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NavyBlue
10-03-2002, 02:00 PM
Hey, guys, I need a little help with this extra credit problem...

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.

And prove the diagonals bisect each other, having the same midpoint?

kermi3
10-03-2002, 02:04 PM
Well...you shouldn't be getting help on a graded assignment to start. But I would look at what happens when you drop an altitude.

NavyBlue
10-03-2002, 02:14 PM
There are 4 coordinates: A = (0,b), B = (-c, 0), C = (a+c,b) D = (a,0)

I tried to find the slopes of diagonal lines AB and BC, but they're not negative reciprocal to each other.

Cshot
10-03-2002, 02:56 PM
That's because AB and BC are not diagonals. AD and BC are. Heheh draw it out man ;)

kermi3
10-03-2002, 02:59 PM
Drop an altitude between the parrallel sides at a 90 degree angle. Do it on your own from there.

NavyBlue
10-03-2002, 03:21 PM
Sorry, guys. I mistyped, not lines AB and BC, but diagonal lines AD and BC. The problem is that I can't really find the slope of BC.

slope of AD = (0 - b) / (a - 0)
= -b/a

slope of BC = (b - 0) / ((a+c) - -c)
= b / a + 2c ?

Aren't the two slopes supposed to be negative recripocal of each others?

kermi3
10-03-2002, 03:27 PM
Look, I'm not going to help you at all anymore....But here's a suggestion...

Do your own work. It's graded, for credit? Don't you think it's cheating to be asking for help?

Fountain
10-03-2002, 03:49 PM
well, it could show initiative

kermi3
10-03-2002, 04:15 PM
You're right it could. However, I think that most teachers would see it as cheating. If it's graded you should do your own work and be proud of it....My opinion...I'm not closing the thread or anything, everyone must make their own choices.

Cshot
10-03-2002, 05:09 PM
If you can prove they are perpendicular, please let me know :D

Cshot
10-03-2002, 05:11 PM
hint hint

kermi3
10-03-2002, 05:29 PM
They aren't parrallel there...he means a parallelogram:

MethodMan
10-03-2002, 05:40 PM
Dont know if this will help but:

From the red lines, if you know they are parallel, then the angles are the same according to the Z pattern.

Cshot
10-03-2002, 05:44 PM
Yes it is. A rectangle is a special kind of parallelogram.


Parallelogram

A quadrilateral whose opposite sides are parallel
The opposite sides of a parallelogram are parallel.
A diagonal of a parallelogram divides it into two congruent triangles.
The opposite sides of a parallelogram are congruent.
The opposite angles of a parallelogram are congruent.
The consecutive angels of a parallelogram are supplementary.
The diagonals of a parallelogram bisect each other.
Five ways to prove a quadrilateral is a Parallelogram
1. A quadrilateral is a parallelogram if its opposite sides are parallel.
2. A quadrilateral is a parallelogram if its opposite sides are congruent.
3. A quadrilateral is a parallelogram if two sides are congruent and parallel.
4. A quadrilateral is a parallelogram if its opposite angles are congruent.
5. A quadrilateral is a parallelogram if its diagonals bisect each other.

Rectangle
An equiangular parallelogram
Rhombus
An equilateral parallelogram
Square
An equilateral and equiangular parallelogram

kermi3
10-03-2002, 06:23 PM
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.

red_baron
10-03-2002, 06:24 PM
try finding the slope of the diaganols

NavyBlue
10-03-2002, 07:14 PM
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?

red_baron
10-03-2002, 07:33 PM
finding the slopes is the easiest way i know of... btw what grade are u in?

moi
10-03-2002, 07:37 PM
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.


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.

Eibro
10-03-2002, 08:16 PM
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.

kermi3
10-03-2002, 08:23 PM
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 think you should answer questions about things that are graded assignments either. Unless teacher doesn't care of course but...

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:


I have not (nor will I) given or recieved unauthorized aide on this assignment

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.

Cshot
10-03-2002, 08:58 PM
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.
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.

Can't think right now but the only 2 parallelograms that have that property is the square and rhombus. Hope I'm right.

Govtcheez
10-04-2002, 06:10 AM
> 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

Cshot
10-04-2002, 10:30 AM
Good point cheez. Been so long since I've taken geometry...(shuts out that part of memory)