General Mathematics Grade 9 -- Find the length of a diagonal of the cuboid

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SUMMARY

The discussion centers on solving for the diagonal of a cuboid given its volume of 40 cm³, surface area of 100 cm², and one edge measuring 2 cm. The user initially struggles with the equations for volume and surface area, leading to confusion over the unknown dimensions. Key equations include the volume formula \( V = L \times W \times H \) and the surface area formula \( SA = 2(LW + LH + WH) \). The correct approach involves solving these equations simultaneously to find the unknown dimensions before applying the diagonal formula \( D = \sqrt{L^2 + W^2 + H^2} \).

PREREQUISITES
  • Understanding of cuboid geometry and properties
  • Familiarity with algebraic manipulation of equations
  • Knowledge of volume and surface area formulas for three-dimensional shapes
  • Ability to solve simultaneous equations
NEXT STEPS
  • Study the derivation and application of the diagonal formula for cuboids
  • Practice solving simultaneous equations with two variables
  • Explore real-world applications of volume and surface area calculations
  • Learn about geometric properties of three-dimensional shapes
USEFUL FOR

Students preparing for mathematics exams, educators teaching geometry, and anyone needing assistance with algebraic problem-solving in three-dimensional contexts.

Physiona
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Homework Statement



A solid cuboid has a volume of 40cm3. The cuboid has a total surface area of 100cm^3. One edge of the cuboid has length 2cm. Find the length of a diagonal of the cuboid. Give your answer correct to 3 significant figures.
This question recently came up on my maths exam today, it was the final question worth 6 marks.

upload_2018-3-13_13-33-38.png


My attempt of the question:
I wrote down the formulas for both Volume and surface area of a cuboid.
I labelled the length in which I didn't know x.
I used LxWxH, and made it equal to 40cm3.

Due to timing, I didn't manage to finish it, it looked fairly easy but i struggled in the exam.
Can someone please help and offer detailed guidance?
(As a side note, this question may not seem entirely difficult at attempts, however I personally got confused in how to approach it, using a logical method. Please do not affirm that it is easy to do, as I know there are many intellectual people here who will find this really simple to do, which is why I come here for homework help and complex questions in which I find generally confusing to get guidance from people. Thank you.)
 

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tn3M2G
I think I attempt it as shown, however, I'm not enitrely sure. Any help would be grateful.
 
Physiona said:
tn3M2G
I think I attempt it as shown, however, I'm not enitrely sure. Any help would be grateful.
You have made a good start by finding a diagram and the equations for the cuboid, but you haven't attempted to solve the equations for the diagonal, yet. Please make an attempt.
 
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tnich said:
You have made a good start by finding a diagram and the equations for the cuboid, but you haven't attempted to solve the equations for the diagonal, yet. Please make an attempt.
The thing is, that I'm struggling to find a decent logical method to attempt it. I have one length of 2cm, and the volume is 40cm3, and the total surface area is 100cm3.
I attempted to do:
2*x*x = 40cm3
2x2 = 40cm3
x2 = 20
x = square root of 20= 4.472135955
However when I do an equation for the surface area I get comepletely different solutions, and then I'm not sure on which one to use for the width and height.
Surface area = 2(LH+WH+LW) (I then labelled the length width and height as base length represented by a letter S)
100cm3 = 2(s*s + s*s + s*s)
100cm3 = 2(3s)2
100cm3 = 6s2
Rearrange for s, is 100/6 and then square rooting the answer, which gave s=4.082482905
Are the formulas correct and am I working on the right line?
 
