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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
	$Removed sector area=((πr^2 )/(360))×120=((πr^2 )/3) Remaining sector area=πr^2 −((πr^2 )/3)=((2πr^2 )/3) slant height of cone l_(cone) =r πR_(cone) l_(c0ne) =((2πr^2 )/3) R_(cone) ×r=((2r^2 )/3) so R_(cone) =((2r)/3) total metalic surface area of cone T.S.A =(πR_(cone) l_(cone) )_(metal sheet) =π×((2r)/3)×r=((2πr^2 )/3)=((2×22×28^2 )/(7×3)) total surface area of cone= (curvedsurface area)_(metal) +(base circular area)_(void) {πR_(cone) l_(cone) }_(metal) +{πR_c ^2 }_(void) =((2πr^2 )/3)+((π×4r^2 )/9) =((6πr^2 +4πr^2 )/9)=((10πr^2 )/9)=((10×22×28^2 )/(7×9))$
	$R e m o v e d s e c t o r a r e a = \frac{π r^{2}}{360} \times 120 = \frac{π r^{2}}{3}$ $R e m a i n i n g s e c t o r a r e a = π r^{2} - \frac{π r^{2}}{3} = \frac{2 π r^{2}}{3}$ $s l a n t h e i g h t o f c o n e l_{c o n e} = r$ $π R_{c o n e} l_{c 0 n e} = \frac{2 π r^{2}}{3}$ $R_{c o n e} \times r = \frac{2 r^{2}}{3} s o R_{c o n e} = \frac{2 r}{3}$ $t o t a l m e t a l i c s u r f a c e a r e a o f c o n e T . S . A$ $= {(π R_{c o n e} l_{c o n e})}_{m e t a l s h e e t}$ $= π \times \frac{2 r}{3} \times r = \frac{2 π r^{2}}{3} = \frac{2 \times 22 \times 28^{2}}{7 \times 3}$ $t o t a l s u r f a c e a r e a o f c o n e =$ ${(c u r v e d s u r f a c e a r e a)}_{m e t a l} + {(b a s e c i r c u l a r a r e a)}_{v o i d}$ ${π R_{c o n e} l_{c o n e}}_{m e t a l} + {π R_{c}^{2}}_{v o i d}$ $= \frac{2 π r^{2}}{3} + \frac{π \times 4 r^{2}}{9}$ $= \frac{6 π r^{2} + 4 π r^{2}}{9} = \frac{10 π r^{2}}{9} = \frac{10 \times 22 \times 28^{2}}{7 \times 9}$
	Commented by byaw last updated on 27/Jan/19

	$p l e a s e D e a r d o e s t h e m e t a l i c c o n e$ $h a v e b a s e a r e a ? I s i t o p e n b a s e o r c l o s e b a s e ?$
	Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jan/19

	$y e s [y o u a r e r i g h t . . . t h e n r e c t i f y t h e q u e s t o n . . d e l e t e$ $t h e w o r d t o t a l s u r f a c e a r e a i n s e r t c u r v e d s u r f a c e$ $a r e a . . .$ $b a s e o f c o n e i s a c i r c l e . . . s o a r e a i s π R_{c}^{2}$ $q u e s t i o n i t s e l f a m b i g o u s o u s . . . (i t h i n k)$ $T . S . A = π R_{c} l_{c} + π R_{c}^{2} . . .$
	Commented by byaw last updated on 27/Jan/19

	$I t i s a p a s t q u e s t i o n$
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