# Two objects, Three views, Identical shapes

## 10 posts in this topic

Posted · Report post

The front, side and top views of two objects give identical appearances.

Object A appears in each case to be a square of side a.

Object B appears in each case to be a circle of diameter a.

If they are made of the same material, can you say which object is heavier?

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Posted · Report post

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Posted · Report post

the density of the material is 1. Then weight=mass. Now can you say which object is heavier?
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Posted (edited) · Report post

What you make us to believe is a cube might be three "squares" of unknown thickness. (Appologies to the purists.)

What you make us to believe is a sphere might be three cylinders of unknown height.

Therefore, the volume cannot be calculated.

Besides, I leave the question open whether "made of the same material" implies the same density.

Edited by harey
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Posted · Report post

By same material I implied same density. Suppose d=1 so volume = weight.

Going a bit further just for fun, which object can weigh the most, and which can weight the least?

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Posted · Report post

The circle is inscribed to the square...

Going a bit further just for fun, which object can weigh the most, and which can weight the least?

That's easy. The circle is inscribed to the square.

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Posted · Report post

What if they were made of a thin film of small thickness d?
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Posted · Report post

The best way is to make them of equal volume ( suppose what I call a "3D square" is a prism?):

v_prism=a*a*d_prism

v_cylinder=pi*a/2*a/2*d_cylinder

=> d_cylinder=(4/pi)*d_prism

While it might work for small d, with growing d, it will became more and more perceptible that the cylinder should be a kind of barrel (or a pancake). The formulae get too complicated for my taste and my possibilities.

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Posted · Report post

I thought of the lightest "cube" as three squares joined at a corner, Volume = 3a2d.

The lightest "sphere" could be a "soap bubble'" Volume = pi a2d.

In that sense, the "cube" could be the lighter of the two.

Just a thought.

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Posted · Report post

Yes, I see. There must be a lot of solutions, like taking a sphere and a cube and drilling holes into the cube until it becames lighter than the sphere - no formula, no misscalculation.

My preferred solution is still that of the three squares and three pankaces in the planes xy, xz and yz.

BTW, the Volume=3a2d-d3 - I have the same oversight in my post n. 8.

I just wonder how often we make false deductions in the real life in the style "three projections=circle => it must be a sphere".

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