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Question Number 53867 by byaw last updated on 26/Jan/19

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jan/19

Removed sector area=((πr^2 )/(360))×120=((πr^2 )/3)  Remaining sector area=πr^2 −((πr^2 )/3)=((2πr^2 )/3)  slant height of cone l_(cone) =r  πR_(cone) l_(c0ne) =((2πr^2 )/3)  R_(cone) ×r=((2r^2 )/3)      so        R_(cone) =((2r)/3)  total metalic  surface area of cone T.S.A  =(πR_(cone) l_(cone) )_(metal sheet)   =π×((2r)/3)×r=((2πr^2 )/3)=((2×22×28^2 )/(7×3))  total surface area of cone=  (curvedsurface area)_(metal) +(base circular area)_(void)   {πR_(cone) l_(cone) }_(metal) +{πR_c ^2 }_(void)   =((2πr^2 )/3)+((π×4r^2 )/9)  =((6πr^2 +4πr^2 )/9)=((10πr^2 )/9)=((10×22×28^2 )/(7×9))

$${Removed}\:{sector}\:{area}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{360}}×\mathrm{120}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${Remaining}\:{sector}\:{area}=\pi{r}^{\mathrm{2}} −\frac{\pi{r}^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{2}\pi{r}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${slant}\:{height}\:{of}\:{cone}\:{l}_{{cone}} ={r} \\ $$$$\pi{R}_{{cone}} {l}_{{c}\mathrm{0}{ne}} =\frac{\mathrm{2}\pi{r}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${R}_{{cone}} ×{r}=\frac{\mathrm{2}{r}^{\mathrm{2}} }{\mathrm{3}}\:\:\:\:\:\:{so}\:\:\:\:\:\:\:\:{R}_{{cone}} =\frac{\mathrm{2}{r}}{\mathrm{3}} \\ $$$${total}\:{metalic}\:\:{surface}\:{area}\:{of}\:{cone}\:{T}.{S}.{A} \\ $$$$=\left(\pi{R}_{{cone}} {l}_{{cone}} \right)_{{metal}\:{sheet}} \\ $$$$=\pi×\frac{\mathrm{2}{r}}{\mathrm{3}}×{r}=\frac{\mathrm{2}\pi{r}^{\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{2}×\mathrm{22}×\mathrm{28}^{\mathrm{2}} }{\mathrm{7}×\mathrm{3}} \\ $$$${total}\:{surface}\:{area}\:{of}\:{cone}= \\ $$$$\left({curvedsurface}\:{area}\right)_{{metal}} +\left({base}\:{circular}\:{area}\right)_{{void}} \\ $$$$\left\{\pi{R}_{{cone}} {l}_{{cone}} \right\}_{{metal}} +\left\{\pi{R}_{{c}} ^{\mathrm{2}} \right\}_{{void}} \\ $$$$=\frac{\mathrm{2}\pi{r}^{\mathrm{2}} }{\mathrm{3}}+\frac{\pi×\mathrm{4}{r}^{\mathrm{2}} }{\mathrm{9}} \\ $$$$=\frac{\mathrm{6}\pi{r}^{\mathrm{2}} +\mathrm{4}\pi{r}^{\mathrm{2}} }{\mathrm{9}}=\frac{\mathrm{10}\pi{r}^{\mathrm{2}} }{\mathrm{9}}=\frac{\mathrm{10}×\mathrm{22}×\mathrm{28}^{\mathrm{2}} }{\mathrm{7}×\mathrm{9}} \\ $$

Commented by byaw last updated on 27/Jan/19

please Dear does the metalic cone  have base area? Is it open base or close base?

$${please}\:{Dear}\:{does}\:{the}\:{metalic}\:{cone} \\ $$$${have}\:{base}\:{area}?\:{Is}\:{it}\:{open}\:{base}\:{or}\:{close}\:{base}? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jan/19

yes[you  are right...then rectify the queston..delete  the word total surface area insert curved surface  area...  base of cone is a circle...so area is πR_c ^2   question it self ambigousous...(i think)  T.S.A=πR_c l_c +πR_c ^2 ...

$${yes}\left[{you}\:\:{are}\:{right}...{then}\:{rectify}\:{the}\:{queston}..{delete}\right. \\ $$$${the}\:{word}\:{total}\:{surface}\:{area}\:{insert}\:{curved}\:{surface} \\ $$$${area}... \\ $$$${base}\:{of}\:{cone}\:{is}\:{a}\:{circle}...{so}\:{area}\:{is}\:\pi{R}_{{c}} ^{\mathrm{2}} \\ $$$${question}\:{it}\:{self}\:{ambigousous}...\left({i}\:{think}\right) \\ $$$${T}.{S}.{A}=\pi{R}_{{c}} {l}_{{c}} +\pi{R}_{{c}} ^{\mathrm{2}} ... \\ $$

Commented by byaw last updated on 27/Jan/19

It is a past question

$${It}\:{is}\:{a}\:{past}\:{question} \\ $$

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