Math Problem

stig · 6 May 2006 at 14:23

ok been trying to do this puzzle all morning and cant seem to see it!

each diagonal is 8cm and u need to find the total area of the square

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*S any ideas?

Arcade Fire · 6 May 2006 at 14:25

Do you know Pythagoras' Law?

a^2 + b^2 = c^2, for right angled triangles?

Burnsy2023 · 6 May 2006 at 14:27

With Pythagoras, you can you it in your head.

Burnsy

William · 6 May 2006 at 14:27

Ohhh its a been a long time, erm.

You do:

(8^2)/2 = Answer. Answer x Answer = area in units.

Edit: whooo thats wrong.

Burbleflop · 6 May 2006 at 14:31

Edit: Having actually read the OP properly, my guess is 32cm^2

theo · 6 May 2006 at 14:33

William said:
Ohhh its a been a long time, erm.

You do:

(8^2)/2 = Answer. Answer x Answer = area in units.

Edit: whooo thats wrong.

i think that's right... using one of the triangles, two of the sides on that triangle are 4cm, using pythagoras we know that the third side of that triangle is the square root of 4^2 + 4^2, which is 'root32', but then considering it's a square the answer would be (root32)^2 which equals 32

ArmyofHarmony · 6 May 2006 at 14:35

4 x (half base x height of one triangle)

Burnsy2023 · 6 May 2006 at 14:36

theo said:
i think that's right... using one of the triangles, two of the sides on that triangle are 4cm, using pythagoras we know that the third side of that triangle is the square root of 4^2 + 4^2, which is 'root32', but then considering it's a square the answer would be (root32)^2 which equals 32

Right sequence, wrong termanology. 4^2 +4^2 = 32, not root 32.

Burnsy

Mattus · 6 May 2006 at 14:37

8^2 = a^2 + b^2, where a=b as it's a square.

64 = 2a^2

32 = a^2.. which is the area of the square if you think about it.

Polarbear · 6 May 2006 at 14:37

It has got an area of 32.

Just use Pythagoras to find the length of one side and square it.

Edit: I'm taking that when you say each diagonal is 8cm you mean from corner of square to corner of square.

Seft · 6 May 2006 at 14:56

32.

You don't even need pythagoras.

Imagine they are four triangles. You could rearrange those four triangles to get two 4x4 squares

VeNT · 6 May 2006 at 15:02

tbh, its quite simple and I'm guessing you did this in classes before being set the homework.
notes = win

Ricochet J · 6 May 2006 at 15:16

Yup its my main pal Pythagoras and his thereom.

His thereom states:
The two shorter sides squared and added together make the longest side squared in a right angled triangle.

Pickers · 6 May 2006 at 15:57

eXSBass said:
Yup its my main pal Pythagoras and his thereom.

His thereom states:
The two shorter sides squared and added together make the longest side squared in a right angled triangle.

It was more along the lines of
"The square of the hypotenuse is equal to the sum of the square of the other two sides"

VeNT · 6 May 2006 at 15:59

Pickers said:
It was more along the lines of
"The square of the hypotenuse is equal to the sum of the square of the other two sides"

or
A^2 + B^2 = C^2

JonC · 6 May 2006 at 16:01

VeNT said:
or
A^2 + B^2 = C^2

Providing that you provide a diagram.....

jonc

DaveyD · 6 May 2006 at 16:04

VeNT said:
or
A^2 + B^2 = C^2

Better to be:

a² + b² = h²

to be a little anal

VeNT · 6 May 2006 at 16:04

DaveyD said:
Better to be:

a² + b² = h²

totaly

carvegio · 6 May 2006 at 16:09

why is it better with h?
:confused:

VeNT · 6 May 2006 at 16:10

carvegio said:
why is it better with h?

cos its the name (or atleast the first letter of) for that side

Pickers said:
for hypotenuse

what he said