Can you solve the volume of a cube with unequal heights?

[ThunderSkunk]

Can you solve the volume of "a cube" with unequal heights?

I have a challenge for someone (which I plan on working on this weekend myself when I have free time from my homework).

Can you calculate the volume of a cube-like shape with four different heights and with perfect square base if you are given the surface area of all sides, including the base, except for its top? The sides of this object would be four trapezoids of which two, and only two, heights each side would be equal (where the two trapezoids meet to create a corner). The angle of the trapezoid to its neighbor and the square base is always 90 degrees.

To get a better idea of what I am taking about, check out the picture I uploaded to this post.

I would be very grateful to whomever can solve this for me. Knowing a formula for this would be very useful for a school project I am creating.

In case you were wondering why I want to know this (for motivational purposes), I am trying to map the the volume of space shaded by irregular objects with a light source coming from a given angle (so I can ask an interesting sunlight competition question for my undergraduate senior project in plant ecology). From what I can tell at this point, field measurements could yield an average volume that would be represented by a shape similar to the one I've described (because I believe I've already figured out a way to find the area of all the sides and the base).

I was just hoping a solution already existed for this (so I don't have to reinvent the wheel). Thanks for taking the time to check out my question.

 

[jetwaterluffy]

Is this actually homework? General maths go in the https://www.physicsforums.com/forumdisplay.php?f=73". If it is homework, you need to attempt it yourself, first.

 

[ThunderSkunk]

No, this is not technically homework and it probably doesn't belong in this section (this is my first post so I am new here). I am trying to map the the volume of space shaded by irregular objects with a light source coming from a given angle (so I can ask an interesting sunlight competition question for my undergraduate senior project in plant ecology). From what I can tell at this point, field measurements could yield an average volume that would be represented by a shape similar to the one I've described. I was just hoping a solution already existed for this (so I don't have to reinvent the wheel).

Thanks for the tip. I'll check out that other section.

 

[phinds]

Your figure appears to have the 4 top points all in the same plane, but I THINK (could be wrong) that this is inconsistent with your statement of the problem which seems to imply that the 4 heights would all be unrelated. Once you pick 3 of the heights, the 4th one is fixed if you want the top to be one plane as you have drawn.

Am I over-interpreting your problem statement?

 

[AlephZero]

Cut the volume into thin slices parallel to one edge of the base.

The volume of one slice = (length) x (average height) x (thickness).

The volume of the whole solid will be (area of base) x (average height). where "average" means (sum of the heights at the four corners)/4.

 

[phinds]

Cut the volume into thin slices parallel to one edge of the base.

The volume of one slice = (length) x (average height) x (thickness).

The volume of the whole solid will be (area of base) x (average height). where "average" means (sum of the heights at the four corners)/4.

Even if all 4 top points are in the same plane, that's going to be a VERY messy and tedius set of calculations and if the 4 points are NOT in the same plane then it doesn't work at all (although a similar, but more complicated one, would).

Conceptually of course, it DOES get you to the answer, but surely there's an easier way?

 

[Mark44]

ThunderSkunk,
It is against forum rules to post the same topic in multiple sections. I am not issuing an warning or infraction this time, as I believe you were acting on the advice of people responding in this thread. I have merged the two threads into this one.

In the future, if you find that you have posted something in the wrong section, click the Report button, and a mentor will take care of moving the thread.

In case you haven't taken a look at the Physics Forums rules, you can see them by clicking Rules at the top of the page, or by clicking this link: https://www.physicsforums.com/showthread.php?t=414380.

 

[vrmuth]

I have a challenge for someone (which I plan on working on this weekend myself when I have free time from my homework).

Can you calculate the volume of a cube-like shape with four different heights and with perfect square base if you are given the surface area of all sides, including the base, except for its top? The sides of this object would be four trapezoids of which two, and only two, heights each side would be equal (where the two trapezoids meet to create a corner). The angle of the trapezoid to its neighbor and the square base is always 90 degrees.

Even if all 4 top points are in the same plane, that's going to be a VERY messy and tedius set of calculations and if the 4 points are NOT in the same plane then it doesn't work at all (although a similar, but more complicated one, would).

Conceptually of course, it DOES get you to the answer, but surely there's an easier way?

