[odinn7]

My daughter is doing homework and is stuck on a question. I think she is looking into it too deeply and is confusing herself because of this.

A cylindrical storage container has a height of 8 inches, a surface area of 1,130.97 in.² and a volume of 2,513.27 in.³ Find the surface area and volume of a similar container whose height is 6 inches.


So maybe I am looking at this wrong. It's been so long since I have done anything like this that I can't recall the way to do it.

My first thought is that since the containers are similar except for height, I am thinking that the smaller container is actually 3/4 the height, surface area and volume of the original 8 inch container.

So... am I correct or am I trying to over simplify this?

[Wondergirl]

That makes sense to me. As a check, work out the surface area and volume for the 6" cylinder, then see if it is 3/4.

[hkstroud]

I would think that objective would be for the student to find the unknown radius using the volume and the height.

volume of a cylinder is pi x radius squared x the height.

pi x R squared x height = 2,513.27
pi x R squared x 8 = 2,513.27
pi x R squared = 2,513.27 / 8
pi x R squared = 314.15875
R squared = 314.15875 / 3.14159
R squared = 99.99992
square root of 99.9992 = 9.9999959
radius of cylinder is 10

pi x radius squared x the height.

3.14159 x 100 x 6 =1884.954

Which just happens to be 3/4 of 2,513.27

Knowing the radius and the new height, the new surface area can be calculated.

The surface area of the 6' cylinder will not be 3/4 of the 8" cylinder. Which is why the surface area was included in the exercise.

Again I would think the objective of the exercise would be for the student of find the unknowns.

[odinn7]

Thanks... though I have to say that was way more complicated than I hoped it would be! Man, she's only in 7th grade! Lol

[hkstroud]

Not complicated conceptional.

"I could find the new volume and surface area if I only knew the radius."

"I can find the radius if I divide the old volume by the height and pi."

Suspect the real purpose of the exercise is to make the student think like that.

Granted I went to an excessive amount of detail.

[odinn7]

You made my head hurt!

[letmethink]

case 1
first cylinder
lets name height as h1,surface area as SA1 and volume as vol1 and radius as r1
h1=8
SA1=1130.97
v1=2513.27
r1=?

formulae for volume of cylinder is pi *r*r*h
v1=pi * r1 * r1 * h1

3.14 * r 1 * r1 * h1 = 2513.27
3.14*r1*r1*8=2513.27
r1*r1=100.05
r1=10


case 2
second cylinder

let us say the height as h2,surface area as sa2,volume as v2 and radius as r2

since the radius of the second is unknown and being height is givens as 6 and it is said both the container are similar
let us equate
volume of the first cylinder=volume of the second cylinder
pi* r1 * r1 * h1 = pi * r2 * r2 * h2
10 * 10 * 8 = r2 * r2 * 6
=r2 * r2 = (100 * 8) / 6
r2=11.54

now calculate surface area and volume
surface area,sa2= 2 * pi * r2 * (r2+h2)=
= 2 * 3.14 * (11.54+6)
=110.15 inch sq
volume v2= pi* r2 * r2 * h2=
3.14 * 11.54 * 11.54 * 6
2508.95 inch cube

hope it helps

[Alty]

This is why I hate math. Just saying.

Odinn, my head is hurting along with yours. :(

I miss the good old days 1 + 1 = 2. That I can handle. Letters and numbers and stars and... isn't this why we have calculators? ;)

[letmethink]

You made my head hurt!

Do you think I am wrong again.sorry I was just trying to help.
When you get the correct answer post it here.
Just feels the need to know the answer

[letmethink]

case 1
first cylinder
lets name height as h1,surface area as SA1 and volume as vol1 and radius as r1
h1=8
SA1=1130.97
v1=2513.27
r1=?

formulae for volume of cylinder is pi *r*r*h
v1=pi * r1 * r1 * h1

3.14 * r 1 * r1 * h1 = 2513.27
3.14*r1*r1*8=2513.27
r1*r1=100.05
r1=10


case 2
second cylinder

let us say the height as h2,surface area as sa2,volume as v2 and radius as r2

since the radius of the second is unknown and being height is givens as 6 and it is said both the container are similiar
let us equate
volume of the first cylinder=volume of the second cylinder
pi* r1 * r1 * h1 = pi * r2 * r2 * h2
10 * 10 * 8 = r2 * r2 * 6
=r2 * r2 = (100 * 8) / 6
r2=11.54

now calculate surface area and volume
surface area,sa2= 2 * pi * r2 * (r2+h2)=
= 2 * 3.14 * (11.54+6)
=110.15 inch sq
volume v2= pi* r2 * r2 * h2=
3.14 * 11.54 * 11.54 * 6
2508.95 inch cube

hope it helps

The surface area of the second cylinder mentioned is incorect as I forgot to multiply the height
Surfacearea is 1271.448
The volume of both the cylinnders will be same, there is slight variation inmy calculation because I neglected the decimal part of the radius of first cylinder r1. Otherwise would have got the correct (I guess)

