try finding the slope of the diaganols
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try finding the slope of the diaganols
Here's a diagram I drew.
A (0, b)----------------C (a+c, b)
B (-c, 0) )----------------D (a, 0)
4 different points on the graph. Point B is in quadrant II. The other points are in quardant I.
I'm trying to find the slopes of AD and BC. If the slopes are negative reciprocal, then they must be perpendicular. Is there another way to prove that the diagonals are perpendicular without using their slopes?
finding the slopes is the easiest way i know of... btw what grade are u in?
you don't because it's not nessecarily true. only rhombusses, a subset of paralellograms where all four sides are the same length, is this nessecarily true for.Quote:
Originally posted by NavyBlue
How do you prove the diagonals of a parallelogram are perpendicular to each other, meaning that the diagonals' slopes are negative reciprocal of each other.
I don't really see getting help as cheating... it's the same as answering questions on the C++ board about programming.
Try #math on DALnet (irc) for your math problems, they're avid about not doing peoples homework for them (kinda like us)... but they're friendly and willing to help if you want help and not just an answer.
I don't think you should answer questions about things that are graded assignments either. Unless teacher doesn't care of course but...Quote:
I don't really see getting help as cheating... it's the same as answering questions on the C++ board about programming.
I don't see how it's not cheating...
Every honor code I've ever been under has been under has been to the effect of:
Now if asking for help on a program without asking for help, or something like this isn't "unauthorized aide" I don't know what is.Quote:
I have not (nor will I) given or recieved unauthorized aide on this assignment
I think the extra credit problem was a trick question. He wants to prove that the diagonals of a parallelogram are always perpendicular. So in this case you show proof by contradiction with one case. If you wanted to prove his statement you would have to prove that all paralelograms, even the "special kind" have this property.Quote:
Bah you win Cshot. But when proving a therom like this you shouldn't use a "special kind". You should use the most generic you have.
Can't think right now but the only 2 parallelograms that have that property is the square and rhombus. Hope I'm right.
> is the square and rhombus.
And since a square is a rhombus, that only leaves one, doesn't it?
The diagonals of a parallelogram are only perpendicular if it's a rhombus, Navy
Good point cheez. Been so long since I've taken geometry...(shuts out that part of memory)