V= \frac{1}{6} (h_{1}+2h_{2}+2h_{3} +h_{4}) a^{2} cubic units, where the h_{4} corresponds to the height of the 4th point which is not in the plane of other 3 points , i hope this will also work if the four points are in the same plane . can anybody check it ?

 

[phinds]

V= \frac{1}{6} (h_{1}+2h_{2}+2h_{3} +h_{4}) a^{2} cubic units, where the h_{4} corresponds to the height of the 4th point which is not in the plane of other 3 points , i hope this will also work if the four points are in the same plane . can anybody check it ?

I have no idea where that equation comes from (and don't know whether it's right or wrong) BUT I find it unlikely that it could be correct since you have a factor of 2 next to 2 of the legs but not next to the other 2. Intuitively it would seem impossible that this could be a correct method.

Just look at a simple degenerative case where all for upper points are at the same height. Now you increase one of the points that you have with a factor of 2 and do the computation. OR you increase one of the points for which you do NOT have a factor of 2, and by the same amount, then do the computation. Clearly you'll get two different answers for the identical figure.

Yep, my intuition was right. Your formula can't work.

 

[DaveC426913]

Guys, if we can assume the top surface is a plane, this problem is trivial.

1] Calculate the volume of the non-cube box formed below the shortest upper corner. (l*w*h_shortest)

2] Calculate the volume of a non-cube box formed from the shortest upper corner to the tallest upper corner. (l*w*(h_tallest - h_shortest))

3] Halve 2]

4] Add 1] and 3].

Presto!

 

[phinds]

Dammit Dave, you're red-shifting on me again

 

[DaveC426913]

Alternately, and even easier:

Make a second identical box, stack it on the first. Now you have a tall box of dimensions l*w*(h_shortest+h_tallest). Divide its volume by 2.

So, (again assuming the top surface is a plane) V = l*w*(h_shortest+h_tallest) / 2

 

[phinds]

:So, (again assuming the top surface is a plane) V = l*w*(h_shortest+h_tallest) / 2

I love it. This is the kind of thing I DO normally think of ... guess I'm a bit slow this morning.

 

[ThunderSkunk]

Wow, DaveC426913! That's brilliant! Thank you all very much! I am very surprised that such an elegant solution was lurking behind this problem. I wonder how long it would have taken me to solve this myself... For me this was certainly not trivial. This will be very helpful. Thank you again.


Also, I am very surprised by how many people on this site were willing to help me out on this. Thank all of you for taking the time to look this over for me.

As for Mark44, sorry about violating the site's rules. I'll give those rules a read and make sure I am not trending on anybody's toes from now on.

 

[Number Nine]

I think I have solution for the case where the points don't lie along a plane. Critique is welcome...

The approach is to take a series of thin slices from the figure and sum their areas. It is trivial to calculate the area of a slice since the heights of each side (call them p and q) clearly lie upon a line and so we can just take A = base*(p+q)/2. Summing the areas gives the volume of the figure...
V = \int_0^b bh(x)\,\mathrm{d}x
Where b is the length of the base and h(x) is a function giving the average height of each slice, and we sum through the entire figure (from 0 to b).

To derive the height function, let u1 and u2 denote the heights of the near and far corners of the left side of the cube; similarly, use v1 and v2 for the right. We can derive the heights of each side at any given slice by passing a straight line p through points u1,u2 and a line q through v1,v2...
p = \frac{u1-u2}{b} x + u1 \qquad q = \frac{v1-v2}{b} x + v1
We then have...
h(x) = \frac{\frac{u1-u2}{b} x + \frac{v1-v2}{b} x + (v1 + u1)}{2}
Finally giving us...
V = b\frac{1}{2}\int_0^b {\frac{u1-u2}{b} x + \frac{v1-v2}{b} x + (v1 + u1)}\,\mathrm{d}x

The approach seems right, but I may have made a stupid error somewhere along the way. Thoughts?

 

[phinds]

I've been away from math long enough that checking EXACTLY what you have would make my head hurt but it certainly sounds like exactly the right approach.

 

[256bits]

To derive the height function, let u1 and u2 denote the heights of the near and far corners of the left side of the cube; similarly, use v1 and v2 for the right. We can derive the heights of each side at any given slice by passing a straight line p through points u1,u2 and a line q through v1,v2...

Can we take that approach of a straight line , as if the surface is not a plane then there is a discontinuity from one corner to the opposite corner. There would have either a peak or a valley depending upon if the fourth corner is above or below the plane. I see in my minds eye a planar surface with a wedge added or missing.

