Can YOU find the length with only one measure?

$BE=2\sqrt{2}$ , since all sides of a square are equal: $CD=BE=2\sqrt{2}$ . Easy. Now, to find AE:

The diameter of this circle is also equal to one of square's diagonals. The length of the diagonal of a square is equal to $s\sqrt{2}$ , where $s$ is a side length. In this case, $s=2\sqrt{2}$ , so we get $2\sqrt{2}*\sqrt{2}=2*2=4$ . So, if $d$ is the diameter, $d=4$ . The radius, $r$ , is equal to $\frac{d}{2}$ , which means that $r=\frac{4}{2}=\frac{2}{1}=2$ . Take the center of the circle (and the square as well), and call it Point F. $FE=2$ , since that would be half of a diagonal. $AF=r=2$ , and Point G is the point of intersection of $AF$ and $ED$ . If we were to find the length of $AG$ , we could create a right triangle with vertices A,G, and E, with $AE$ as the hypotenuse! $AG=r-GF$ . Since it is half of a side (this should be easy to visualize), $GF=\frac{2\sqrt{2}}{2}=\sqrt{2}$ . Therefore, $AG=2-\sqrt{2}$ . Leave that there. Now, $EG=GF$ since they are both halves of sides, so $EG=\sqrt{2}$ . Now, we can use the Pythagorean Theorem to solve for $AE$ ! $\sqrt{2} ^ 2+(2-\sqrt{2}) ^ 2=AE$ . If we were to plug this into the variables in the denominator of the problem, we find that it fits perfectly! For $BE$ , $2\sqrt{2}=\sqrt{8}$ , so:

• $a=8$
• $b=2$
• $c=2$
• $d=2$
• $k=2$

Plugging these variables into the equation for y, we get

$y=2^8-(2^2)^2=256-4^2=256-16=240$

Plugging our new value into the next equation, we find

$\frac{240}{8}=\frac{30}{1}$

Since the GCF of the numbers in that second fraction is 1, that means that

• $f=30$
• $g=1$

Plug THESE into the last equation:

$x=30-1=29$

Therefore, our answer is $\boxed{29}$ !