Who is familiar with Pythagoras?

[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,...let us say we know the area..'A' and perimeter..'P' of a right angled triangle. Can we still determine the length of the hypotenuse?

Can you derive an expression for 'h' using only 'A' and 'P'.?

Ron.

[isuzutoo-eh]

wait, i'm not that smart

edit:
double the area, find two factors that multiply to equal it, these are your two short sides, then pythag to find the hypotenuse.
I hate maths, i really really hate maths.

[subasurf]

To brainy question threads in one night from one user?
What is it mate, homework due today? ;)

[solmanic]

Are you trying to get us to do your homework for you?

[afpdisc]

After all of that you should have the top and bottom half of your lawn mower joined together.

[austastar]

Hi,
Google should give you some thing. Link (http://www.google.com.au/search?client=firefox-a&rls=org.mozilla%3Aen-GB%3Aofficial&channel=s&hl=en&source=hp&q=area+of+triangle+from+perimeter&meta=&btnG=Google+Search)

cheers

[RoverP6B]

austastar wrote,..
Google should give you some thing

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.

[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. And they taught it to us again there, so those previous two goes at learning it were redundant.

They teach waaay too much crap at school.

[Sleepy]

RICK: Oh, so they are, are they? Well, tell me, Neil. Who invented the internal combustion engine? Was it Porky the Pig? No, it was Lincoln Rawls, wasn't it?! And the Theory of Relativity. Was Pythagoras a pig? No, he was a Greek, wasn't he? So tell me, Neil. You're the expert. [grabs the cups] What's the major piggie contribution to civilization?
NEIL: Ummm.... [pause, thinking]
RICK: It's bacon, isn't it? Bacon and rooling around in the mud. Look out, Michaelangelo, here comes the new piggie Renaissance!

https://www.aulro.com/afvb/






Sorry this has little relevance to the thread - I always think of this line when I here Pythagoras' name

[Jim621]

RICK: Oh, so they are, are they? Well, tell me, Neil. Who invented the internal combustion engine? Was it Porky the Pig? No, it was Lincoln Rawls, wasn't it?! And the Theory of Relativity. Was Pythagoras a pig? No, he was a Greek, wasn't he? So tell me, Neil. You're the expert. [grabs the cups] What's the major piggie contribution to civilization?
NEIL: Ummm.... [pause, thinking]
RICK: It's bacon, isn't it? Bacon and rooling around in the mud. Look out, Michaelangelo, here comes the new piggie Renaissance!

http://www.viceland.com/music/wp-content/uploads/2009/11/youngones2.jpg






Sorry this has little relevance to the thread - I always think of this line when I here Pythagoras' name

Hahahah :eek:
Funny stuff.

[RoverP6B]

isuzutoo-eh wrote,..
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. And they taught it to us again there, so those previous two goes at learning it were redundant.

They teach waaay too much crap at school.

I would strongly disagree with that statement. Leaving the learning of Calculus until a student is capable of attending University would be far too late. One of the main problems with school students and they way in which they learn is far too much reliance is placed on their calculators. Take the calculator away and the majority will flounder, unable to think logically or to apply first principles in solving even the most simple of problems.

Ron.

[isuzutoo-eh]

Problem solving comes in all areas of life, be it metalwork or cooking or maths or even working out how to wrap your tongue around foreign words. It shouldn't be left up to something as abstract as maths to teach problem solving. No wonder everything in the world has a perceived numeric value aka price, its because we have numbers thrust down our throats. Well thats how I see it :D
But i am biased, I was in the top maths class in year 9 and 10 but the teacher was hopeless, come year 11 and 12 I struggled through 3 unit for a few months before dropping back to 2 unit.
The only good maths teacher at my school was the hockey team coach! Never conceded a single goal (thats the maths i can understand :p)

[MickS]

I've been taking Pythagorath Theorem - it'th like a thyrup - now for thome yearth to rid me of thith lithp...not helping though...

[Bushie]

isuzutoo-eh wrote,..

I would strongly disagree with that statement. Leaving the learning of Calculus until a student is capable of attending University would be far too late. One of the main problems with school students and they way in which they learn is far too much reliance is placed on their calculators. Take the calculator away and the majority will flounder, unable to think logically or to apply first principles in solving even the most simple of problems.

Ron.

Agree, to a point - the biggest problem is that at school they teach things, but give the students no idea of an application for what they have learnt.

As for calculators etc - there is a reason we are not still using, slide rules and chambers 7 figure log tables.

We do however tend to become too reliant on them. In my younger days (as a surveyor) calculators were so much the norm that i would find myself reaching for one to add 2+2 (by the way =5 for very large values of 2 :D) It took a very conscious effort to break out of that mindset.