I did not rigourously go through your derivation, but is the discontinuity taken into account?

 

[DaveC426913]

Can we take that approach of a straight line , as if the surface is not a plane then there is a discontinuity from one corner to the opposite corner. There would have either a peak or a valley depending upon if the fourth corner is above or below the plane. I see in my minds eye a planar surface with a wedge added or missing.

I did not rigourously go through your derivation, but is the discontinuity taken into acoount?

Yeah, it doesn't work if the top surface is not a plane. Each slice would be a five-sided shape, possibly convex, possibly concave.

Then again, if it is a plane, using calculus would be using a bulldozer to excavate ... a ... tea ... cup*.

*subtle reference. just watched it again the other night

 

[Number Nine]

Now that you mention it, it does seem obvious that there would be some sort of discontinuity on the top surface. It seems the integration approach is right out, then.

I'm starting to like this problem; I'm going to start trying out some properties of curved surfaces. Surely we can simply fit a surface to the 4 points?

 

[DaveC426913]

Now that you mention it, it does seem obvious that there would be some sort of discontinuity on the top surface. It seems the integration approach is right out, then.

I'm starting to like this problem; I'm going to start trying out some properties of curved surfaces. Surely we can simply fit a surface to the 4 points?

It's not curved; it's simply folded along a diagonal. The top surface is still planes, there's just two of them.

 

[phinds]

Here's an example. The one on the right is just the one on the left with an additional slice taken out of it and butted up against the first one.

I just realized that at this angle, it doesn't look as though all 4 heights are different, but they are.

 

[DaveC426913]

And I just realized that, if the top does not form a plane, we don't have enough information to answer this question. Which means all proposed answers in this thread for the case of a non-plane top are wrong). (They make an assumption we can't make.)

For any given configuration of 4 unequal heights, there are two ways it could be divided, resulting in two different volumes. We don't know which it is.

(It was your image that twigged me. You folded it transversely; I had always mentally been folding it longitudinally.)

 

[phinds]

For any given configuration of 4 unequal heights, there are two ways it could be divided, resulting in two different volumes. We don't know which it is.

Clever catch Dave

 

[vrmuth]

look at a simple degenerative case where all for upper points are at the same height. Now you increase one of the points that you have with a factor of 2 and do the computation. OR you increase one of the points for which you do NOT have a factor of 2, and by the same amount, then do the computation. Clearly you'll get two different answers for the identical figure.
Yep, my intuition was right. Your formula can't work.

sorry it's actually \frac{1}{6}a^{2}(h1+2h2+h3+2h4) where 'a' is the side of the base which is a square and i meant that the points h2 and h4 makes the edge on the top surface , i got this formula two days ago

 

[logics]

It's not curved; it's simply folded along a diagonal. The top surface is still planes, there's just two of them.

hi Dave, couldn't top be just any shape, how do we figure out?

 

[vrmuth]

And I just realized that, if the top does not form a plane, we don't have enough information to answer this question. Which means all proposed answers in this thread for the case of a non-plane top are wrong). (They make an assumption we can't make.)

For any given configuration of 4 unequal heights, there are two ways it could be divided, resulting in two different volumes. We don't know which it is.

Particular cases of my formula:
1) if the four points in the top are in a plane , they satisfy the following
a) shortest and the tallest are always on the opposite corners

b) h1+h3 = h2+h4 ( any proof ?)
(here either one pair will be the shortest and tallest )
so suppose h1 and h3 are the shortest and the tallest then
\frac{1}{6}a^{2}(h1+2h2+h3+2h4) = \frac{1}{6}a^{2}(h1+h3+2(h2+h4))
replacing (h2+h4) by ( h1+h3) we have
\frac{1}{2}a^{2}(h1+h3) which is your formula
2) when all the heights are equal to h , then its \frac{1}{6}a^{2}(6h) = ha^{2}
out of those two ways one is preferable , did you know which one it is ? :)

 

[vrmuth]

Then again, if it is a plane, using calculus would be using a bulldozer to excavate ... a ... tea ... cup*.

*subtle reference. just watched it again the other night

yea i used the BULLDOZER, let's see who uses "just the fingers"
then i will share the proof later

 

[phinds]

hi Dave, couldn't top be just any shape, how do we figure out?

