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How can I find the volume of a rectangle from its surface area? [closed]

0

$begingroup$

The area of the base of an open rectangular box is 120cm². The front has an area of 96cm² and the side has an area of 100cm². What is the volume of the box?

So I was given the surface area of three of the sides of the rectangular prism and I'm trying to figure out how to get the width, height and length only using the surface areas provided to me. I've done some trial and error but no luck. Is there any way I can go about this differently? Thanks!

Area of rectangle is length x height. Volume of rectangular prism is length x width x height.

geometry

share|cite|improve this question

edited Mar 17 at 5:20

scratch

asked Mar 17 at 5:12

scratchscratch

82

$endgroup$

closed as off-topic by Alex Provost, mrtaurho, José Carlos Santos, Leucippus, Shailesh Mar 18 at 0:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alex Provost, mrtaurho, José Carlos Santos, Leucippus, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. How do you calculate the area of a rectangle and the volume of a rectangular box?
  $endgroup$
  – J. W. Tanner
  Mar 17 at 5:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Area of rectangle is length x height. Volume of rectangular prism is length x width x height.
  $endgroup$
  – scratch
  Mar 17 at 5:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Please add your thoughts to the question and put some effort in!
  $endgroup$
  – Rhys Hughes
  Mar 17 at 5:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @scratch you should edit your question and add that information
  $endgroup$
  – Rhys Hughes
  Mar 17 at 5:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @scratch: Put it this way. Let's say the lengths of sides of the box are $a$, $b$, and $c$. What do you then know about $a$, $b$, and $c$ from the areas given, and what is the volume of the box?
  $endgroup$
  – J. W. Tanner
  Mar 17 at 5:23
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

The area of the base of an open rectangular box is 120cm². The front has an area of 96cm² and the side has an area of 100cm². What is the volume of the box?

So I was given the surface area of three of the sides of the rectangular prism and I'm trying to figure out how to get the width, height and length only using the surface areas provided to me. I've done some trial and error but no luck. Is there any way I can go about this differently? Thanks!

Area of rectangle is length x height. Volume of rectangular prism is length x width x height.

geometry

share|cite|improve this question

edited Mar 17 at 5:20

scratch

asked Mar 17 at 5:12

scratchscratch

82

$endgroup$

closed as off-topic by Alex Provost, mrtaurho, José Carlos Santos, Leucippus, Shailesh Mar 18 at 0:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alex Provost, mrtaurho, José Carlos Santos, Leucippus, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math Stack Exchange. How do you calculate the area of a rectangle and the volume of a rectangular box?
  $endgroup$
  – J. W. Tanner
  Mar 17 at 5:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Area of rectangle is length x height. Volume of rectangular prism is length x width x height.
  $endgroup$
  – scratch
  Mar 17 at 5:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. Please add your thoughts to the question and put some effort in!
  $endgroup$
  – Rhys Hughes
  Mar 17 at 5:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @scratch you should edit your question and add that information
  $endgroup$
  – Rhys Hughes
  Mar 17 at 5:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @scratch: Put it this way. Let's say the lengths of sides of the box are $a$, $b$, and $c$. What do you then know about $a$, $b$, and $c$ from the areas given, and what is the volume of the box?
  $endgroup$
  – J. W. Tanner
  Mar 17 at 5:23
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

The area of the base of an open rectangular box is 120cm². The front has an area of 96cm² and the side has an area of 100cm². What is the volume of the box?

So I was given the surface area of three of the sides of the rectangular prism and I'm trying to figure out how to get the width, height and length only using the surface areas provided to me. I've done some trial and error but no luck. Is there any way I can go about this differently? Thanks!

Area of rectangle is length x height. Volume of rectangular prism is length x width x height.

geometry

share|cite|improve this question

edited Mar 17 at 5:20

scratch

asked Mar 17 at 5:12

scratchscratch

82

$endgroup$

share|cite|improve this question

edited Mar 17 at 5:20

scratch

asked Mar 17 at 5:12

scratchscratch

82

share|cite|improve this question

edited Mar 17 at 5:20

scratch

asked Mar 17 at 5:12

scratchscratch

82

asked Mar 17 at 5:12

scratchscratch

82

asked Mar 17 at 5:12

scratchscratch

82

2 Answers
2

active

oldest

votes

2

$begingroup$

Say the sides of the box are $a, b,$ and $c$.

