Page 1 of 2
Math Problem
Posted:
Sat Dec 20, 2008 9:44 pmby .Yunoko
Okay, so my brother ask me this math problem today, and he showed me an example and everything, and he told me that it is unsolvable. But wants to know more about it; Anyways this is what it sorta looks like.
And it gives you like the width and height of the sides or something and you have to find like the length of the inside box or something I am not sure. But anyways, does anybody know the name of this problem?
Posted:
Sat Dec 20, 2008 11:24 pmby crait
If he only gave you the length and width of the outside boxes, then yes, it would be impossible to answer because you wouldn't have enough information to solve it. If you were given a length of the inside box then you would be able to or if you were given the distance between the corners of the outside rectangle and corners of the inside rectangle, then yeah, you'd be able to. but like this, no.
Posted:
Sat Dec 20, 2008 11:34 pmby .Yunoko
Yea, I know that aha, but he is here anymore, so I will not know until I see him again, also even with the length and width I I believe it is still impossible but I don't know. Anyways Do you happen to know the name of this problem.
Posted:
Sun Dec 21, 2008 7:18 amby ace
Actually....I think we had a problem similar to this in calculus. But since it's Christmas break my brain is fried and I can't even remember how to find the hypotenuse using the degrees of the angles >.< But I think you would still need to kno the length of the little triangles on the bottom left and top right. So it isn't impossible, just not enough information is given.
So yea, I don't know what the name of the question is though :p But I mean it technically isn't impossible because you could leave information out of any problem and then it wouldn't be solvable.
Posted:
Mon Dec 22, 2008 2:33 amby DarkPacMan77
if the rectangle left "right triangles" in the corners where the rectangle is longer at and you knew the measurements of the outsides of the overall rectangle, you can get the hypotenuse of the longer sides of each triangle (which is the answer).
-DarkPacMan77-
Posted:
Mon Dec 22, 2008 8:28 amby staxx
it gives you the width and height of the outside box or the inside box? im assuming you mean the outside, it has to do with trigonometry as pacman was explaining. But yes it is solvable, just been at least 4 years since i did this kind of crap in college.
Posted:
Mon Dec 22, 2008 10:22 amby Puncharger
DarkPacMan77 wrote:if the rectangle left "right triangles" in the corners where the rectangle is longer at and you knew the measurements of the outsides of the overall rectangle, you can get the hypotenuse of the longer sides of each triangle (which is the answer).
-DarkPacMan77-
*Babble and Drool.
Posted:
Mon Dec 22, 2008 4:16 pmby DarkPacMan77
It's just the Pythagorean theorem. But you need to know the measurements of the sides and also how much space the "small triangles" take up against the outsides of the rectangles. If you knew that info, you could solve it. Without that info, you can't.
-DarkPacMan77-
Posted:
Tue Dec 23, 2008 11:09 amby ace
DarkPacMan77 wrote:It's just the Pythagorean theorem. But you need to know the measurements of the sides and also how much space the "small triangles" take up against the outsides of the rectangles. If you knew that info, you could solve it. Without that info, you can't.
-DarkPacMan77-
Just like I said, you need to kno the width and height of the "small" triangles. It's not impossible, theres just not enough information given :p
Posted:
Tue Dec 23, 2008 8:04 pmby crait
Which is what I said in my first post.
Posted:
Tue Dec 23, 2008 9:15 pmby light_alistor
yes yes your all very intelligent ::claps::
Posted:
Tue Dec 23, 2008 10:29 pmby .Yunoko
Okay so that's the math problem with the information and you have to find x, so can you explain in detail exactly how to figure this out.
Posted:
Wed Dec 24, 2008 1:40 pmby TheOriginalToxxy
simple the awnser is.......dont do the question
Posted:
Wed Dec 24, 2008 2:44 pmby zumiuran
The only way I would be able to solve it is if Y = W as shown in the picture below
but that doesnt look like the case so I have no idea how to solve it
Posted:
Wed Dec 24, 2008 5:14 pmby .Yunoko
zumiuran wrote:The only way I would be able to solve it is if Y = W as shown in the picture below
but that doesnt look like the case so I have no idea how to solve it
Well, I drew that drawing so its not the best but it should fit perfect.