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- Nov 27th 2007, 04:27 PMMacleefFinding the value of "x" inside a square
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- Nov 27th 2007, 06:44 PMDivideBy0
This can be solved using simultaneous equations. Please refer to the diagram for the labelling.

From the diagram we can see (by Pythagoras' Theorem):

$\displaystyle a^2+b^2 = 7^2$ ...[1]

$\displaystyle b^2 + c^2 = 35^2$ ...[2]

$\displaystyle c^2+d^2 = 49^2$ ...[3]

$\displaystyle a^2+d^2 = x^2$ ...[4]

Subtracting the first from the second,

$\displaystyle c^2 - a^2 = 35^2 - 7^2$

$\displaystyle c^2 - a^2 = 1176$ ...[5]

Then, subtracting [5] from [3],

$\displaystyle a^2 + d^2 = 49^2 - 1176$

So we have $\displaystyle x^2 = 49^2 - 1176=1225$

$\displaystyle x = 35$ - Nov 28th 2007, 03:01 PMMacleef
Wow...I'm so confused -_-

- Nov 28th 2007, 04:33 PMDivideBy0
Ok, could you please tell me which part confused you? Do you understand the diagram, have you learnt Pythagoras' Theorem?

- Nov 28th 2007, 04:48 PMMacleef
I don't get why $\displaystyle a^2 + b^2 = 7^2$. If this is the Pythagorean theorem, how did you come to that equation? I mean, when I physically look at the diagram, those 2 sides, "a" and "b" are not in the same triangle... I don't know if I'm clearly explaining myself here...

http://i2.tinypic.com/7147pet.jpg

http://i14.tinypic.com/6sh1ilc.jpg

What happens to the other 2 sides that aren't labelled? - Nov 28th 2007, 04:51 PMDivideBy0
You can make them be in the same triangle :P, since what you have here is a rectangle, the opposite sides are equal.

- Nov 28th 2007, 04:54 PMMacleef
Thanks, that clears up everything, I understand it now

And I think what I did previously is kind of similar to the process you posted