\right) \\ $$$${A}_{\mathrm{1}} =\pi{r}_{\mathrm{1}} ^{\mathrm{2}} ,{A}_{\mathrm{2}} =\pi{r}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}{h}\left(\pi{r}_{\mathrm{1}} ^{\mathrm{2}} +\pi{r}_{\mathrm{1}} ^{\mathrm{2}} +\sqrt{\pi{r}_{\mathrm{1}} ^{\mathrm{2}} \pi{r}_{\mathrm{2}} ^{\mathrm{2}} }\right) \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{3}}{h}\pi\left({r}_{\mathrm{1}} ^{\mathrm{2}} +{r}_{\mathrm{2}} ^{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{2}} \right) \\ $$
Commented by abbb last updated on 10/May/25
$$ \\ $$$${please}\:{deeply}\:\:{proof}\:{step}\:{by}\:{step} \\ $$$$\:\:{using}\:{diagram} \\ $$$$ \\ $$
Commented by abbb last updated on 10/May/25
$${please}\:{help}\:{me}\:{its}\:{my}\:{assignment} \\ $$
Answered by mr W last updated on 10/May/25
Commented by mr W last updated on 10/May/25
![h_2 =h_1 +h (A_1 /A_2 )=((πr_1 ^2 )/(πr_2 ))=((r_1 /r_2 ))^2 =((h_1 /h_2 ))^2 ⇒h_1 =h_2 (√(A_1 /A_2 )) h_2 −h_1 =h h_2 (1−(√(A_1 /A_2 )))=h ⇒h_2 =(h/(1−(√(A_1 /A_2 ))))=((√A_2 )/( (√A_2 )−(√A_1 )))×h ⇒h_1 =((√(A_1 /A_2 ))/(1−(√(A_1 /A_2 ))))×h=((√A_1 )/( (√A_2 )−(√A_1 )))×h V=V_2 −V_1 =((A_2 h_2 )/3)−((A_1 h_1 )/3) =(((A_2 (√A_2 )−A_1 (√A_1 ))h)/(3((√A_2 )−(√A_1 )))) =(([((√A_2 ))^3 −((√A_1 ))^3 ]h)/(3((√A_2 )−(√A_1 )))) =((((√A_2 )−(√A_1 ))[((√A_2 ))^2 +(√(A_2 A_1 ))+((√A_1 ))^2 ]h)/(3((√A_2 )−(√A_1 )))) =((h(A_1 +A_2 +(√(A_1 A_2 ))))/3)](https://www.tinkutara.com/question/Q220273.png)
$${h}_{\mathrm{2}} ={h}_{\mathrm{1}} +{h} \\ $$$$\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }=\frac{\pi{r}_{\mathrm{1}} ^{\mathrm{2}} }{\pi{r}_{\mathrm{2}} }=\left(\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }\right)^{\mathrm{2}} =\left(\frac{{h}_{\mathrm{1}} }{{h}_{\mathrm{2}} }\right)^{\mathrm{2}} \:\Rightarrow{h}_{\mathrm{1}} ={h}_{\mathrm{2}} \sqrt{\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }} \\ $$$${h}_{\mathrm{2}} −{h}_{\mathrm{1}} ={h} \\ $$$${h}_{\mathrm{2}} \left(\mathrm{1}−\sqrt{\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }}\right)={h}\: \\ $$$$\Rightarrow{h}_{\mathrm{2}} =\frac{{h}}{\mathrm{1}−\sqrt{\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }}}=\frac{\sqrt{{A}_{\mathrm{2}} }}{\:\sqrt{{A}_{\mathrm{2}} }−\sqrt{{A}_{\mathrm{1}} }}×{h} \\ $$$$\Rightarrow{h}_{\mathrm{1}} =\frac{\sqrt{\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }}}{\mathrm{1}−\sqrt{\frac{{A}_{\mathrm{1}} }{{A}_{\mathrm{2}} }}}×{h}=\frac{\sqrt{{A}_{\mathrm{1}} }}{\:\sqrt{{A}_{\mathrm{2}} }−\sqrt{{A}_{\mathrm{1}} }}×{h} \\ $$$${V}={V}_{\mathrm{2}} −{V}_{\mathrm{1}} =\frac{{A}_{\mathrm{2}} {h}_{\mathrm{2}} }{\mathrm{3}}−\frac{{A}_{\mathrm{1}} {h}_{\mathrm{1}} }{\mathrm{3}} \\ $$$$\:\:=\frac{\left({A}_{\mathrm{2}} \sqrt{{A}_{\mathrm{2}} }−{A}_{\mathrm{1}} \sqrt{{A}_{\mathrm{1}} }\right){h}}{\mathrm{3}\left(\sqrt{{A}_{\mathrm{2}} }−\sqrt{{A}_{\mathrm{1}} }\right)} \\ $$$$\:\:=\frac{\left[\left(\sqrt{{A}_{\mathrm{2}} }\right)^{\mathrm{3}} −\left(\sqrt{{A}_{\mathrm{1}} }\right)^{\mathrm{3}} \right]{h}}{\mathrm{3}\left(\sqrt{{A}_{\mathrm{2}} }−\sqrt{{A}_{\mathrm{1}} }\right)} \\ $$$$\:\:=\frac{\left(\sqrt{{A}_{\mathrm{2}} }−\sqrt{{A}_{\mathrm{1}} }\right)\left[\left(\sqrt{{A}_{\mathrm{2}} }\right)^{\mathrm{2}} +\sqrt{{A}_{\mathrm{2}} {A}_{\mathrm{1}} }+\left(\sqrt{{A}_{\mathrm{1}} }\right)^{\mathrm{2}} \right]{h}}{\mathrm{3}\left(\sqrt{{A}_{\mathrm{2}} }−\sqrt{{A}_{\mathrm{1}} }\right)} \\ $$$$\:\:=\frac{{h}\left({A}_{\mathrm{1}} +{A}_{\mathrm{2}} +\sqrt{{A}_{\mathrm{1}} {A}_{\mathrm{2}} }\right)}{\mathrm{3}} \\ $$
Commented by abbb last updated on 10/May/25
$${thank}\:{u}\: \\ $$