Sum of all 12 edges of a brick is 60inch and surface area is 144sq. inch?

Question

What is the length of spacial diagonal of the brick?

Answer 1

9 in.
We only have one unknown. It is not necessary to know L, W and D, only the root of the sum of their squares.
L + W + D = 15
2(LW + WD + LD) = 144
(L + W + D)^2 = L^2 + W^2 + D^2 + 2(LW + WD + LD)
225 = (diagonal)^2 + 144
(diagonal)^2 = 81
diagonal = 9

Answer 2

Wow...I think I might like this one. Let's start with some equations, using x,y,z as the length of the three different types of sides:
4(x+y+z)=60
2(xy+xz+yz)=144
These easily simplify to
x+y+z=15
xy+xz+yz=72.

But this is two equations with three unknowns! Now, we don't necessarily need to solve this, so perhaps it's possible to find what we need--sqrt(x^2 + y^2 + z^2). However, I can't seem to find a way to even retrieve this bit of information, at least not easily.

Perhaps this is more of a guess-and-check. Using only integers, I wrote a quick calculator program to run through the possibilites that match the first equation, then to output those combinations that also fulfilled the second equation. I got 3,6,6 and 4,4,7. The corresponding spacial diagonals have length 9 in either case.

I'm rather disappointed that I couldn't think of a better way around this problem, but I would believe that 9 will be your answer for any brick matching the requirements...I'll look into it further, unless someone else gets to the answer before I do.

EDIT: Ah, kirchwey gets it. Very nicely done. I was trying to manipulate the two equations to get the diagonal formula and hadn't realized that adding the two gives you a perfect square. Bravo.

Answer 3

The brick has L, W, and D. There are 4 of each length of edges, so the equation would be:

4xL + 4xW + 4xD = 60in

The surface area is equal to the sum of all faces, but since there are two of each of the opposite faces the formula simplifies to:

2x(LxW + WxD + DxL) = 144in^2

So now you have two equations and three unknown variables (L, W, D). So this problem cannot be solved without additional information.

Answer 4

There is more than one answer, depending on what the dimensions of the brick are.

Answer 5

Ain't no way but the Kirchwey!