Then we have $ab=120, ac=96$, and $bc=100$ cm$^2$.

The volume is $abc$ cm$^3$.

Note that $120times96times100$ = $abtimes actimes bc = (abc)^2, $ so

$abc=sqrt{120times96times100}=sqrt{4times30times16times6times100}=$

$2times 4times 10timessqrt{5times6times6}=80times6sqrt5=480sqrt5$ cm$^3$.

share|cite|improve this answer

answered Mar 17 at 5:27

J. W. TannerJ. W. Tanner

4,0611320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! So the answer must be in this case 1073.312629 cm3.
  $endgroup$
  – scratch
  Mar 17 at 5:38
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes (approximately)
  $endgroup$
  – J. W. Tanner
  Mar 17 at 5:39
  

  
  
  
  

add a comment | 

1

$begingroup$

Let the length, height, width be $L, H, W$ respectively.

Then you have:

$$LW=120to L=frac{120}{W}$$
$$LH=96to L=frac{96}{H}$$
$$WH=100$$

Notice from the first two that: $$frac{120}{W}=frac{96}{H}tofrac WH = frac{120}{96}=frac 54$$

Use this with the third statement ($WH=100$) to find $W$ and $H$. Then use either of the first two to find $L$, armed with this information.

share|cite|improve this answer

answered Mar 17 at 5:27

Rhys HughesRhys Hughes

7,0801630

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For reference, I achieve that $$H=sqrt{80}=4sqrt5; W=5sqrt 5; L=frac{24}{5}sqrt5$$
  $endgroup$
  – Rhys Hughes
  Mar 17 at 5:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so we agree $HWL=480sqrt 5$
  $endgroup$
  – J. W. Tanner
  Mar 17 at 15:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed we do, I actually didn't realise that your method (I.e. one so direct) was a possibility, so it was nice to see.
  $endgroup$
  – Rhys Hughes
  Mar 17 at 17:35
  

  
  
  
  

add a comment | 

2

$begingroup$

Say the sides of the box are $a, b,$ and $c$.

Then we have $ab=120, ac=96$, and $bc=100$ cm$^2$.

The volume is $abc$ cm$^3$.

Note that $120times96times100$ = $abtimes actimes bc = (abc)^2, $ so

$abc=sqrt{120times96times100}=sqrt{4times30times16times6times100}=$

$2times 4times 10timessqrt{5times6times6}=80times6sqrt5=480sqrt5$ cm$^3$.

share|cite|improve this answer

answered Mar 17 at 5:27

J. W. TannerJ. W. Tanner

4,0611320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! So the answer must be in this case 1073.312629 cm3.
  $endgroup$
  – scratch
  Mar 17 at 5:38
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes (approximately)
  $endgroup$
  – J. W. Tanner
  Mar 17 at 5:39
  

  
  
  
  

add a comment | 

2

$begingroup$

Say the sides of the box are $a, b,$ and $c$.

Then we have $ab=120, ac=96$, and $bc=100$ cm$^2$.

The volume is $abc$ cm$^3$.

Note that $120times96times100$ = $abtimes actimes bc = (abc)^2, $ so

$abc=sqrt{120times96times100}=sqrt{4times30times16times6times100}=$

$2times 4times 10timessqrt{5times6times6}=80times6sqrt5=480sqrt5$ cm$^3$.

share|cite|improve this answer

answered Mar 17 at 5:27

J. W. TannerJ. W. Tanner

4,0611320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! So the answer must be in this case 1073.312629 cm3.
  $endgroup$
  – scratch
  Mar 17 at 5:38
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes (approximately)
  $endgroup$
  – J. W. Tanner
  Mar 17 at 5:39
  

  
  
  
  

add a comment | 

$begingroup$

Say the sides of the box are $a, b,$ and $c$.