You COULD construct a figure that would meet the side-height requirements and so forth and that had a bubble top, but the point is that the OP is CLEARLY talking about a top with a plane or planes.

 

[DaveC426913]

You COULD construct a figure that would meet the side-height requirements and so forth and that had a bubble top, but the point is that the OP is CLEARLY talking about a top with a plane or planes.

And we can assume we're looking for the minimum - or at least the simplest - surface. If we want to allow non-minimum surfaces, then the sky's the limit. The top surface could look like a statue of Goofy.

 

[logics]

I was referring to this:

. Each slice would be a five-sided shape, possibly convex, possibly concave.

we don't have enough information to answer this question. For any given configuration of 4 unequal heights, there are two ways it could be divided, resulting in two different volumes. We don't know which it is.

That's right, but the two solids are complementary, if you put (as you showed earlier) one on top of the other they fit. If you are considering two planes [not arbitrary convex/concave shape], the slice is plus or minus, but volume is the same.
does problem boil down to: calculate the volume of a triangular wedge.

 

[256bits]

I was referring to this:



That's right, but the two solids are complementary, if you put (as you showed earlier) one on top of the other they fit. If you are considering two planes [not arbitrary convex/concave shape], the slice is plus or minus, but volume is the same.
does problem boil down to: calculate the volume of a triangular wedge.

find the volume of the wedge and either subtract or add the wedge volume depending upon whether or not the 4th point is above or below the planar surface,
See post 17

Assuming corners of unequal heights and the top formed by the intersection of planes:
Case 1: The top formed by the sides intersected by a 6th plane - solved by Dave by symmetry, by Vmuth analytically, and Number Nine by calculus.

Case 2: top formed by the intersection of 2 planes. The line of intersection runs from one corner to the opposite, and forms a ridge or a valley.

Case 3: top formed by the intersection of 3 planes - The line of intersection in Case 2 is cut somewhere along its length.

Case 4: top formed from the intersection of 4 planes. The line of intersection from Case 2 is cut in at 2 points.

That's it for planes having on their surface at least 2 points from the corners.

 

[256bits]

sorry it's actually \frac{1}{6}a^{2}(h1+2h2+h3+2h4) where 'a' is the side of the base which is a square and i meant that the points h2 and h4 makes the edge on the top surface , i got this formula two days ago

Your formula was correct, except you failed to mention which point h1 ... h4 were at which corner.

 

[256bits]

Particular cases of my formula:
1) if the four points in the top are in a plane , they satisfy the following
a) shortest and the tallest are always on the opposite corners

b) h1+h3 = h2+h4 ( any proof ?)

Proof? Since the top surface is a plane, it intersects 2 parrallel side planes forming equal sized triangles, with one (bottom) side of the triangle parrallel to the base. need I go on...

 

[vrmuth]

Proof? Since the top surface is a plane, it intersects 2 parrallel side planes forming equal sized triangles, with one (bottom) side of the triangle parrallel to the base. need I go on...

yes , could you elaborate ?

 

[vrmuth]

Case 3: top formed by the intersection of 3 planes - The line of intersection in Case 2 is cut somewhere along its length.

Case 4: top formed from the intersection of 4 planes. The line of intersection from Case 2 is cut in at 2 points.

That's it for planes having on their surface at least 2 points from the corners.

There are only 4 vertices on the top of this shape , intersection of 3 planes will give rise to 5th point which in turn will give 5th height so that is a different figure , i think that's the next problem to be discussed in this thread, i like to try to find that volume also

 

[logics]

* a) shortest and the tallest are always on the opposite corners... b) h1+h3 = h2+h4 ( any proof ?)

reductio ad absurdum [\rightarrow\leftarrow] and symmetry give you an instant proof by contradiction. I do not know if there is another proof, but surely it would be complex:
Axis of rotation can be: space diagonal or face diagonal.
A plane rotating on a space diagonal [h₁,h₃] divides a cuboid (V=a²h₃) into two symmetrical, complementary solids [h₂ = h₃-h₄]: V₁= a²h₃/2 (case 1, h: 0,2,8,6 + 8,6,0,2)
Two such planes delimit a (V_w= 1/6...) wedge: V₂ = V₁ \pm V_w (case 2, h: 0,2,8,5 + 8,6,0,3, h_w =1)
If we are to exclude concave, convex or goofy surfaces, other cases are determined by: additional plane on axis [h₂, h₄], planes rotating on face diagonal axis etc.
The problem says: given the "surface areas" of all sides: we can deduce the values of, but we need to know the order of h _1,2,3,4 and the shape of the top, else if * we can only solve the trivial case 1, brilliantly solved by Dave: V = a²h₁₊₃/ 2

Edit: V = a²h_(0-1)+3/2

 

[vrmuth]

If we are to exclude concave, convex or goofy surfaces,other cases are determined by:additional plane on axis [h₂, h₄], planes rotating on face diagonal axis etc.

