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Question Number 45753 by Tawa1 last updated on 16/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Oct/18
	$for cylnder volume =πR^2 H when base is circle of area πR^2 but here volume is sector aresfor 30^o ×height requiref volume is sector area=((πR^2 )/(2π))×(π/6) [(π/6)=30^(o]) volume =((πR^2 )/(2π))×(π/6)×H=((πR^2 H)/(12))=((3.14×7^2 ×5)/(12)) totsl surface area=((πR^2 )/(12))+((πR^2 )/(12))+2×{5×7}+ ((2πRH)/(2π))×(π/6) note...total surcace area= two sectlor area of30^o +two rectangle area+ one curve area =(((πR^2 )/(12))×2)+(5×7×2)+((2πRH)/(2π))×(π/6)$
	$f o r c y l n d e r v o l u m e = π R^{2} H w h e n b a s e i s c i r c l e$ $o f a r e a π R^{2}$ $b u t h e r e v o l u m e i s s e c t o r a r e s f o r 30^{o} \times h e i g h t$ $r e q u i r e f v o l u m e i s$ $s e c t o r a r e a = \frac{π R^{2}}{2 π} \times \frac{π}{6} [\frac{π}{6} = 30^{o]}$ $v o l u m e = \frac{π R^{2}}{2 π} \times \frac{π}{6} \times H = \frac{π R^{2} H}{12} = \frac{3.14 \times 7^{2} \times 5}{12}$ $t o t s l s u r f a c e a r e a = \frac{π R^{2}}{12} + \frac{π R^{2}}{12} + 2 \times {5 \times 7} +$ $\frac{2 π R H}{2 π} \times \frac{π}{6}$ $n o t e . . . t o t a l s u r c a c e a r e a =$ $t w o s e c t l o r a r e a o f 30^{o} + t w o r e c t a n g l e a r e a +$ $o n e c u r v e a r e a$ $= (\frac{π R^{2}}{12} \times 2) + (5 \times 7 \times 2) + \frac{2 π R H}{2 π} \times \frac{π}{6}$
	Commented by Tawa1 last updated on 16/Oct/18

	$God bless you sir$
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