<p>If a square has a side of length x+4 and a diagonal of length x+8, what is the value of x?</p>
<p>I'm kind of stuck and I feel so stupid because this is a Pythagorean Theorem question and I can't seem to get the answer doing that. I plugged in numbers and got the right answer but how would you do this?</p>
<p>Okay, we know that a^2 + b^2 = c^2. We have a square with a diagonal in the middle, looking something like this.
<strong>x<em>+</em>4</strong>
|\ |
| \ |
| \ x | x
| \ + | +
| \ 8 | 4
| \ |
| \ |
</p>
<p>So you get (x+4)^2 + (x+4)^2 = (x+8)^2</p>
<p>Then solve for x:
x^2 + 8x + 16 + x^2 + 8x + 16 = x^2 + 16x + 64
x^2 + 0x - 32 = 0
x^2 - 32 = 0
x^2 = 32
x = √32
x = 4√2</p>
<p>This is a tough problem, especially since there are no answer choices.</p>
<p>Note that since the sides of a square are congruent, each triangle formed is an isosceles right triangle. This is the same as a 45, 45, 90 right triangle. So we need x + 8 = √2 (x+4). So we have</p>
<p>x + 8 = x√2+4√2
x√2-x = 8 – 4√2
x(√2-1) = 8 – 4√2
x = (8 – 4√2)/( √2-1) ~ 5.65685</p>
<p>So grid in 5.65 or 5.66</p>