rajathpanta wrote:

Hi Mike, Unable to visualize...

OK, it's a bit challenging to explain visualization via internet text, but here goes.

For this problem, it doesn't matter at all whether we are dealing with the skeleton frame or a solid box --- the math is exactly the same either way, so I am going to discuss a box, because that's easier to visualize.

Imagine we have a box sitting on a table --- say a carton around the size of a half-gallon of milk or orange juice, but flat at the top instead of peaked. This box has 12 edges. Any rectangular solid in the world has 12 edges. I strongly recommend you get a physical box of anything, as long as its fully rectangular, and trace your finger along the 12 edges to convince yourself of this fact. ----- An "edge" is the "corner line" where two flat faces meet --- any rectangular solid has six

faces, the flat surfaces where they print stuff, and twelve edges, the lines where two faces meet, and six vertices, the corners where three edges come together.

Now, back to our box sitting on the table. First, we are going to take account of the four edges that are touching the table, the four on the bottom, surround the square on the bottom. There are four lengths of x, for a grand total of 4x for the lengths on the bottom.

Attachment:

box top & bottom.JPG [ 27.01 KiB | Viewed 705 times ]
Now, consider the top of the box, the four edges that surround the square on the top, which is congruent to the bottom and parallel to it. This also has four lengths of x, so this also has a grand total length of 4x. If we add the lengths of the bottom and the lengths of the top together, that is 4x + 4x = 8x of wire needed to surround those two faces.

Now, we have to consider the four remaining edges. These edges have different lengths from the other eight. Unlike the bottom, which sits on the table, and the top, which is parallel to the table, these last four edges are perpendicular to the surface of the table. The make a right angle with the surface of the table and move vertically away from the plane of the table. These each have a length of the height --- the height is unknown, so I will just call it "h."

Attachment:

box sides.JPG [ 16.45 KiB | Viewed 706 times ]
There are four of these edges, each with length h, so that's a total length of 4h for these final four edges.

If we add all twelve edge lengths, we get 8x + 4h = 480, which, as I explained above, leads to h = 120 - 2x.

Does this make more sense now?

Mike

_________________

Mike McGarry

Magoosh Test Prep