Last edited:
Physiona said:
The thing is, that I'm struggling to find a decent logical method to attempt it. I have one length of 2cm, and the volume is 40cm3, and the total surface area is 100cm3.
I attempted to do:
2*x*x = 40cm3
2x2 = 40cm3
x2 = 20
x = square root of 20= 4.472135955
However when I do an equation for the surface area I get comepletely different solutions, and then I'm not sure on which one to use for the width and height.
Surface area = 2(LH+WH+LW) (I then labelled the length width and height as base length represented by a letter S)
100cm3 = 2(s*s + s*s + s*s)
100cm3 = 2(3s)2
100cm3 = 6s2
Rearrange for s, is 100/6 and then square rooting the answer, which gave s=4.082482905
Are the formulas correct and am I working on the right line?
You seem to be making an incorrect assumption. If all sides are length ##s=2cm##, then the volume would be ##s^3=8cm^3##. The problem statement says one edge is 2cm. Not all of them.
 
tnich said:
You seem to be making an incorrect assumption. If all sides are length ##s=2cm##, then the volume would be ##s^3=8cm^3##. The problem statement says one edge is 2cm. Not all of them.
I don't understand your point. I haven't made that assumption. I have labelled the unknown lengths as x for the first section involving the volume and then s for the second section involving surface area. My main issue is that I end up with two solutions, and I can't be able to interpret that into the diagonal formula to use. This is because I'm missing the other two lengths; I have only been provided with one and volume and total surface area. I'm presuming I need to find the other two lengths which match to the volume AND surface area.
 
Physiona said:
I don't understand your point. I haven't made that assumption. I have labelled the unknown lengths as x for the first section involving the volume and then s for the second section involving surface area. My main issue is that I end up with two solutions, and I can't be able to interpret that into the diagonal formula to use. This is because I'm missing the other two lengths; I have only been provided with one and volume and total surface area. I'm presuming I need to find the other two lengths which match to the volume AND surface area.
Physiona said:
I don't understand your point. I haven't made that assumption. I have labelled the unknown lengths as x for the first section involving the volume and then s for the second section involving surface area. My main issue is that I end up with two solutions, and I can't be able to interpret that into the diagonal formula to use. This is because I'm missing the other two lengths; I have only been provided with one and volume and total surface area. I'm presuming I need to find the other two lengths which match to the volume AND surface area.
But you are using the same variable ##s## to represent length, width and height. Suppose you use the variable names suggested by the figure, ##L, W## and ##H##? Try writing the equations for surface area and volume in terms of those variables.
 
tnich said:
But you are using the same variable ##s## to represent length, width and height. Suppose you use the variable names suggested by the figure, ##L, W## and ##H##? Try writing the equations for surface area and volume in terms of those variables.
Surface area = 2(LW+WH+HL)
Volume= 2*L*H
what difference is there? That I'm not assuming they're the same right?
How do I progress on from there?
 
Physiona said:
Surface area = 2(LW+WH+HL)
Volume= 2*L*H
what difference is there? That I'm not assuming they're the same right?
How do I progress on from there?
Right.
In the formula for volume you have assumed ##W=2##. You also know the volume and the surface area.
 
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  • #10
tnich said:
Right.
In the formula for volume you have assumed ##W=2##. You also know the volume and the surface area.
Yes I do know that. Will I make it equal to each other? 40cm3 = 2*L*H I have two Unknown coefficients though.
 
  • #11
Physiona said:
Yes I do know that. Will I make it equal to each other? 40cm3 = 2*L*H I have two Unknown coefficients though.
Yes, you have two unknowns, but you also have the equation for surface area. What can you do with that?
 
  • #12
Do I make it equal to each other?
2*L*H = 2(2*H+L*2+L*H)
What do I do with the actual total surface area of 100cm3 and volume of 40cm3?
 
  • #13
Physiona said:
Do I make it equal to each other?
2*L*H = 2(2*H+L*2+L*H)
What do I do with the actual total surface area of 100cm3 and volume of 40cm3?
Why would the surface area be equal to the volume? You have an equation for the surface area and a value for the surface area. What can you do with them?
 
