A room is 12' X 8' East/WEst walls. North/South Walls 10' X 8'. A spider is 5' up and 2' in from SE corner, a fly is 4' up and 4' in from NW corner. How far must the spider go to catch the fly? Need a math equation to solve.
2007-01-04 01:45:37 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
This should be fairly simple to calculate once you figure out their coordinates in 3D space. However I dont understand what you mean by "in". Do you mean along the wall? At a 45 degree angle to the walls?
After trying to remember my vector geometry I realized you were probably not talking about the distance between the spider and fly(10.19 ft assuming 45 degree angle to the wall) but rather the distance the spider will have to travel on the wall to get to the spider. So go with the answer below mine.
2007-01-04 01:54:39 · answer #1 · answered by E 5 · 0⤊ 0⤋
I'm assuming spider and fly on opposite 12' walls. Spider crawls down 5', crawls along the hypotinuse of a 6' x 10' x ?' triangle (a^2 + b^2 = c^2, 36 + 100 =c^2, c = 11.7') and then must crawl up 4'. A formula??? You gotta' start making up symbols for distances in from corners, etc., etc., etc. Good luck: This is easier to solve than formulate! - the spider has 20.7' feet to travel.
2007-01-04 02:02:27 · answer #2 · answered by Richard S 6 · 0⤊ 0⤋
Draw the box. Put the numbers on the box. Draw the path of the spider to the fly. Identify the triangles from your diagram. Use the Pythagorean Theorem to solve.
2007-01-04 01:49:04 · answer #3 · answered by Jerry P 6 · 1⤊ 0⤋