# Thread: Finding the value of "x" inside a square

1. ## Finding the value of "x" inside a square

Solved

2. 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 $\displaystyle x^2 = 49^2 - 1176=1225$

$\displaystyle x = 35$

3. Wow...I'm so confused -_-

4. Ok, could you please tell me which part confused you? Do you understand the diagram, have you learnt Pythagoras' Theorem?

5. Originally Posted by DivideBy0
Ok, could you please tell me which part confused you? Do you understand the diagram, have you learnt Pythagoras' Theorem?
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...

What happens to the other 2 sides that aren't labelled?

6. You can make them be in the same triangle :P, since what you have here is a rectangle, the opposite sides are equal.

7. 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