the altitude of a circular cylinder is increased 6 times and the base area is decreased one ninth of its value. the factor by which lateral surface area of cylinder increases is ?
Answers
Answered by
60
since lateral surface area= 2πrh
when the given increases and decreases happens....
lateral surface area=2π(r/9)(6h) =2πrh×6/9 =2πrh×2/3
hence ... lateral surface area is 2/3rd of old one
hence it is decreased by 1/3rd
or we can also say increased by factor of -1/3
and these 2 can be our required answers
when the given increases and decreases happens....
lateral surface area=2π(r/9)(6h) =2πrh×6/9 =2πrh×2/3
hence ... lateral surface area is 2/3rd of old one
hence it is decreased by 1/3rd
or we can also say increased by factor of -1/3
and these 2 can be our required answers
Answered by
62
Let H and R represent the final radius and height of the cylinder
Given,
H = 6h(ii)
Also,
(Base area)f = 1/9th*(Base area)i
TTR2 = 1/9*TTr2
R2 = 1/9r2
R = r/3(ii)
Initial C.S.A. = 2TTrh
Final C.S.A. = 2TTRH = 2TT(r/3)(6h) = 4TTrh
So, C.S.A. gets doubled i.e. increases by a factor of 2
Given,
H = 6h(ii)
Also,
(Base area)f = 1/9th*(Base area)i
TTR2 = 1/9*TTr2
R2 = 1/9r2
R = r/3(ii)
Initial C.S.A. = 2TTrh
Final C.S.A. = 2TTRH = 2TT(r/3)(6h) = 4TTrh
So, C.S.A. gets doubled i.e. increases by a factor of 2
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