Thread: Who is familiar with Pythagoras?

WedWon wrote,..

My Turn My Turn !!!!

So if

a<squared> + b<squared> = c<squared>
and
A(area) = ab/2
then 2A = ab
and 4A =2ab

(Remember these things)

P(perimeter) = a + b + c
P - c = a + b
(P - c)<squared> = (a + b)<squared>
(p - c)<squared> = a<squared> = b<squared> +2ab

then bringing in those things we remembered above:

(P - c)<squared> = c<squared> + 4A
P<squared> - c<squared> -2Pc = c<squared> + 4A
cancel out the c<squared>s

P<squared> -2Pc = 4A
-2Pc = 4A - P<squared>
2Pc = P<squared> - 4A

c = P/2 - 2A/P

Hello Jason,

Yes indeed... Very well done. Your answer is spot on!!

Doing mathematics problems is excellent exercise for our brains,..an investment in our long term health you might say, so if you would like to try your hand at some more, keep watching out and I'll post some more.

Ron.

Originally Posted by RoverP6B

We all know or at least should know that Pythagoras, the ancient Greek Philosopher and Mathematician ( 570 - 495 BC ) proved the relationship between the three sides of a right angled triangle

I will ask my Mother In Law. She was probably on a first name basis with him. Or at least knew his parents...

Originally Posted by RoverP6B

We all know or at least should know that Pythagoras, the ancient Greek Philosopher and Mathematician ( 570 - 495 BC ) proved the relationship between the three sides of a right angled triangle,..the square of the hypotenuse being equal to the sum of the squares of the remaining two sides.

But what if we don't have this information,...

I assumed that because "we don't have this information" that Pythagoras theorem could not be used in the derivation ...

No wonder I was getting nowhere ...

Originally Posted by RoverP6B

Disco_owner wrote,..

Hello Disco_owner,

You have introduced a number of unknows, where as the question requires that you define the length of the hypotenuse only in terms of Perimeter 'P' and Area 'A', which as the question says you already know. In other words,..given the area and perimeter as 2 numerical values,...what is the length of the hypotenuse? The expression will only contain 'P' and 'A' how ever many times is necessary to give the exact length of the hypotenuse.

Ron.

ah huh , my solution has provided a relationship which includes all side on any triangle with respect to Height , but I now see your original question was pertaining to a right angle triangle .

Originally Posted by isuzutoo-eh

Talking about learning the skills and knowing how to use them...
When learning differentiation/integration at school, my teachers in year 11 and 12 didn't know or couldn't tell us what it was for. Then I learnt it at uni, again the lecturers couldn't tell us what it was for. It wasn't until studying 2nd year modern (or was it classical?) physics that any of it was used. .

If that is the case then your teachers weren't very good. However I find it hard to believe that a university level maths lecturer wouldn't know the practical applications of integrals and derivatives.

There are billions of practical applications - i.e. - how would you determine the volume of the droplet in the top left corner of the link?
http://www.nasaimages.org/luna/servl...9~58592~162436

For another question, what is the integral of x^x?

Ok you open the back of your Disco.
You retrieve the emergency triangle and turn it on and then you place it behind the broken down toyota.

Gadgets AKA Tim

Originally Posted by RoverP6B

austastar wrote,..

Thanks for that, but I already know how to do it,..I am only trying to encourage thought, and some people like to try and find solutions to problems.

As an example you might have a geometrical problem where the distance between two points if required and another distance is given. Drawing a scale diagram could be used to obtain a solution, but that to me is a pointless exercise,...I would much rather find the solution by algebraic means,..much more satisfying to do it that way.

The problem in itself is not difficult,..the mathematics required is known to the majority of school students by the time they reach the end of year 10. The problem lies in knowing how to use that knowledge,..you have the tools, but do you know how to use them?

Ron.

Ron to my mind it's all bollocks I was crap at maths and had a maths teacher who didn't care.
I got A'S & B's in most of my chosen subjects at school, but sadly just didn't sit the maths exam as I didn't understand it beyond basic arithmatic, In adult life I had to teach myself how to do costings and how to work out gross profits for menu planning in my job as a Chef to which end I was succesful.
It is sad for all those kids who don't get satisfaction in the classroom because a teacher isn't up to it much.
Sadly I will never be an engineer
Andy.

isuzurover wrote,..

If that is the case then your teachers weren't very good. However I find it hard to believe that a university level maths lecturer wouldn't know the practical applications of integrals and derivatives.

There are billions of practical applications - i.e. - how would you determine the volume of the droplet in the top left corner of the link?
http://www.nasaimages.org/luna/servl...9~58592~162436

For another question, what is the integral of x^x?

Hello Ben,

I agree, the University lecturers would most certainly know practical applications for differentiation and integration.

The question that you posed,...the integral of x^x....no doubt integrating wrt x,...must say much easier if it wasn't.

You can express x^x as e^xlnx and integrating xlnx is straight forward using integration by parts,...but that of course is not the question,...hmmm is x^x integrable? Have a funny feeling about this one.

Ron.