A 12 by 25 by 36 cm cereal box is lying on the floor on one of its 25 by 36 cm faces. An ant, located at one of the bottom corners of the box, must crawl along the outside of the box to reach the opposite bottom corner. What is the length of the shortest such path?
Note: The ant can walk on any of the five faces of the box, except for the bottom face, which is flush in contact with the floor. It can crawl along any of the edges. It cannot crawl under the box.
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so if i read this correstly, the ant is on one corner and wants to get to the corner horizontal of it..... the fastest way isto walk on the crease of the box untill you get to the other corner. Simple and straight forward!lol
ReplyDeleteIris
Well, before measuring I thought that the fastest way to get from one corner to the opposite corner was to climb to the top corner, walk diagonally across the top, and climb back down. But, after measuring, I proved myself wrong. I measured (using my model) the diagonal across and between the top corners, which was 44 cm, and added the height of the box, 12 cm, and then the other side of the box, another 12 cm, to get 68 cm. But, it was quite a high box (higher than any other cereal box i've heard of), so, I thought it might be faster to not climb up, but to go around the bottom of the box. I added 25+36=61 (The sides of the box), which is lower than 68, so it is faster to not climb up and to go around the bottom instead.
ReplyDeletewhat i did was the ant climbed up(12cm),then across(36cm), then forward(25cm) which brings you to the other corner. the shortest path is 36+25+12=73cm. i believe that it was the fastest way to go.
ReplyDelete~Adele~
I think that the shortest path that the ant can crawl is, first, going to the opposite top corner, then the opposite bottom corner along the 2 diagonals. To figure out the distance, I used Pythagoras' theorem which states that the length of the hypotenuse in a right angle triangle is the square route of the base squared + the height squared. Therefore the equation for the shortest path (the sum of the 2 diagonals)(p), p = the square route(12 (height) squared + 36 (length) squared) plus the square route of (12 (height)squared + 25 (width) squared) = 37.95 + 27.73 cm (square routed already)= 65.68 cm.
ReplyDeleteI agree with DAVIN to go around the box. For 25cm + 36cm would equal 61 cm and the ant would reach there the fastest.
ReplyDeleteThe simplest way would be for the ant to walk the 25cm side, then walk the 36cm side and reach the corner. 36+25=61. That is the shortest way. The other way would be to walk the height of the box (12cm), walk the 25cm side, then walk the 36cm side, and the walk the height down (12). That would be 12cm+12cm+25cm+36cm=85cm. That is considerably shorter than 61. I might have understood the question wrong because it just seems to straightforward. But that is my final anwswer.
ReplyDeleteI agree with Iris, the fastest way to go is horizontal, so, I made the bottom of my box 25 cm, so the ant just walks the 25cm.
ReplyDelete