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2

$begingroup$

A math class is working with similar figures. Mr. Kole told the class that the surface area of the cones are 180 and 320 square units, and the volume of the smaller cone is 151 cubic units. He challenged his class to find the volume of the bigger cone. What is the volume of the bigger cone?

I tried to use the proportion 320/180 = x/151 to solve the problem. But my answer did not match the correct answer, which is 358 cubic units.

How would you solve this problem?

volume

share|cite|improve this question

edited 46 mins ago

orangebull

asked 8 hours ago

orangebullorangebull

2044 bronze badges

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Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Surface areas of similar figures are in proportion to the squares of corresponding lines; volumes are in proportion to the cubes.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are cones open bases or closed?
  $endgroup$
  – Mohammad Riazi-Kermani
  8 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

A math class is working with similar figures. Mr. Kole told the class that the surface area of the cones are 180 and 320 square units, and the volume of the smaller cone is 151 cubic units. He challenged his class to find the volume of the bigger cone. What is the volume of the bigger cone?

I tried to use the proportion 320/180 = x/151 to solve the problem. But my answer did not match the correct answer, which is 358 cubic units.

How would you solve this problem?

volume

share|cite|improve this question

edited 46 mins ago

orangebull

asked 8 hours ago

orangebullorangebull

2044 bronze badges

New contributor

orangebull is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Surface areas of similar figures are in proportion to the squares of corresponding lines; volumes are in proportion to the cubes.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are cones open bases or closed?
  $endgroup$
  – Mohammad Riazi-Kermani
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

A math class is working with similar figures. Mr. Kole told the class that the surface area of the cones are 180 and 320 square units, and the volume of the smaller cone is 151 cubic units. He challenged his class to find the volume of the bigger cone. What is the volume of the bigger cone?

I tried to use the proportion 320/180 = x/151 to solve the problem. But my answer did not match the correct answer, which is 358 cubic units.

How would you solve this problem?

volume

share|cite|improve this question

edited 46 mins ago

orangebull

asked 8 hours ago

orangebullorangebull

2044 bronze badges

New contributor

orangebull is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 46 mins ago

orangebull

asked 8 hours ago

orangebullorangebull

2044 bronze badges

New contributor

orangebull is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 46 mins ago

orangebull

asked 8 hours ago

orangebullorangebull

2044 bronze badges

New contributor

orangebull is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

orangebullorangebull

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New contributor

orangebull is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

orangebullorangebull

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3 Answers
3

active

oldest

votes

3

$begingroup$

Since the cones are similar the ratio of their areas is equal to the square of corresponding lines. (For example, the height or the radius of the base.) The ration of those lines is $$a=sqrt320over180=frac43$$ The ratio of the volume is $a^3$ so the volume of the large cone is $$left(frac43right)^3cdot151$$

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

24.9k44 gold badges1616 silver badges4141 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

Well you could use the formulas for the are and volume of cones to figure everything out and I guess that will be enough to solve the problem. However the problem is much more simple than that.

The area scales as a square of the radius $Apropto r^2$ and the volume as the cube $Vpropto r^3$, then the quantity
$$
fracA^3V^2
$$
is an invariant of similar cones (or any other figure for that mater). And we have
$$
frac320^3x^2 = frac180^3151^2
$$

This should be alright, however I am getting $x=358$ which doesn't seem to be the answer you are getting. Could you please check the numbers?

share|cite|improve this answer

answered 8 hours ago

David JaramilloDavid Jaramillo

54422 silver badges88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Your answer 358 is right. I made a typo.
  $endgroup$
  – orangebull
  44 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But why did you put $fracA^3V^2$? Shouldn't it have been $fracA^2V^3$?
  $endgroup$
  – orangebull
  42 mins ago
  
  

  
  
  
  

add a comment
 | 

1

$begingroup$

No, surface area is not directly proportional to volume.

I would try to construct a cone with surface area $180$ and volume $151$ using the formulas for surface area and volume, in particular keeping track of the ratio between the cone's height and radius.

Then I would construct a cone with surface area $320$ and the same ratio between its height and radius. Finally, apply the volume formula to this larger cone.

share|cite|improve this answer

answered 8 hours ago

7903766279037662

6781212 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  All you need is the ratio of the areas. The shape of the cone doesn't matter.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz Really? Aren't there infinitely many cones with surface area $320$ but different volumes, depending on their heights and radii? And only one of these will be similar to the smaller cone?
  $endgroup$
  – 79037662
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess I assumed the cones being similar was relevant information.
  $endgroup$
  – 79037662
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is relevant. That why all you need is the ratio of the areas.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz At the first glance I think the ratio alone is not sufficient.
  $endgroup$
  – callculus
  8 hours ago
  
  

  
  
  
  

 | 
show 2 more comments

Your Answer

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3

$begingroup$

Since the cones are similar the ratio of their areas is equal to the square of corresponding lines. (For example, the height or the radius of the base.) The ration of those lines is $$a=sqrt320over180=frac43$$ The ratio of the volume is $a^3$ so the volume of the large cone is $$left(frac43right)^3cdot151$$

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

24.9k44 gold badges1616 silver badges4141 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

Since the cones are similar the ratio of their areas is equal to the square of corresponding lines. (For example, the height or the radius of the base.) The ration of those lines is $$a=sqrt320over180=frac43$$ The ratio of the volume is $a^3$ so the volume of the large cone is $$left(frac43right)^3cdot151$$

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

24.9k44 gold badges1616 silver badges4141 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Since the cones are similar the ratio of their areas is equal to the square of corresponding lines. (For example, the height or the radius of the base.) The ration of those lines is $$a=sqrt320over180=frac43$$ The ratio of the volume is $a^3$ so the volume of the large cone is $$left(frac43right)^3cdot151$$

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

24.9k44 gold badges1616 silver badges4141 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

saulspatzsaulspatz

24.9k44 gold badges1616 silver badges4141 bronze badges

answered 8 hours ago

saulspatzsaulspatz

24.9k44 gold badges1616 silver badges4141 bronze badges

answered 8 hours ago

saulspatzsaulspatz

24.9k44 gold badges1616 silver badges4141 bronze badges

2

$begingroup$

Well you could use the formulas for the are and volume of cones to figure everything out and I guess that will be enough to solve the problem. However the problem is much more simple than that.

