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

[tnich]

To make it balanced with the same coefficient. Right? As you multiply both sides by a..

Say you have a number. Whatever you decide to call it, it's the same number, right? So whether you call it 3 or you call it x, it's the same number. That's what it means when we say $x=3$. If you take that number and multiply it by a, then you get the same result whether you call it $x$ or 3. So you can say $ax = 3a$. The point is, when you have an equation, if you want both sides to stay equal, you can't multiply one side by a number (or a variable) without multiplying the other side by the same number. In the same way, if you want to maintain equality, you can't multiple one term in an equation by a number without multiplying all of the other terms in the equation on both sides by the same number.
So now, take your equation 4H + 80/H + 40 = 100 and multiply by H.

To make it balanced with the same coefficient. Right? As you multiply both sides by a..

 

[Physiona]

Say you have a number. Whatever you decide to call it, it's the same number, right? So whether you call it 3 or you call it x, it's the same number. That's what it means when we say $x=3$. If you take that number and multiply it by a, then you get the same result whether you call it $x$ or 3. So you can say $ax = 3a$. The point is, when you have an equation, if you want both sides to stay equal, you can't multiply one side by a number (or a variable) without multiplying the other side by the same number. In the same way, if you want to maintain equality, you can't multiple one term in an equation by a number without multiplying all of the other terms in the equation on both sides by the same number.
So now, take your equation 4H + 80/H + 40 = 100 and multiply by H.

I'm sorry but I did do that, and you haven't gave feedback on it. (Refer to post 44)

 

[tnich]

I'm sorry but I did do that, and you haven't gave feedback on it. (Refer to post 44)

You have not written the equation, yet. Multiply both sides of 4H + 80/H + 40 = 100 by H and show me what you get.

 

[tnich]

You have not written the equation, yet. Multiply both sides of 4H + 80/H + 40 = 100 by H and show me what you get.

You should get a single equation as a result.

 

[Physiona]

I get 4H² + 80 + 40H = 100H

 

[tnich]

I get 4H² + 80 + 40H = 100H

Great! Now solve that for H.

 

[Physiona]

Okay thank you!
So do I collect like terms from both side:
4H² + 80 = 100H - 40H
4H²+80= 60H


Hang on, can I solve it as a quadratic?
In the form of:
4H²-60H+80=0

 

[tnich]

Okay thank you!
So do I collect like terms from both side:
4H² + 80 = 100H - 40H
4H²+80= 60H


Hang on, can I solve it as a quadratic?
In the form of:
4H²-60H+80=0

Go ahead.

 

[Physiona]

I end up with two solutions of:
13.520797289396
1.4792027106039
Is this correct?

 

[tnich]

I end up with two solutions of:
13.520797289396
1.4792027106039
Is this correct?

I think so. Now that you have H, how are you going to find L?

 

[Physiona]

Substitute it in the equation L.
(There's two numbers for H however..)

 

[tnich]

Substitute it in the equation L.
(There's two numbers for H however..)

Try them both and see what you get.

 

[Physiona]

For H₁= 13.520797289..
I got L₁= 1.479202711

For H₂= 1.4792027106039
I got L₂= 13.52079729

 

[tnich]

For H₁= 13.520797289..
I got L₁= 1.479202711

For H₂= 1.4792027106039
I got L₂= 13.52079729

Notice anything odd/interesting about that?

 

[Physiona]

Notice anything odd/interesting about that?

Yes they are similar!
I presume I use one set of them into substituting it in the diagonal length formula.
Thank you so much, you have literally taught me this question in hours which my teacher can't even teach in 3 years!
I give you all my thanks, and thank you for guiding me till the very last.

 

[Chestermiller]

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

There's a much easier way of solving this. Divide both sides of both equations by 2 to obtain:
$$2H+2L+LH=50\tag{1}$$
$$LH=20\tag{2}$$Subtract Eqn.2 from Eqn. 1 to yield: $$2(L+H)=30\tag{3}$$or$$L+H=15\tag{4}$$Square this equation to obtain:
$$(L+H)^2=L^2+2LH+H^2=225\tag{5}$$ Subtract 2 times Eqn. 2 from this equation to obtain:$$L^2+H^2=225-40=185\tag{6}$$Add $W^2=4$ to this equation to obtain:
$$L^2+W^2+H^2=185+4=189\tag{7}$$

 

[Ray Vickson]

Yes they are similar!
I presume I use one set of them into substituting it in the diagonal length formula.
Thank you so much, you have literally taught me this question in hours which my teacher can't even teach in 3 years!
I give you all my thanks, and thank you for guiding me till the very last.

Not only are they similar, they just swap the variables. After all, YOU know what you mean by length and height, but the model does not! It just knows that one of them should be about 1.48 and the other should be about 13.52

 

[Physiona]

Not only are they similar, they just swap the variables. After all, YOU know what you mean by length and height, but the model does not! It just knows that one of them should be about 1.48 and the other should be about 13.52

Yes thank you for your guidance. As said before, this question came in my exam and I did struggle as I judged the question really quickly into thinking it's hard, however after multiple attempts and solutions I figured it. Thank you for the support though.

