Two cubes, each with integral side lengths, have a combined volume equal to the total of the lengths of their edges. How big are the cubes? [If you find a result by 'trial and error' you'll need to prove you have found all possible solutions.]
What fractions can you find between the square roots of 56 and 58?
Is the mean of the squares of two numbers greater than, or less
than, the square of their means?
It is definitely impossible to have a tetrahedron which has edge
lengths of 10, 20, 30, 40, 50 and 60 units.
This is because to produce a triangle two sides have to be
bigger (when added together) than the third. So this tetrahedron
won't work because the 10 is too small to make a triangle. e.g. 10,
10 will not go with any other two numbers in the above list.
However it is possible to make a tetrahedron which has lengths
10, 20, 25, 45, 50 and 60 units. The reason is that the numbers are
big and small enough to fit the rule of triangles (the sum of two
sides bigger than the third) Here is my net of this tetrahedron to
explain my answer (not to scale!)
So all four faces are fine, so it can make a tetrahedron.