anonymous
  • anonymous
Cylinder help..if the volume of a cylinder is 340 how do i find the perimeter, surface aria, and the hight
Mathematics
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anonymous
  • anonymous
Cylinder help..if the volume of a cylinder is 340 how do i find the perimeter, surface aria, and the hight
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
You know that the Surface Area equation of a cylinder right? which is:\[SA = 2\pi r^2 + 2\pi r H\] and the volume is : \[V = \pi r^2 H\] Perimeter is : \[P = 2\pi R\] all you have to do is find a relationship between all of them and you'll be able to figure out the height, perimeter and SA , so you'll get : Notice that the volume is = pi r ^2 h and the perimeter is = 2pi r so the SA can be written as : \[SA = 2(340)H + PH\]
anonymous
  • anonymous
are you sure that that's the only given?
anonymous
  • anonymous
yep i am givin x 1-a2 i think that is the radious and i have to find all of those for numbers 1-12

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anonymous
  • anonymous
x = 1- a2?
anonymous
  • anonymous
you need more given, I doubt that the only given is the volume lol
anonymous
  • anonymous
ok so the whole thing is a project we have to figure out what is best for making a box..we are given x values 1-12 and the volume of 340 cubic inches 1-12 i believe are the radices
anonymous
  • anonymous
i need to find the hight. the surface aria..and the perimeter of a cylinder
anonymous
  • anonymous
with that info
anonymous
  • anonymous
Alright, pick one of the radiuses and plug it in the Volume's equation to find H :) see if the number makes sense.
anonymous
  • anonymous
ok so i take piXr squared and get the hight?
anonymous
  • anonymous
for example .3.14X1 squared?
anonymous
  • anonymous
V = pi r^2 H , you can plug in any value or R, then find H, and after finding H you can plug it in the perimeter, and when you find the perimeter you'll be able to find the SA, the new equation that I have wrote up there ^_^
anonymous
  • anonymous
as far as I can go lol, since I don't have enough given
anonymous
  • anonymous
im not getting how you find h when you need it in the equation
anonymous
  • anonymous
alright, we have the volume right? :)
anonymous
  • anonymous
\[V = 340 \]
anonymous
  • anonymous
you have values of x, which are considered as the radius right?
anonymous
  • anonymous
so you can take R = 4 for example and plug it in the following equation:\[V = \pi r^2h\] so we have R and V, we can solve for H, right?
anonymous
  • anonymous
lol, are you getting the idea now? :)
anonymous
  • anonymous
\[340=\pi(4) squared ?\]
anonymous
  • anonymous
actually it's :\[H = (340)/(8\pi)\] and calculate it, that ofcourse if you choose your radius to be 4 ^_^ clear?
anonymous
  • anonymous
wait it's 32 pi instead of 8 pi I forgot to square it lol!
anonymous
  • anonymous
ok so i take x=340/r is this correct...if not i give up for the night and am going to bed..i am sure i am making this 100X harder than it actually is
anonymous
  • anonymous
and that will give me the hight ?
anonymous
  • anonymous
don't give up lol, alright, let's try once more! WE have x values, right?
anonymous
  • anonymous
yes those values= the radius...so if the virst x value is 1 how do i get the hight?
anonymous
  • anonymous
it's simple, just take your time and concentrate with me
anonymous
  • anonymous
Okay so x = r = 1 we have given: R = 1 V = 340 find H= ?
anonymous
  • anonymous
yep thats what i need to do...now how?
anonymous
  • anonymous
The equation is : \[V = \pi r^2 h\]
anonymous
  • anonymous
plug in the given in the equation and solve for H : 340 = pi (1)^2 H H = 340/pi H = 108.23 in ^_^
anonymous
  • anonymous
we have found the height now :)
anonymous
  • anonymous
now we need to find the perimeter right?
anonymous
  • anonymous
are you following me lol? =P wakey wakey~
anonymous
  • anonymous
hold on
anonymous
  • anonymous
okay ^_^
anonymous
  • anonymous
ok so the hight for a radios of 1 is 108.2 correct
anonymous
  • anonymous
excellent!
anonymous
  • anonymous
ok so now perimeter
anonymous
  • anonymous
now, we can find the perimeter right? The equation is : P = 2pi R we have R = 1, solve :)
anonymous
  • anonymous
ok so i take \[2piX(1)squared\]
anonymous
  • anonymous
i got 6.28
anonymous
  • anonymous
lol no, the equation is :\[P = 2 \pi R\] substitute R in the equation and you'll get P ^_^
anonymous
  • anonymous
lol yeah , that's right :)
anonymous
  • anonymous
now, we can find the SA right?
anonymous
  • anonymous
ok so now the SA
anonymous
  • anonymous
The equation is :\[SA = 2\pi r^2 + 2\pi r h\] we have R = 1 and H = 108.2 substitute the given and solve ^_^
anonymous
  • anonymous
i got 686.1
anonymous
  • anonymous
very good !
anonymous
  • anonymous
you have solved the question using R = 1 ^_^ did you get the idea now? :)
anonymous
  • anonymous
ok let me make sure i can get this for 2 now and then ill let you go..if you dont mind
anonymous
  • anonymous
good morning
anonymous
  • anonymous
hi
anonymous
  • anonymous
alright, it's the same idea blackford, but different values for R ^_^, good morning andy
anonymous
  • anonymous
same way of solving , good luck! ^_^
anonymous
  • anonymous
i got 120.8
anonymous
  • anonymous
for?
anonymous
  • anonymous
the hight with a radios of 2
anonymous
  • anonymous
hmm it should be 27.06 (340)/(4pi) = H H = 27.06 in
anonymous
  • anonymous
and andy going quite lol
anonymous
  • anonymous
oops i see i think..i just realized you want h by itself
anonymous
  • anonymous
yep ^_^ since we want to find it
anonymous
  • anonymous
ok for a R of 3 i got 12.02 for the hight..correct?
anonymous
  • anonymous
excellent ^_^
anonymous
  • anonymous
ok i think i am getting it now
anonymous
  • anonymous
you got the idea now :)
anonymous
  • anonymous
you can proceed on your own right?
anonymous
  • anonymous
for now thank you :)
anonymous
  • anonymous
np ^_^
anonymous
  • anonymous
i got a new one
anonymous
  • anonymous
alright
anonymous
  • anonymous
now for a square pyramid. this time i need the hight,perimeter, SA, and lateral aria
anonymous
  • anonymous
volume still 340 and same x values 1-12
anonymous
  • anonymous
All you have to do is write down the new equations for the square pyramid
anonymous
  • anonymous
i think that x= base length now
anonymous
  • anonymous
Volume is : \[V = (b^2h)/3\] Surface Area is : \[SA = 2bs + b^2\] and perimeter : \[P = (\S-B)(1/2)(l)\] find the relationship and solve ^_^ same idea as the previous one :)
anonymous
  • anonymous
P = (S-B)(1/2)(L) *
anonymous
  • anonymous
oh and the lateral area is = PL/2 where P = perimeter and L = slant height ^_^
anonymous
  • anonymous
ok so 340=1squared(h)/3
anonymous
  • anonymous
oh yeah i want the slant hight not lateral aria
anonymous
  • anonymous
so \[340=1^2(h).....340=1/3 =.33 340/.33..correct or no\]
anonymous
  • anonymous
340(3) = h h = 1020
anonymous
  • anonymous
lol, I see what's your problem, you got the idea, but have trouble with calculation, take your time answering the questions :) step at a time ^_^

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