[letmethink]

if u want to be more accurate the updated answer is below.

case 1

first cylinder
lets name height as h1,surface area as SA1 and volume as vol1 and radius as r1
h1=8
SA1=1130.97
v1=2513.27
r1=?

formulae for volume of cylinder is pi *r*r*h
v1=pi * r1 * r1 * h1

3.14157 * r 1 * r1 * h1 = 2513.27
3.14157*r1*r1*8=2513.27
r1*r1=100.05
r1=10.000028


case 2
second cylinder

let us say the height as h2,surface area as sa2,volume as v2 and radius as r2

since the radius of the second is unknown and being height is givens as 6 and it is said both the container are similar
(means both the cylinders will be equal in volume only different in surface area)
let us equate
volume of the first cylinder=volume of the second cylinder
pi* r1 * r1 * h1 = pi * r2 * r2 * h2
10.00028 * 10.00028 * 8 = r2 * r2 * 6
=r2 * r2 = (100 * 8) / 6
r2=11.547038

now calculate surface area and volume of second cylinder

surface area,sa2= 2 * pi * r2 * (r2+h2)=
= 2 * 3.14157 *11.547038 (11.547038+6)
= 1273.0667 inch sq

volume v2= pi* r2 * r2 * h2=
3.14157* 11.547038 * 11.547038 * 6
2513.2702 inch cube

[hkstroud]

Disagree with you interpretation of the problem.

Unless you interpret "similar" as meaning of the same diameter, the problem cannot be solved.

volume of the first cylinder=volume of the second cylinder

Not stated in the problem.

You cannot use the volume of the 6" to compute the diameter or radius of the 6" cylinder because computing its volume is one of the things you are task to do.

[letmethink]

Disagree with you interpretation of the problem.

Unless you interpret "similar" as meaning of the same diameter, the problem cannot be solved.


Not stated in the problem.

You cannot use the volume of the 6" to compute the diameter or radius of the 6" cylinder because computing its volume is one of the things you are task to do.

I am really sorry if my suggestions are wrong.
Try to post the real solution when u find the answer so it may help some others too.

[odinn7]

i am really sorry if my suggestions are wrong.
try to post the real solution when u find the answer so it may help some others too.

No need to apologize.

Anyway, he already did post the solution in post #3.

I'm closing this as I feel it was answered already and the assignment has already been done and turned in days ago.

EDIT- Ok, so I reopened it for ebaines who has the answer.

[ebaines]

Thanks for repoening the thread Odinn7. Please forgive my "inaccurate" rating above (and all it's misspellings) - I tried to edit or delete but I can't.

I'm afraid that all previous answers to the OP's question are incorrect. The issue is that the term "similar" in geometry refers to objects that have similar proportions and angles but are of different size. So given that the second cylinder has a height that is 3/4 the first, since the second cylinder is "similar" to the first its radius is also 3/4 that of the first.

There are two ways to proceed to answer the OP's question. First is the brute force way - calculate the radius of the first cylinder (which is 10 inches, as others have shown), then the radius of the second cylinder is 3/4 of that or 7.5 inches, and you can calculate the surface area and volume as you now know it's height and radius. Note that this technique does not take advantage of the fact that you are given the surface area of the first cylinder.

The second, and easier way to solve this is to use the principle that if you change all principal dimensions of a 3-dimensional object by a factor of 'x,' the new object's surface area is x^2 times the first, and its volume is x^3 times the first. So here x=3/4, and you can answer the question without ever having to determine the radius of either the first or second cylinder - just multiply the given surface area of the first cylinder by (3/4)^2, and the given volume of the first cylinder by (3/4)^3. Done!

I suggest that the OP and his 7th grade daughter try solving this problem using both techniques, so as to drive home the principle of how changing dmensions on a 3-D object affects its surface area and volume. Hope this helps!

[odinn7]

That all makes sense. Thanks for taking the time to post this. My daughter already turned it in but when I get home, I will show her this and we will go over it again.

[Wondergirl]

I suggest that the OP and his 7th grade daughter try solving this problem using both techniques

Odinn7 IS the OP, and that's why he could close the thread.