# Did you know that a right triangle can be found using only one side?

There is nothing new to your method. All you do is to choose the length of the hypotenuse (here: 5) plus one additional length in your triangle (here: 1.8). You use Pythagoras in an implicit way, as pwsnafu showed. You find three values c,a,b which satisfy c^2=a^2+b^2 by construction.

In the same way, you could directly choose another side length:
5 and 2 give sqrt(5^2-2^2)=sqrt(21) as third side.
5 and 3 give sqrt(5^2-3^2)=sqrt(16)=4 as third side.
...
If there is nothing new about my method then why do you not see "c^2 = cx + cy" in the text books? Or anything pertaining to c = x + y? Exactly! Because it is new. 5 * (1.8+3.2) = 5^2. So the sqrt of 5x = one side of the triangle. This is completely new.

pwsnafu
Pythagoras can't come close to an offer like that.
Except in post #19, dx demonstrated how to use Pythagoras to do it.

Edit: I was wondering where I had seen dx's drawing when he first posted, but now I remember. It was a question on my year 9 maths mid-year exam. Strange how you remember these things.

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mfb
Mentor
If there is nothing new about my method then why do you not see "c^2 = cx + cy" in the text books?
It is a more complicated way to write the usual equations.

You can replace "5" with "545349-545344". Nobody does, as it does not add anything new.

With Pythagoras method you have no choice but to consider all the right triangles from a given hypotenuse.
You don't have to, but you have the freedom to do so.
Your "method" is just one possible way to pick some specific triangle from the set of all triangles (with that hypotenuse). You can do the same by choosing a, or h, or anything else in the triangle.

Except in post #19, dx demonstrated how to use Pythagoras to do it.

Edit: I was wondering where I had seen dx's drawing when he first posted, but now I remember. It was a question on my year 9 maths mid-year exam. Strange how you remember these things.
You can use Pythagoras for double checking. example sqrt 5x = 3. sqrt 5y = 4.

c^2 = a^2 + b^2
5^2 - 3^2 = b^2
16 = b^2
4 = b so yes Pythagoras confirms that the triangle with sides 3, 4, and 5 make a right triangle but what are x and y?

sqrt 5x = 3
5x = 9
x = 9/5
x = 1.8

sqrt 5y = 4
5y = 16
y = 16/5
y= 3.2 simply algebra finds x and y not Pythagoras theorem.

c = x + y
c = 1.8 + 3.2
c = 5 its square! without Pythagoras.

And for the record my circles formed from my rational pi equation inscribe within my squares perfectly.

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pwsnafu
You can use Pythagoras for double checking.
Your expression is equivalent to Pythagoras. You even said so yourself. So really, you're just doing the same thing twice.

but what are x and y?
Why should we care? simply algebra finds x and y not Pythagoras theorem.
Again, your formula is equivalent to Pythagoras, so you are using it indirectly.

And for the record my circles formed from my rational pi equation inscribe within my squares perfectly.
I have no idea what you mean. Please use clear English.

Edit: Anyway, to summarise
Did you know that a right triangle can be found using only one side?
If we are given the length of one side and the length of a line segment from that side, we can determine a unique right angle triangle. Yes we knew that. It is basic geometry and is equivalent to Pythagoras.

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It is a more complicated way to write the usual equations.

You can replace "5" with "545349-545344". Nobody does, as it does not add anything new.

You don't have to, but you have the freedom to do so.
Your "method" is just one possible way to pick some specific triangle from the set of all triangles (with that hypotenuse). You can do the same by choosing a, or h, or anything else in the triangle.
A line makes up one side of the triangle what are the other two sides? boo who I don't have enough information?

okay x = 3 and it is 1/5th of the line what are the other two sides? boo who I don't have enough information?

but yes you do!
c = 5x
c = 5 * 3
c = 15
y = 15 - 3
y = 12
c = x + y
15 = 3 + 12
c^2 = cx + cy
side a = sqrt cx = sqrt 45
side b = sqrt cy = sqrt 180
side c = 15

double checking with Pythagoras:
c^2 = a^2 + b^2
225 = 45 + 180 yes the triangle that I just designed is a right triangle using my own methods. But using my methods at least we now know what the right triangle looks like.