 

[Physiona]

There's a much easier way of solving this. Divide both sides of both equations by 2 to obtain:
$$2H+2L+LH=50\tag{1}$$
$$LH=20\tag{2}$$Subtract Eqn.2 from Eqn. 1 to yield: $$2(L+H)=30\tag{3}$$or$$L+H=15\tag{4}$$Square this equation to obtain:
$$(L+H)^2=L^2+2LH+H^2=225\tag{5}$$ Subtract 2 times Eqn. 2 from this equation to obtain:$$L^2+H^2=225-40=185\tag{6}$$Add $W^2=4$ to this equation to obtain:
$$L^2+W^2+H^2=185+4=189\tag{7}$$

Oh okay. May I ask how do you obtain the "Subtract 2 times Eqn. 2"? Do you multiply LH =20 by 2, and then subtract?

 

[Chestermiller]

Oh okay. May I ask how do you obtain the "Subtract 2 times Eqn. 2"? Do you multiply LH =20 by 2, and then subtract?

Sure.

 

[tnich]

Yes they are similar!
I presume I use one set of them into substituting it in the diagonal length formula.
Thank you so much, you have literally taught me this question in hours which my teacher can't even teach in 3 years!
I give you all my thanks, and thank you for guiding me till the very last.

You're welcome. I'm glad you were able to work the problem all the way through! What value did you get for d?

 

[Chestermiller]

You're welcome. I'm glad you were able to work the problem all the way through! What value did you get for d?

$$\sqrt{189}$$

 

[Mark Hughes]

Here's the answer! Its a tricky question
Assume length L width W and depth D. Let's assign the 2cm we know to the width, so W = 2 (it doesn't matter which dimension you choose)

Now substitute into the formulae for Volume and Area.

Volume = 40 = WLD = 2LD (using W=2)
LD = 20, Let's call this equation 1

Area = 100 = 2(WL + WD + DL) = 4L + 4D + 2DL (using W=2)
2L + 2D + DL = 50, Let's call this equation 2

Now, substitute for D = 20/L from equation 1, into equation 2

We get, 2L + 2*20/L + 20/L * L = 50 which is the same as, 2L + 40/L -30 = 0, or L +20/L -15 = 0

Multiply all terms by L to get a quadratic, L^2 -15L +20 = 0

Solving this quadratic gives L = 13.52, and L = 1.48. (The second value is actually the value for D since (remember equation 1) D=20/L D=20/13.52 = 1.48)

So we have W = 2, L = 13.52, D = 1.48 The last part is difficult to understand. The diagonal is the one that cuts through the cuboid, from (front top left) to (rear bottom right) if you like

By Pythagoras on two triangles we get that the diagonal D = SQRT(L^2 + W^2 + D^2)

So, D = SQRT ( 182.8 + 4 + 2.19) = 13.7 to 3 SF

PLEASE NOTE, THIS IS THE LAST AND PROBABLY THE MOST DIFFICULT QUESTION ON THE PAPER.
It is designed to sort out the Grade 9ers, and anyone aiming for a lower grade would probably find it extremely difficult!
Hope this answer helps

 

[Chestermiller]

Here's the answer! Its a tricky question
Assume length L width W and depth D. Let's assign the 2cm we know to the width, so W = 2 (it doesn't matter which dimension you choose)

Now substitute into the formulae for Volume and Area.

Volume = 40 = WLD = 2LD (using W=2)
LD = 20, Let's call this equation 1

Area = 100 = 2(WL + WD + DL) = 4L + 4D + 2DL (using W=2)
2L + 2D + DL = 50, Let's call this equation 2

Now, substitute for D = 20/L from equation 1, into equation 2

We get, 2L + 2*20/L + 20/L * L = 50 which is the same as, 2L + 40/L -30 = 0, or L +20/L -15 = 0

Multiply all terms by L to get a quadratic, L^2 -15L +20 = 0

Solving this quadratic gives L = 13.52, and L = 1.48. (The second value is actually the value for D since (remember equation 1) D=20/L D=20/13.52 = 1.48)

So we have W = 2, L = 13.52, D = 1.48The last part is difficult to understand. The diagonal is the one that cuts through the cuboid, from (front top left) to (rear bottom right) if you like

By Pythagoras on two triangles we get that the diagonal D = SQRT(L^2 + W^2 + D^2)

So, D = SQRT ( 182.8 + 4 + 2.19) = 13.7 to 3 SF

PLEASE NOTE, THIS IS THE LAST AND PROBABLY THE MOST DIFFICULT QUESTION ON THE PAPER.
It is designed to sort out the Grade 9ers, and anyone aiming for a lower grade would probably find it extremely difficult!
Hope this answer helps

Was there something wrong with the answers the other responders gave? This seems much more involved.