  • #14
tnich said:
Why would the surface area be equal to the volume? You have an equation for the surface area and a value for the surface area. What can you do with them?
Yes I know, however I have unknowns,
So you'd make the surface area equation equal to the answer along with the Unknown coefficients.
2(2*H+L*2+L*H) = 100cm3
And the volume:
2*L*H =40cm3
 
  • #15
Hang on, would I solve simultaneously?
By multiplying out the brackets and then solving simultaneously..?
 
  • #16
Physiona said:
Yes I know, however I have unknowns,
So you'd make the surface area equation equal to the answer along with the Unknown coefficients.
2(2*H+L*2+L*H) = 100cm3
And the volume:
2*L*H =40cm3
Good! Now you have two equations in two unknowns. How will you solve that?
 
  • #17
So for surface area: it'd equal
4H + 4L + 2LH = 100cm3
And for volume it'd equal
2LH = 40cm3
Is that right? But for volume, I don't have a coefficient like I do in the surface area; of 4H and 4L..
 
  • #18
Physiona said:
So for surface area: it'd equal
4H + 4L + 2LH = 100cm3
And for volume it'd equal
2LH = 40cm3
Is that right? But for volume, I don't have a coefficient like I do in the surface area; of 4H and 4L..
So far, so good. How do you usually go about solving two equations in two unknowns?
 
  • #19
For these kinds, I've not entirely come across before. Usually, they would both have the same coefficients, and I'd make one term equal and subtract all the terms from one equation from the other. Then I'd rearrange and substitute the value back into the simplest equation to then figure out the other Unknown. Here I have only 2LH which is common in both terms. I don't have 4L and 4H.
 
  • #20
Unless I substitute 2LH into the surface area equation... I'd still end up with two unknowns..
 
  • #21
Physiona said:
For these kinds, I've not entirely come across before. Usually, they would both have the same coefficients, and I'd make one term equal and subtract all the terms from one equation from the other. Then I'd rearrange and substitute the value back into the simplest equation to then figure out the other Unknown. Here I have only 2LH which is common in both terms. I don't have 4L and 4H.
The usual way to proceed in a case like this is to take one of the equations and solve it for one of the variables. In this case, I would suggest solving the volume equation for L in terms of H. In other words, get L by itself on one side of the equation. Then you can substitute for L in the other equation.
 
  • #22
Physiona said:
Unless I substitute 2LH into the surface area equation... I'd still end up with two unknowns..
You could do that, and that would give you a nice equation to solve for L in terms of H, which you could then substitute into the volume equation. This way might be easier than the one I proposed in post #19.
 
  • #23
tnich said:
The usual way to proceed in a case like this is to take one of the equations and solve it for one of the variables. In this case, I would suggest solving the volume equation for L in terms of H. In other words, get L by itself on one side of the equation. Then you can substitute for L in the other equation.
If I use this method, by rearranging the second one would it equal L = 20/H, (if I'm rearranging for L) and then substitute it into the first equation?
 
  • #24
Physiona said:
If I use this method, by rearranging the second one would it equal L = 20/H, (if I'm rearranging for L) and then substitute it into the first equation?
Right.
 
  • #25
The main issue is the Unknowns. It's making me frustrated on where to put them in use.
So if I insert L into the first equation it'll equal:
4H + 4*(20/H) + 2(20/H) * H = 100
 
  • #26
Physiona said:
The main issue is the Unknowns. It's making me frustrated on where to put them in use.
So if I insert L into the first equation it'll equal:
4H + 4*(20/H) + 2(20/H) * H = 100
Great! Now you have an equation you can solve for H.
 
  • #27
How do I consider solving 4(20/H).. Is it 80/H?
So from there
4H + 80/H + 40/H * H = 100
 
  • #28
Physiona said:
How do I consider solving 4(20/H).. Is it 80/H?
So from there
4H + 80/H + 40/H * H = 100
Keep going.
 
  • #29
And then it'll equal:
4H + 120/H * H = 100.
The H will cancel, leaving:
4H + 120 = 100
 
  • #30
I'm going to end up with a negative.. 120-100= -20..
 

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