stapel

Super Moderator
Staff member
What are your thoughts? What have you tried? How far have you gotten?

Eliz.

Dr.Peterson

Elite Member
I'd suggest you assign variables to the length, width, and height, and write some equations. Then write an expression for the length of AG in terms of those variables, and see what you can do. (There's a very simple answer; you don't need to solve for all the variables.)

greentree.23

New member
I'd suggest you assign variables to the length, width, and height, and write some equations. Then write an expression for the length of AG in terms of those variables, and see what you can do. (There's a very simple answer; you don't need to solve for all the variables.)
What i’ve done so far is:
Let width = x = AB = DC = GH = FE
Let length = y = EH = AD = BC = FG
Let height = z = AE = DH =BF = CG.
AH^2 = 6^2 = z^2 + y^2 using Pythag.
AG^2 = ?
AC^2 = x^2 + y^2
AF^2 = 4^2 = ?
And I’m stuck from here

Dr.Peterson

Elite Member
What i’ve done so far is:
Let width = x = AB = DC = GH = FE
Let length = y = EH = AD = BC = FG
Let height = z = AE = DH =BF = CG.
AH^2 = 6^2 = z^2 + y^2 using Pythag.
AG^2 = ?
AC^2 = x^2 + y^2
AF^2 = 4^2 = ?
And I’m stuck from here
You should have three equations, for AC, AF, and AH. I don't know why you didn't finish AF and AC.

Then, to write an equation for AG, observe that ACG is a right triangle.

Then start putting things together.

greentree.23

New member
Now I have AG^2 = 3^2 + z^2 with AG being the hypotenuse - I don't know what to do from here?
You should have three equations, for AC, AF, and AH. I don't know why you didn't finish AF and AC.

Then, to write an equation for AG, observe that ACG is a right triangle.

Then start putting things together.

The Highlander

Full Member
What i’ve done so far is:
Let width = x = AB = DC = GH = FE
Let length = y = EH = AD = BC = FG
Let height = z = AE = DH =BF = CG.
AH^2 = 6^2 = z^2 + y^2 using Pythag.
AG^2 = ?
AC^2 = x^2 + y^2
AF^2 = 4^2 = ?
And I’m stuck from here
Having defined your variables, why didn't you use them?

based on this diagram...

you could have said....

x² + y² = 3² $$\displaystyle \implies$$ x² + y² = 9
x² + z² = 4² $$\displaystyle \implies$$ x² + z² = 16
y² + z² = 6² $$\displaystyle \implies$$ y² + z² = 36

Let us call AG h (since it is the hypotenuse of the right-angled triangle ACG) then: h² = 3² + z²

But let us now also call
AC "a" (so that a = 3 and a² = 9) but note too then that a² = x² + y²

Therefore, h² = a² + z² $$\displaystyle \implies$$ h² = x² + y² + z²

It is possible to manipulate the three equations above to arrive at an algebraic answer for h (Hint: start by adding them together) but it should be noted, however, that using the equations to actually solve for x, y & z leads to a contradiction which means that this prism is a physical impossibility!

Notwithstanding that fact, all you are asked to do is arrive at a "value" for h (AG) so I suppose you can just proceed as suggested to arrive at one.

Steven G

Elite Member
This doesn't seem to be possible.
I will not show my work as I want the OP to solve this problem.

greentree.23

New member
This doesn't seem to be possible.
I will not show my work as I want the OP to solve this problem.
I'm not too sure on how to manipulate the other equations to solve side AG using Pythagoras. 3^2 + z^2 = AG^2 and I'm stuck from here?

The Highlander

Full Member
This doesn't seem to be possible.
I will not show my work as I want the OP to solve this problem.
It is possible to solve it algebraically (body diagonal of a cuboid) but the prism just can't exist with those dimensions.

greentree.23

New member
It is possible to solve it algebraically (body diagonal of a cuboid) but the prism just can't exist with those dimensions.
Can you give a hint on how to solve it algebraically?

Steven G

Elite Member
It is possible to solve it algebraically (body diagonal of a cuboid) but the prism just can't exist with those dimensions.
You lost me. If the prism can't exist, then how can you find the diagonal of the prism, you know, the one that doesn't exist??

The Highlander

Full Member
You lost me. If the prism can't exist, then how can you find the diagonal of the prism, you know, the one that doesn't exist??
I can't tell you that for four days or stapel will delete me!

However, (as suggested above) add the equations & simplify.

(I presume that (like me) you found x2 to be negative hence your comment: "
This doesn't seem to be possible." but using the formula for the body diagonal of a cuboid* does produce a value for the length of AG
despite the impossibility of a prism with the given dimensions. )

* See Post
#6.

blamocur

Senior Member
I agree with @The Highlander:
$2x^2 = -11 = (x^2+y^2) + (x^2 + y^2) - (y^2+z^2) = 9+16-36$

Steven G

Elite Member
I can't tell you that for four days or stapel will delete me!

but using the formula for the body diagonal of a cuboid* does produce a value for the length of AG despite the impossibility of a prism with the given dimensions. )
I didn't know that stapel had the ability to delete people. I'm glad that I've been nice to her.

How can you find the length of AG, A and G being points in a rectangular prism, if the rectangular prism doesn't exist?

I'm not saying that you're wrong but am saying that I don't understand how you can be correct.

Sometimes I am completely confused with math and other times I just simply love math.

blamocur

Senior Member
How can you find the length of AG, A and G being points in a rectangular prism, if the rectangular prism doesn't exist?
It exists, but it isn't real

Steven G

Elite Member
It exists, but it isn't real
I won't blame/hate you, but I am not a fan of math today. It exists, but it isn't real

Dr.Peterson

Elite Member
How can you find the length of AG, A and G being points in a rectangular prism, if the rectangular prism doesn't exist?

I'm not saying that you're wrong but am saying that I don't understand how you can be correct.

Sometimes I am completely confused with math and other times I just simply love math.
I have long found it interesting that there are some problems like this for which there is a shortcut to the solution that bypasses work that shows whether there is in fact a solution at all, and just finds the solution -- even when it does not really exist. Unfortunately, I have never made a list of such problems, so I can never point to one.

In this case, we can just add the three equations $x^2+y^2=9\\x^2+z^2=16\\y^2+z^2=36$ together, and divide by 2, to get $x^2+y^2+z^2=30.5$ which provides an answer, but since we never actually solved for the individual variables, we don't realize (at least, I did not) that all we've proved is that if there is a solution, that would be it.

Sometimes existence has to be proved separately!