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A-solid-cone-of-height-56cm-and-a-basem-diaeter-of-56cm-is-formed-from-al-cylindrica-drum-equal-in-height-andt-diameer-with-the-cone-Find-the-surfacee-ara-of-the-remaining-part-of-the-drum-if-the




Question Number 180191 by cherokeesay last updated on 08/Nov/22
  A solid cone of height 56cm and a basem  diaeter of 56cm is formed from al  cylindrica drum equal in height andt  diameer with the cone. Find the surfacee  ara of the remaining part of the drum.    (if the answer is accompanied by a   diagramit will be perfect and I thank youo  fr it).
$$ \\ $$$$\mathrm{A}\:\mathrm{solid}\:\mathrm{cone}\:\mathrm{of}\:\mathrm{height}\:\mathrm{56cm}\:\mathrm{and}\:\mathrm{a}\:\mathrm{basem} \\ $$$$\mathrm{diaeter}\:\mathrm{of}\:\mathrm{56cm}\:\mathrm{is}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{al} \\ $$$$\mathrm{cylindrica}\:\mathrm{drum}\:\mathrm{equal}\:\mathrm{in}\:\mathrm{height}\:\mathrm{andt} \\ $$$$\mathrm{diameer}\:\mathrm{with}\:\mathrm{the}\:\mathrm{cone}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{surfacee} \\ $$$$\mathrm{ara}\:\mathrm{of}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{drum}. \\ $$$$ \\ $$$$\left(\mathrm{if}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{accompanied}\:\mathrm{by}\:\mathrm{a}\:\right. \\ $$$$\mathrm{diagramit}\:\mathrm{will}\:\mathrm{be}\:\mathrm{perfect}\:\mathrm{and}\:\mathrm{I}\:\mathrm{thank}\:\mathrm{youo} \\ $$$$\left.\mathrm{fr}\:\mathrm{it}\right). \\ $$
Answered by Acem last updated on 09/Nov/22
As i got that we formed a cone from  a cylindric drum... right? at this case the surface   consists of three surfaces:   upper base, external side and internal one   S= 0.25πD^2  + πDh + 0.5πD (√(0.25D^2 +h^2 ))   S= πD (0.25D+ h+ 0.25 (√(D^2 + 4 h^2 )))   S≈ 1.782 m^2
$${As}\:{i}\:{got}\:{that}\:{we}\:{formed}\:{a}\:{cone}\:{from} \\ $$$${a}\:{cylindric}\:{drum}…\:{right}?\:{at}\:{this}\:{case}\:{the}\:{surface} \\ $$$$\:{consists}\:{of}\:{three}\:{surfaces}: \\ $$$$\:{upper}\:{base},\:{external}\:{side}\:{and}\:{internal}\:{one} \\ $$$$\:{S}=\:\mathrm{0}.\mathrm{25}\pi{D}^{\mathrm{2}} \:+\:\pi{Dh}\:+\:\mathrm{0}.\mathrm{5}\pi{D}\:\sqrt{\mathrm{0}.\mathrm{25}{D}^{\mathrm{2}} +{h}^{\mathrm{2}} } \\ $$$$\:{S}=\:\pi{D}\:\left(\mathrm{0}.\mathrm{25}{D}+\:{h}+\:\mathrm{0}.\mathrm{25}\:\sqrt{{D}^{\mathrm{2}} +\:\mathrm{4}\:{h}^{\mathrm{2}} }\right) \\ $$$$\:{S}\approx\:\mathrm{1}.\mathrm{782}\:{m}^{\mathrm{2}} \\ $$$$\: \\ $$
Commented by Acem last updated on 08/Nov/22
Commented by Acem last updated on 08/Nov/22
But with turning machine nothing remains but   iron filings
$${But}\:{with}\:{turning}\:{machine}\:{nothing}\:{remains}\:{but} \\ $$$$\:{iron}\:{filings} \\ $$$$\: \\ $$
Commented by cherokeesay last updated on 09/Nov/22
thank you so much sir.
$${thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$

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