Math Problem

spirit · 6 May 2006 at 17:31

lol as usual i missed all the anarchy

Nitefly · 6 May 2006 at 17:31

I'm unaware where pythag was needed in that question...? :confused:

spirit · 6 May 2006 at 17:34

Nitefly said:
I'm unaware where pythag was needed in that question...?

how else you gonna solve it? Easiest way by far is to go

let x = side of square

2x^2 = 8^2
x^2 = 32

which is also the area?

TO OP: did you do the SAT this morning?

soz bout colours n such if that makes sense

Rotty · 6 May 2006 at 17:36

stig said:
Math Problem

and not that good with English too :eek:

Nitefly · 6 May 2006 at 17:37

spirit said:
how else you gonna solve it? Easiest way by far is to go

let x = side of square

2x^2 = 8^2
x^2 = 32

which is also the area?

TO OP: did you do the SAT this morning?

actually i see a quicker way, two seconds ill up a pic

Rofl. If you know that the diagonal is 8 therefore the angles in the centre are 90 degrees, you can use the base as 4. Pythag was not needed at ALL for that question....

As far as I see... (4x4)+(4x4)=32

Ricochet J · 6 May 2006 at 17:39

Yeh you can get in the unnesseceties and even use trigonometry and calculus. In maths, there is usually more than one way to do things

spirit · 6 May 2006 at 17:52

Nitefly said:
Rofl. If you know that the diagonal is 8 therefore the angles in the centre are 90 degrees

That makes no sense, so are you implying in a square with a diagonal of 9 that they dont have right angles in the diagonal ?

Whether thats easier or not depends on the person

You could also say taht its a 45-45 triangle, therefore the side of the square is 8 / root(2) .. and the area is therefore (8/root(2))^2

isn't it suprising that theres a million and one ways to do a simple problem ?

what is easier depends on the eye of the beholder..

Nitefly · 6 May 2006 at 18:00

spirit said:
That makes no sense, so are you implying in a square with a diagonal of 9 that they dont have right angles in the diagonal ?

Don't know what you mean, if its a square, where the diagonals meet gives four 90 degree segments.

Ah wait, what I typed made no sense! Phaaaaaa you know what i meant!

Forgive meee!

Inquisitor · 6 May 2006 at 18:02

With regards to the variable name business, a, o and h tend to be used when using trig with right-angled triangles, otherwise, the convention is to use a, b, and c for sides and A, B, and C for angles with triangles.