Then we have $ab=120, ac=96$, and $bc=100$ cm$^2$.

The volume is $abc$ cm$^3$.

Note that $120times96times100$ = $abtimes actimes bc = (abc)^2, $ so

$abc=sqrt{120times96times100}=sqrt{4times30times16times6times100}=$

$2times 4times 10timessqrt{5times6times6}=80times6sqrt5=480sqrt5$ cm$^3$.

share|cite|improve this answer

answered Mar 17 at 5:27

J. W. TannerJ. W. Tanner

4,0611320

$endgroup$

share|cite|improve this answer

answered Mar 17 at 5:27

J. W. TannerJ. W. Tanner

4,0611320

answered Mar 17 at 5:27

J. W. TannerJ. W. Tanner

4,0611320

answered Mar 17 at 5:27

J. W. TannerJ. W. Tanner

4,0611320

1

$begingroup$

Let the length, height, width be $L, H, W$ respectively.

Then you have:

$$LW=120to L=frac{120}{W}$$
$$LH=96to L=frac{96}{H}$$
$$WH=100$$

Notice from the first two that: $$frac{120}{W}=frac{96}{H}tofrac WH = frac{120}{96}=frac 54$$

Use this with the third statement ($WH=100$) to find $W$ and $H$. Then use either of the first two to find $L$, armed with this information.

share|cite|improve this answer

answered Mar 17 at 5:27

Rhys HughesRhys Hughes

7,0801630

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For reference, I achieve that $$H=sqrt{80}=4sqrt5; W=5sqrt 5; L=frac{24}{5}sqrt5$$
  $endgroup$
  – Rhys Hughes
  Mar 17 at 5:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so we agree $HWL=480sqrt 5$
  $endgroup$
  – J. W. Tanner
  Mar 17 at 15:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed we do, I actually didn't realise that your method (I.e. one so direct) was a possibility, so it was nice to see.
  $endgroup$
  – Rhys Hughes
  Mar 17 at 17:35
  

  
  
  
  

add a comment | 

1

$begingroup$

Let the length, height, width be $L, H, W$ respectively.

Then you have:

$$LW=120to L=frac{120}{W}$$
$$LH=96to L=frac{96}{H}$$
$$WH=100$$

Notice from the first two that: $$frac{120}{W}=frac{96}{H}tofrac WH = frac{120}{96}=frac 54$$

Use this with the third statement ($WH=100$) to find $W$ and $H$. Then use either of the first two to find $L$, armed with this information.

share|cite|improve this answer

answered Mar 17 at 5:27

Rhys HughesRhys Hughes

7,0801630

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For reference, I achieve that $$H=sqrt{80}=4sqrt5; W=5sqrt 5; L=frac{24}{5}sqrt5$$
  $endgroup$
  – Rhys Hughes
  Mar 17 at 5:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  so we agree $HWL=480sqrt 5$
  $endgroup$
  – J. W. Tanner
  Mar 17 at 15:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed we do, I actually didn't realise that your method (I.e. one so direct) was a possibility, so it was nice to see.
  $endgroup$
  – Rhys Hughes
  Mar 17 at 17:35
  

  
  
  
  

add a comment | 

$begingroup$

Let the length, height, width be $L, H, W$ respectively.

Then you have:

$$LW=120to L=frac{120}{W}$$
$$LH=96to L=frac{96}{H}$$
$$WH=100$$

Notice from the first two that: $$frac{120}{W}=frac{96}{H}tofrac WH = frac{120}{96}=frac 54$$

Use this with the third statement ($WH=100$) to find $W$ and $H$. Then use either of the first two to find $L$, armed with this information.

share|cite|improve this answer

answered Mar 17 at 5:27

Rhys HughesRhys Hughes

7,0801630

$endgroup$

share|cite|improve this answer

answered Mar 17 at 5:27

Rhys HughesRhys Hughes

7,0801630

answered Mar 17 at 5:27

Rhys HughesRhys Hughes

7,0801630

answered Mar 17 at 5:27

Rhys HughesRhys Hughes

7,0801630