this question was so what time consuming but easy one

as triangle ACD has one side as diameter of circle so angle C=90

from that diameter \(AD^2=(1/2)^2 + (1/2)^2 \)
AD=(1/√2)
radius=(1/2√2)
so each side of square is = (1/√2)

and

as BG=4AE=√2
Draw a line from B which passes through center of circle and reaches point R and BR=(1/√2)

now triangle BGR is right triangle
\(BG^2 = GR^2 + BR^2 \)
GR=√(3/2)
GF=GR-FR
=\sqrt{(3/2)}-(1/2√2)
= (2√3-1)/2√2
multiply √2 to numerator and denaminator
= (2√6-√2)/4
so option D is correct

"if you like the explanation please press kudos please"

Hi guys,

I don´t understand why we can assume B is forming a right triangle with the base of the square. Technically, with no more information given, B could not be forming the 90º triangle with the base, since it could be slightly away from the centre of the square. hence we could not determine the length of GF. I do understand the logic of the problem, but I would like to understand what can we assume and what we can´t. Honestly, the fact that option E is on the table is confusing.

A vertical line dropped from point B will only pass through the center C of the Circle if B is the midpoint of the side of the square, but in the problem it is nowhere mentioned that B is the midpoint of the side.

Carcass pls help explain

It may help to first redraw the figure by simply rotating triangle $A C D$ about the center of the circle so that $A D$ will be vertical. This is acceptable, because we aren't changing any lengths or angles except to create a right triangle $A O G$, as shown:




Now, let's start with the one length we were given. Since $\(A C=C D\)$, triangle $A C D$ is an equilateral right triangle, or a 45-45-90 triangle (referring to the angle measures). In an equilateral right triangle, the hypotenuse is $\(\sqrt{2}\)$ times the length of either other side, so $\(A D=\frac{\sqrt{2}}{2}\)$.

In the figure, $A D$ is the diameter of the circle, and $\(A E\)$ is a radius of the circle. Thus, $\(A E\)$ is half $\(A D\)$, or $\(A E=\frac{\sqrt{2}}{4}\)$. Also, the square side length equals the diameter, and $\(D F\)$ is half a side of the square, so $\(D F=\frac{\sqrt{2}}{4}\)$, too.

The problem states that $\(B G=4 A E\), so \(B G=\sqrt{2}\)$.

We now have two of the side lengths for the right triangle we created:





By Pythagorean Theorem, \(\sqrt{(\sqrt{2})^2-(\frac{\sqrt{2}}{2})^2 \)\(= \sqrt{2-\frac{2}{4}}=\sqrt{\frac{3}{2}}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}\)

We are looking for $G F$, which is simply $\(G D-D F\)$. Since $\(D F=\frac{\sqrt{2}}{4}\), GF\(=\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{4}=\frac{2\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{2 \sqrt{6}-\sqrt{2}}{4}\)$.

The correct answer is D .