Consider a cube of side 100cm. Consider two diaganolly opposite points (the main body diagnol) as A and B. The cube is kept flat on the ground, and is filled with water upto a 40cm height. Now, the cube is tilted such that AB is vertical. What's the height of the water now?

Question

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I figured that the water would form a triangular pyramid shape at the base of the titled cube, the volume of which would be (h*B/3), B being the area of the triangular base. The volume can be found to be 120000cm^3

I'm unable to figure out how to proceed then, as the volume equation has two unknown variables. Further, would the water rise to such a point where it would no longer form a triangular pyramid?

Answer 1

It's not really a pyramid, since it has no vertex. Its top is an edge of the original cube, rather than a point.

Now, looking at the triangular cross-section, it is an isosceles right triangle. If its altitude is h (the depth of the water), then its base is 2h, making its area h^2.

Thus, the volume is 100h^2

So, now we solve

100h^2 = 40*10^2 = 400000
h^2 = 4000
h = 20√10

Answer 2

sorry Steve, it IS a pyramid...with either A or B as the vertex

the base is an equilateral triangle consisting of three of the diagonals of faces of the cube

the top of the cube is an identical pyramid

the complexity is the section between the bases of the pyramids

since the areas of the ends of this complex section are equal (pyramid bases), it is reasonable to assume that the cross-sectional area is uniform...so the volume is proportional to the thickness (depth)

find the volume of one pyramid

the volume of the complex section is the cube volume, minus the two pyramids

calculate how much depth of the complex section must be added to a pyramid volume to achieve the 40% of the cube volume

add the depth to the height of the pyramid to find the height of the water

Answer 3

Good catch, Scott. I had not read the description carefully. Anyway, maybe you gave him something to work with, eh?