I don't understand what you are saying here , and i don't know what's face diaonal and how a plane rotating about it. explanation pls ...

 

[logics]

I don't understand what you are saying here , and i don't know what's face diaonal and how a plane rotating about it. explanation pls ...

you're welcome, vrmuth: the best explanation is a picture http://wikipedia.org/wiki/Space_diagonal"

 

[DaveC426913]

reductio ad absurdum [\rightarrow\leftarrow] and symmetry give you an instant proof by contradiction. I do not know if there is another proof, but surely it would be complex:
Axis of rotation can be: space diagonal or face diagonal.
A plane rotating on a space diagonal [h₁,h₃] divides a cuboid (V=a²h₃) into two symmetrical, complementary solids [h₂ = h₃-h₄]: V₁= a²h₃/2 (case 1, h: 0,2,8,6 + 8,6,0,2)
Two such planes delimit a (V_w= 1/6...) wedge: V₂ = V₁ \pm V_w (case 2, h: 0,2,8,5 + 8,6,0,3, h_w =1)
If we are to exclude concave, convex or goofy surfaces, other cases are determined by: additional plane on axis [h₂, h₄], planes rotating on face diagonal axis etc.
The problem says: given the "surface areas" of all sides: we can deduce the values of, but we need to know the order of h _1,2,3,4 and the shape of the top, else if * we can only solve the trivial case 1, brilliantly solved by Dave: V = a²h₁₊₃/ 2

I do not know if your "logics" incorporates this but as you can see by my earlier analysis, the diagonal's endpoints are not actually known. They could be from h₁ to h₃ or from h₂ to h₄ - even on the same object.

i.e. any solution is going to have to spit out two equally possible answers, from which the OP will have to choose which matches his setup.

See diagram in post 22.

 

[logics]

... given the surface area of all sides, ... The sides of this object would be four trapezoids of which two, and only two, heights each side would be equal (where the two trapezoids meet to create a corner

I do not know if your "logics" incorporates this ...i.e. any solution is going to have to spit out two equally possible answers, from which the OP will have to choose which matches his setup..

if ...

OP meticolously formulated the problem and gives the impression he would give Areas in an ordered sequence. If gives him the benefit of doubt, if we get an ordered sequence A_2,3,4,1 we can solve any rotation on space diagonal. Now that you know it is so, your logics , will easily lead you to the solution. I imagine you want the satisfaction of finding it yourself.

 

[vrmuth]

Two such planes delimit a (V_w= 1/6...) wedge: V₂ = V₁ \pm V_w (case 2, h: 0,2,8,5 + 8,6,0,3, h_w =1)

Calculating the volume of the wedge separately may seem to be a good logic but it will complicate the problem because even if you deduce all h and their order , the order of the heights won't decide whether is there a wedge or a valley on the top surface

 

[DaveC426913]

OP meticolously formulated the problem and gives the impression he would give Areas in an ordered sequence. If gives him the benefit of doubt, if we get an ordered sequence A_2,3,4,1 we can solve any rotation on space diagonal. Now that you know it is so, your logics , will easily lead you to the solution. I imagine you want the satisfaction of finding it yourself.

No. You are not understanding the import of my assertion. You do not have a unique solution yet.

Even with all corners precisely defined in an ordered sequence, and making no change to them of any kind, it is still possible to construct the box two ways after the fact. you have two choices for the space diagonal and you cannot specify them. Only the OP can.

You could physically build the entire structure, floor and four walls, including precisely the heights of all four corners, made out of steel bars - yet you still do not have enough information to determine the shape of the top, and thus determine the volume.

Please see attached greatly clarified diagram. This should make it obvious.

 

[phinds]

Even with all corners precisely defined in an ordered sequence, and making no change to them of any kind, it is still possible to construct the box two ways after the fact.

