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

[Physiona]

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.

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.)

 

[Physiona]

I think I attempt it as shown, however, I'm not enitrely sure. Any help would be grateful.

 

[tnich]

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.

 

[Physiona]

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?

 

[tnich]

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.

 

[Physiona]

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.

 

[tnich]

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.

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.

 

[Physiona]

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?

 

[tnich]

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.

 

[Physiona]

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.

 

[tnich]

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?

 

[Physiona]

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?

 

[tnich]

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?

 

[Physiona]

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

 

[Physiona]

Hang on, would I solve simultaneously?
By multiplying out the brackets and then solving simultaneously..?

 

[tnich]

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?

 

[Physiona]

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..

 

[tnich]

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?

 

[Physiona]

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.

 

[Physiona]

Unless I substitute 2LH into the surface area equation... I'd still end up with two unknowns..

 

[tnich]

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.

 

[tnich]

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.

 

[Physiona]

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?

 

[tnich]

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.

 

[Physiona]

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

 

[tnich]

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.

 

[Physiona]

How do I consider solving 4(20/H).. Is it 80/H?
So from there
4H + 80/H + 40/H * H = 100

 

[tnich]

How do I consider solving 4(20/H).. Is it 80/H?
So from there
4H + 80/H + 40/H * H = 100

Keep going.

 

[Physiona]

And then it'll equal:
4H + 120/H * H = 100.
The H will cancel, leaving:
4H + 120 = 100

 

[Physiona]

I'm going to end up with a negative.. 120-100= -20..

 

[tnich]

And then it'll equal:
4H + 120/H * H = 100.
The H will cancel, leaving:
4H + 120 = 100

Oops. What is $\frac {120} H * H$
Sorry, I meant what is $\frac {40} H * H$

 

[Physiona]

Oops. What is $\frac {120} H * H$
Sorry, I meant what is $\frac {40} H * H$

Oh, is it 40H/H?

 

[tnich]

Is it 120H/H?

Sorry, I messed that up. Let's go back to post #27 where you say correctly that

4H + 80/H + 40/H * H = 100

What is $\frac {40} H * H$?

 

[Physiona]

Yes would it equal 40H/H?

 

[tnich]

Yes would it equal 40H/H?

OK, what is $\frac H H$? It's not a trick question.