I posted this earlier and received a reply - but the reply doesn't make sense. I drew up a picture of what was posted, but the figure comes out to finding the hypotenuse of a triangle that is 16x14 - not 16x10. What am I not seeing?
Original problem:
There is a room 10 feet high by 10 feet wide by 10 feet in length. On one wall, there is an electrical outlet 1 foot from the floor and centered five feet from each side of the wall. On the opposite wall, one foot from the ceiling and centered five feet from each side of the wall, is an electrical box. How would you run a wire from the box to the outlet, attached to either the floor, ceiling or walls, that is less than 20 feet in length?
I have tried making a box to represent the room but can not figure out how the wire, running from box to outlet, could be less than 20 feet no matter how I run it.
Steve's answer:
Math -word problem - Steve, Sunday, November 9, 2014 at 1:04pm
Consider the room. The front wall is square ABCD, and the back wall is EFGH.
The front outlet (P) is a foot from the floor (AB), and the back outlet (Q) is a foot from the ceiling (GH).
A line directly up and across and down is 20 feet. But, you can do the following: Unfold the room and draw a new diagram, with
square ABCD
on its right is the wall BCFG
above that is the ceiling CGDH
to its right is the far wall, GHEF.
Now draw a line PQ. Its length is
√(16^2+10^2) = 18.86
My answer back:
Math -word problem - Mary, Sunday, November 9, 2014 at 2:47pm
I made your configuration with sheets of paper 10" x 10". I got the 16" leg, but I don't get the 10 foot height. I have the walls/ceiling in position similar to a backwards z. But from one foot over to the right from F, if I go straight up to form one of the legs, I get, up to G (one foot to the right), 9', plus 5' more to reach the box at Q, as it would have to travel half of the width of the wall. 9' + 5' =14', and the hypotenuse where the wire would lay would be slightly more than 21'. What am I looking at wrongly?
Mary, your problem is a variation of the famous
"spider and fly" problem.
Here is a webpage that has the same concept, you have an electrical outlet and box instead of spider and fly.
http://mathworld.wolfram.com/SpiderandFlyProblem.html
Use Steve's instructions and try to match them with the diagrams on the webpage.
Of course , also change the numbers.
Hi Reiny - I looked at the picture and redrew my walls, but it still doesn't come out for walls that are 10' in each direction. I noticed that the ceiling and floor in the picture had a longer dimension than the side heights, so I took ceiling and floor and extended them two feet at a time and recalculated the wire each time. At 20' in length, the wire that is stretched as in the picture finally becomes longer than the straight wire up and over the ceiling and back down. At this point, the legs of the triangle are 22' and 20', and the hypoteneuse (wire length) = 29.7', .3' less than the 30' of stringing it straight overhead. What do you think? Thanks!!
He earned $22.50 working on Saturday. Working 8am-2pm how much did he earn?
To understand the solution provided by Steve, let's break it down step by step:
1. According to the problem, the front wall of the room is square ABCD, and the back wall is square EFGH. The room is 10 feet high, wide, and long.
2. Steve suggests unfolding the room and drawing a new diagram to get a better visualization. This means unfolding the square ABCD to form the wall BCFG, the ceiling CGDH, and the far wall GHEF.
3. In this new diagram, Steve draws a line from the front outlet (P) to the back outlet (Q), which represents the wire we need to run. The length of this line is what we are trying to find.
4. To find the length of the line PQ, Steve applies the Pythagorean theorem. The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
5. In this case, the line PQ is the hypotenuse, and the two legs of the triangle are 16 feet and 10 feet. So, using the Pythagorean theorem, we have:
PQ^2 = 16^2 + 10^2
PQ^2 = 256 + 100
PQ^2 = 356
6. Taking the square root of both sides, we get:
PQ = √(356)
PQ ≈ 18.86 feet
According to Steve's calculations, the length of the wire needed is approximately 18.86 feet, which is less than 20 feet as required by the problem.
Now, regarding your confusion, it seems you are considering the wrong measurements for the height of the room. The problem states that the room is 10 feet high, not 12 feet high as you are assuming. Double-checking the measurements should clear up the discrepancy.