Yep, With the single exception of the case where all 4 upper points are on the same plane.

EDIT: Hm ... not even sure why I posted that Dave, since I KNOW you already know that. That's one of the problems of being a speed typist with no mental bladder control.

 

[DaveC426913]

Yep, With the single exception of the case where all 4 upper points are on the same plane.

EDIT: Hm ... not even sure why I posted that Dave, since I KNOW you already know that. That's one of the problems of being a speed typist with no mental bladder control.

Right. OP specified that no two heights are the same.

 

[vrmuth]

The problem says: given the "surface areas" of all sides: we can deduce the values of, but we need to know the order of h _1,2,3,4 and the shape of the top, else if * we can only solve the trivial case 1

the diagonal's endpoints are not actually known. They could be from h₁ to h₃ or from h₂ to h₄ - even on the same object.

i.e. any solution is going to have to spit out two equally possible answers, from which the OP will have to choose which matches his setup.

here is the proof
assume all the h are known , and the base be a rectangle with dimension a and b , divide the figure into two volumes by a vertical plane along the diagonal in the top (say h2h4)
lets take anyone (say with h1,h2 and h4 ) let x,y axes be along two sides that make right angle in the base and z be along h1 take a series of thin vertical slices parallel to x-axis with thickness dy , its face is a trapezium , ( see the picture for its dimensions )
this volume is given by \frac{a(b-y)}{2b} \left[ \frac{(2h4-h1-h2)y}{b}+ (h1+h2) \right] dy integrating this from 0 to b gives V_{1}=\frac{ab(h1+h2+h4)}{6} since the other volume is a similar shape with heights h2 , h3 and h4 its volume is given by V_{2}=\frac{ab(h2+h3+h4)}{6} hence the total volume V=V1+V2=\frac{ab(h1+2h2+h3+2h4)}{6} where the heights h2 and h4 are the end points of the diagonal in the top surface , the choice of heights take care of whether is there wedge or valley

 

[phinds]

Right. OP specified that no two heights are the same.

Yeah, but that doesn't keep them from being in the same plane, as you pointed out in post #10 when you solved the original problem.

 

[DaveC426913]

Yeah, but that doesn't keep them from being in the same plane

Apologies. Correct.

 

[DaveC426913]

here is the proof
assume all the h are known , and the base be a rectangle with dimension a and b , divide the figure into two volumes by a vertical plane along the diagonal in the top (say h2h4)
lets take anyone (say with h1,h2 and h4 ) let x,y axes be along two sides that make right angle in the base and z be along h1 take a series of thin vertical slices parallel to x-axis with thickness dy , its face is a trapezium , ( see the picture for its dimensions )
this volume is given by \frac{a(b-y)}{2b} \left[ \frac{(2h4-h1-h2)y}{b}+ (h1+h2) \right] dy integrating this from 0 to b gives V_{1}=\frac{ab(h1+h2+h4)}{6} since the other volume is a similar shape with heights h2 , h3 and h4 its volume is given by V_{2}=\frac{ab(h2+h3+h4)}{6} hence the total volume V=V1+V2=\frac{ab(h1+2h2+h3+2h4)}{6} where the heights h2 and h4 are the end points of the diagonal in the top surface , the choice of heights take care of whether is there wedge or valley

Yes, there is a solution, provided the OP makes one more decision - whether the roof is concave or convex (i.e. where he chooses to build the diagonal).

Any proofs not given this piece of information will have to spit out two answers.

 

[vrmuth]

[/itex]

Yes, there is a solution, provided the OP makes one more decision - whether the roof is concave or convex (i.e. where he chooses to build the diagonal).

Any proofs not given this piece of information will have to spit out two answers.

yes you are right, the orientation of the diagonal decides whether the top is concave or convex. But since its meaningless to talk about the volume and its formula when the shape of one surface is not specified , the diagonal should be known ,
then if the heights h_{2,4} correspond to the diagonal , where is the another answer ?

 

[phinds]

[/itex]

yes you are right, the orientation of the diagonal decides whether the top is concave or convex. But since its meaningless to talk about the volume and its formula when the shape of one surface is not specified , the diagonal should be known ,
then if the heights h_{2,4} correspond to the diagonal , where is the another answer ?

What are you on about? IF the diagonal is known then there is NOT "another" solution. The statement that there has to be two solutions is based on not knowing which diagonal is being used so you have to solve for both.