anonymous 5 years ago Cylinder help..if the volume of a cylinder is 340 how do i find the perimeter, surface aria, and the hight

1. 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 : $SA = 2(340)H + PH$

2. anonymous

are you sure that that's the only given?

3. 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

4. anonymous

x = 1- a2?

5. anonymous

you need more given, I doubt that the only given is the volume lol

6. 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

7. anonymous

i need to find the hight. the surface aria..and the perimeter of a cylinder

8. anonymous

with that info

9. anonymous

Alright, pick one of the radiuses and plug it in the Volume's equation to find H :) see if the number makes sense.

10. anonymous

ok so i take piXr squared and get the hight?

11. anonymous

for example .3.14X1 squared?

12. 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 ^_^

13. anonymous

as far as I can go lol, since I don't have enough given

14. anonymous

im not getting how you find h when you need it in the equation

15. anonymous

alright, we have the volume right? :)

16. anonymous

$V = 340$

17. anonymous

you have values of x, which are considered as the radius right?

18. 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?

19. anonymous

lol, are you getting the idea now? :)

20. anonymous

$340=\pi(4) squared ?$

21. anonymous

actually it's :$H = (340)/(8\pi)$ and calculate it, that ofcourse if you choose your radius to be 4 ^_^ clear?

22. anonymous

wait it's 32 pi instead of 8 pi I forgot to square it lol!

23. 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

24. anonymous

and that will give me the hight ?

25. anonymous

don't give up lol, alright, let's try once more! WE have x values, right?

26. anonymous

yes those values= the radius...so if the virst x value is 1 how do i get the hight?

27. anonymous

it's simple, just take your time and concentrate with me

28. anonymous

Okay so x = r = 1 we have given: R = 1 V = 340 find H= ?

29. anonymous

yep thats what i need to do...now how?

30. anonymous

The equation is : $V = \pi r^2 h$

31. anonymous

plug in the given in the equation and solve for H : 340 = pi (1)^2 H H = 340/pi H = 108.23 in ^_^

32. anonymous

we have found the height now :)

33. anonymous

now we need to find the perimeter right?

34. anonymous

are you following me lol? =P wakey wakey~

35. anonymous

hold on

36. anonymous

okay ^_^

37. anonymous

ok so the hight for a radios of 1 is 108.2 correct

38. anonymous

excellent!

39. anonymous

ok so now perimeter

40. anonymous

now, we can find the perimeter right? The equation is : P = 2pi R we have R = 1, solve :)

41. anonymous

ok so i take $2piX(1)squared$

42. anonymous

i got 6.28

43. anonymous

lol no, the equation is :$P = 2 \pi R$ substitute R in the equation and you'll get P ^_^

44. anonymous

lol yeah , that's right :)

45. anonymous

now, we can find the SA right?

46. anonymous

ok so now the SA

47. 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 ^_^

48. anonymous

i got 686.1

49. anonymous

very good !

50. anonymous

you have solved the question using R = 1 ^_^ did you get the idea now? :)

51. anonymous

ok let me make sure i can get this for 2 now and then ill let you go..if you dont mind

52. anonymous

good morning

53. anonymous

hi

54. anonymous

alright, it's the same idea blackford, but different values for R ^_^, good morning andy

55. anonymous

same way of solving , good luck! ^_^

56. anonymous

i got 120.8

57. anonymous

for?

58. anonymous

the hight with a radios of 2

59. anonymous

hmm it should be 27.06 (340)/(4pi) = H H = 27.06 in

60. anonymous

and andy going quite lol

61. anonymous

oops i see i think..i just realized you want h by itself

62. anonymous

yep ^_^ since we want to find it

63. anonymous

ok for a R of 3 i got 12.02 for the hight..correct?

64. anonymous

excellent ^_^

65. anonymous

ok i think i am getting it now

66. anonymous

you got the idea now :)

67. anonymous

you can proceed on your own right?

68. anonymous

for now thank you :)

69. anonymous

np ^_^

70. anonymous

i got a new one

71. anonymous

alright

72. anonymous

now for a square pyramid. this time i need the hight,perimeter, SA, and lateral aria

73. anonymous

volume still 340 and same x values 1-12

74. anonymous

All you have to do is write down the new equations for the square pyramid

75. anonymous

i think that x= base length now

76. 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 :)

77. anonymous

P = (S-B)(1/2)(L) *

78. anonymous

oh and the lateral area is = PL/2 where P = perimeter and L = slant height ^_^

79. anonymous

ok so 340=1squared(h)/3

80. anonymous

oh yeah i want the slant hight not lateral aria

81. anonymous

so $340=1^2(h).....340=1/3 =.33 340/.33..correct or no$

82. anonymous

340(3) = h h = 1020

83. 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 ^_^