Who is familiar with Pythagoras?

isuzutoo-eh wrote,..

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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.

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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.

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)

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

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Originally Posted by RoverP6B

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.

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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

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

Ron.

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

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

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


http://www.calculateme.com/cArea/are...ase-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:

Disco_owner wrote,..

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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)

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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.

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