What is your answer to the question? Personally I feel that you still don't have enough information if all you are limited to Pythagoras theorem or the law of sine's!

I don't get this. If you are given one side, then there are infinitely many right triangles which have that as a side. So why would you expect to get only one solution? Why would we care about a formula that gives only one solution?

What you take for granted is that you already have a right triangle to begin with for your exercises.

What you cannot do is calculate a triangle from a single side. Thank-you for allowing me to show you how I can do that using my own methods at that.

Borek
Mentor
What you cannot do is calculate a triangle from a single side. Thank-you for allowing me to show you how I can do that using my own methods at that.
OK, so if you are right, you will have no problem solving my problem.

I have a triangle with hypotenuse length equal to 10. Give me the other two sides of the triangle I have on mind.

I don't get this. If you are given one side, then there are infinitely many right triangles which have that as a side. So why would you expect to get only one solution? Why would we care about a formula that gives only one solution?
Those are two good questions to ask of yourself from for the answer.

OK, so if you are right, you will have no problem solving my problem.

I have a triangle with hypotenuse length equal to 10. Give me the other two sides of the triangle I have on mind.
Frankly how do you know that your other two sides are correct? Show me your design work since you have admitted having a solution on your mind. Did you use any of my methods or not?.

Frankly how do you know that your other two sides are correct? Show me your design work since you have admitted having a solution on your mind. Did you use any of my methods or not?.
Let me ask you this instead. At what point do you want your hypotenuse separated into x and y?

I cannot believe people are giving you so much greif for this Crazy Horse. As what you found (while not a new rule) is a simple beautiful association.

The mathematical law is of course priceless (although already discovred) but great credit to you for rediscovering.

Here I shall explain. If you are given a hypotenus, and you "assume that you have a right angle" (this part is key) these are all the clues you need for a variant of the proof tequnique known as (Side Angle Side/ Side Angle Angle). Its obvious that for a given hypotenus there can only be one right side angle with respect to the hypotenus.

Briefely put the tecnique invovles alwasy constructing a trialnge of of a line (assuming its the hypotenus) with an angle of 90 degrees. There is only one way this can possibly be done which is why crazy horse alwasy gets the right results.

I would consider this more a tecnique (an extremely usefull and overlooked); but it really isn't a new way or another proof etc; just a very intelligent and well posed method by carefully using the asumptions you can give a problem (the 90 degree ange) and hypotenus length to incredibly simplify a problem; basically what heavyside did to maxwell.

Great Job Crazy Horse if you have any other great ideas let me know I always give good open minded scientific feedback :D

mfb
Mentor
okay x = 3 and it is 1/5th of the line what are the other two sides? boo who I don't have enough information?

but yes you do!
Yes of course you can calculate it, and that is nothing new.
c=15
a=sqrt(cx)=sqrt(45)
b=sqrt(c(c-x))=sqrt(180)

Personally I feel that you still don't have enough information if all you are limited to Pythagoras theorem or the law of sine's!
You can apply Pythagoras twice, or use a formula which is the result of that application. It is called "Kathetensatz" (literally: "theorem of the cathetus") in German, I don't know how it is called in English.

Fredrik
Staff Emeritus
Gold Member
Closed for moderation.

Borek
Mentor
Frankly how do you know that your other two sides are correct?
Once we finish I will gladly give you my result and you will be able to check by yourself, using Pythagorean theorem, that they are correct.

Show me your design work since you have admitted having a solution on your mind. Did you use any of my methods or not?.
I am not going to show them - you said you can calculate just by knowing hypotenuse, it is time that you prove you can do it.

If you need more information, it means you can't solve the problem.