The area scales as a square of the radius $Apropto r^2$ and the volume as the cube $Vpropto r^3$, then the quantity
$$
fracA^3V^2
$$
is an invariant of similar cones (or any other figure for that mater). And we have
$$
frac320^3x^2 = frac180^3151^2
$$

This should be alright, however I am getting $x=358$ which doesn't seem to be the answer you are getting. Could you please check the numbers?

share|cite|improve this answer

answered 8 hours ago

David JaramilloDavid Jaramillo

54422 silver badges88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Your answer 358 is right. I made a typo.
  $endgroup$
  – orangebull
  44 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But why did you put $fracA^3V^2$? Shouldn't it have been $fracA^2V^3$?
  $endgroup$
  – orangebull
  42 mins ago
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

Well you could use the formulas for the are and volume of cones to figure everything out and I guess that will be enough to solve the problem. However the problem is much more simple than that.

The area scales as a square of the radius $Apropto r^2$ and the volume as the cube $Vpropto r^3$, then the quantity
$$
fracA^3V^2
$$
is an invariant of similar cones (or any other figure for that mater). And we have
$$
frac320^3x^2 = frac180^3151^2
$$

This should be alright, however I am getting $x=358$ which doesn't seem to be the answer you are getting. Could you please check the numbers?

share|cite|improve this answer

answered 8 hours ago

David JaramilloDavid Jaramillo

54422 silver badges88 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Your answer 358 is right. I made a typo.
  $endgroup$
  – orangebull
  44 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But why did you put $fracA^3V^2$? Shouldn't it have been $fracA^2V^3$?
  $endgroup$
  – orangebull
  42 mins ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

Well you could use the formulas for the are and volume of cones to figure everything out and I guess that will be enough to solve the problem. However the problem is much more simple than that.

The area scales as a square of the radius $Apropto r^2$ and the volume as the cube $Vpropto r^3$, then the quantity
$$
fracA^3V^2
$$
is an invariant of similar cones (or any other figure for that mater). And we have
$$
frac320^3x^2 = frac180^3151^2
$$

This should be alright, however I am getting $x=358$ which doesn't seem to be the answer you are getting. Could you please check the numbers?

share|cite|improve this answer

answered 8 hours ago

David JaramilloDavid Jaramillo

54422 silver badges88 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

David JaramilloDavid Jaramillo

54422 silver badges88 bronze badges

answered 8 hours ago

David JaramilloDavid Jaramillo

54422 silver badges88 bronze badges

answered 8 hours ago

David JaramilloDavid Jaramillo

54422 silver badges88 bronze badges

1

$begingroup$

No, surface area is not directly proportional to volume.

I would try to construct a cone with surface area $180$ and volume $151$ using the formulas for surface area and volume, in particular keeping track of the ratio between the cone's height and radius.

Then I would construct a cone with surface area $320$ and the same ratio between its height and radius. Finally, apply the volume formula to this larger cone.

share|cite|improve this answer

answered 8 hours ago

7903766279037662

6781212 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  All you need is the ratio of the areas. The shape of the cone doesn't matter.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz Really? Aren't there infinitely many cones with surface area $320$ but different volumes, depending on their heights and radii? And only one of these will be similar to the smaller cone?
  $endgroup$
  – 79037662
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess I assumed the cones being similar was relevant information.
  $endgroup$
  – 79037662
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is relevant. That why all you need is the ratio of the areas.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz At the first glance I think the ratio alone is not sufficient.
  $endgroup$
  – callculus
  8 hours ago
  
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

No, surface area is not directly proportional to volume.

I would try to construct a cone with surface area $180$ and volume $151$ using the formulas for surface area and volume, in particular keeping track of the ratio between the cone's height and radius.

Then I would construct a cone with surface area $320$ and the same ratio between its height and radius. Finally, apply the volume formula to this larger cone.

share|cite|improve this answer

answered 8 hours ago

7903766279037662

6781212 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  All you need is the ratio of the areas. The shape of the cone doesn't matter.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz Really? Aren't there infinitely many cones with surface area $320$ but different volumes, depending on their heights and radii? And only one of these will be similar to the smaller cone?
  $endgroup$
  – 79037662
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess I assumed the cones being similar was relevant information.
  $endgroup$
  – 79037662
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is relevant. That why all you need is the ratio of the areas.
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saulspatz At the first glance I think the ratio alone is not sufficient.
  $endgroup$
  – callculus
  8 hours ago
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

No, surface area is not directly proportional to volume.

I would try to construct a cone with surface area $180$ and volume $151$ using the formulas for surface area and volume, in particular keeping track of the ratio between the cone's height and radius.

Then I would construct a cone with surface area $320$ and the same ratio between its height and radius. Finally, apply the volume formula to this larger cone.

share|cite|improve this answer

answered 8 hours ago

7903766279037662

6781212 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

7903766279037662

6781212 bronze badges

answered 8 hours ago

7903766279037662

6781212 bronze badges

answered 8 hours ago

7903766279037662

6781212 bronze badges