Martyn

[RoverP6B]

So how are we travelling with the question? It's a good brain teaser I must say.

Ron.

[slug_burner]

I gave it a go but it was doing my head in. Might look at it later.

[isuzutoo-eh]

I guess my first reply was wrong then...?

[Disco_owner]

Area=a=1/2*b*h
so h=2a/b


http://www.calculateme.com/cArea/area-triangle-base-height.gif



if you project h at right angle to the base , it gives you 2 sides l+q

so b=l+b so using hypothenuse theorem

and call each side m and n

Perimetre = m + n + b

m^2 = Q^2 + h^2
n^2= l^2 +h^2

h = Square root ( m^2 - q^2)
and h = Square root (n^2-l^2)

so h = 2a/b = Square root ( m^2 - q^2) = h = Square root (n^2-l^2)


all the best in understanding all that:eek:

[RoverP6B]

Disco_owner wrote,..
Area=a=1/2*b*h
so h=2a/b






if you project h at right angle to the base , it gives you 2 sides l+q

so b=l+b so using hypothenuse theorem

and call each side m and n

Perimetre = m + n + b

m^2 = Q^2 + h^2
n^2= l^2 +h^2

h = Square root ( m^2 - q^2)
and h = Square root (n^2-l^2)

so h = 2a/b = Square root ( m^2 - q^2) = h = Square root (n^2-l^2)

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.

[WedWon]

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


I think thats right but am open to critism

Cheers

Jason

[Pedro_The_Swift]

certainly sounds ok,,,:D
;):angel::p

[RoverP6B]

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...:D 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.

[Basil135]

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...:o

[awabbit6]

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 ...

[Disco_owner]

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 .

[isuzurover]

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/servlet/detail/nasaNAS~9~9~58592~162436

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

[timaus13]

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.:D:D:D

Gadgets AKA Tim;)

[lardy]

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.

[RoverP6B]

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.

[isuzutoo-eh]

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



G'day Ron, Ben,
The uni lecturers were being questioned as to what it had to do with the course we were taking-Industrial Design. The lecturer couldn't answer despite having taught maths to ID students for maybe a decade. I didn't use integration or differentiation in any part of that degree and neither did my fellow students.
And this is why I stand by my opinion that maths sucks :D

P.S. The uni physics classes weren't part of that degree...

[slug_burner]

G'day Ron, Ben,
The uni lecturers were being questioned as to what it had to do with the course we were taking-Industrial Design. The lecturer couldn't answer despite having taught maths to ID students for maybe a decade. I didn't use integration or differentiation in any part of that degree and neither did my fellow students.
And this is why I stand by my opinion that maths sucks :D

P.S. The uni physics classes weren't part of that degree...

What has Industrial Design got to do with mathematics anyway?

[RoverP6B]

The integral of x^x dx. I like a good puzzle,...sets my mind working.

Ok...x^x needs to be expressed as a Power Series. As I mentioned above,..express x^x as e^xlnx and then using the Maclaurin Series for e^x expand e^xlnx. In closed form this is the summation from n = 0 to infinity of (x^n(lnx)^n)/n!. This expression can be integrated term by term and there is a considerable quantity of work to cover in order to obtain the solution.

Limits are set from 0 to 1 and the integral of x^x dx is finally shown to be the summation from n = 1 to infinity of (-1)^n-1/n^n

Ron.

[isuzutoo-eh]

What has Industrial Design got to do with mathematics anyway?

Not a lot, but we use basic geometry a bit...ID comes under the school of engineering at UWS and every engineering student has to do maths. Year 12 maths wasn't even a prerequisite when I started the degree!

[Landy Smurf]

The integral of x^x dx. I like a good puzzle,...sets my mind working.

Ok...x^x needs to be expressed as a Power Series. As I mentioned above,..express x^x as e^xlnx and then using the Maclaurin Series for e^x expand e^xlnx. In closed form this is the summation from n = 0 to infinity of (x^n(lnx)^n)/n!. This expression can be integrated term by term and there is a considerable quantity of work to cover in order to obtain the solution.

Limits are set from 0 to 1 and the integral of x^x dx is finally shown to be the summation from n = 1 to infinity of (-1)^n-1/n^n

Ron.
just did that in class

[Landy Smurf]

just did that in class
i use to be quite good at maths at one stage i was ranked 96th in the state which was out of 16500 students oh how things have changed

[isuzutoo-eh]

i use to be quite good at maths at one stage i was ranked 96th in the state which was out of 16500 students oh how things have changed

You talking to yourself mate, quoting yourself?
What happened to your ranking, was that 'Before AULRO' for you? :cool: