We are given a circle with center A. We are given that $BE = 34$ and $CD = 26$. We need to find the length of $AF$. We know that $AF$ is perpendicular to $CD$.

1. Problem Description

We are given a circle with center A. We are given that $BE = 34$ and $CD = 26$. We need to find the length of $AF$. We know that $AF$ is perpendicular to $CD$.

2. Solution Steps

First, we can find the radius of the circle. Since $BE$ is a chord that passes through the center $A$, it is a diameter. Thus, the radius $r$ is half of $BE$, so $r = BE/2 = 34/2 = 17$.

Since $AF$ is perpendicular to the chord $CD$, it bisects the chord $CD$. Therefore, $CF = CD/2 = 26/2 = 13$.

Now, we can consider the right triangle $ACF$. We have $AC = r = 17$ and $CF = 13$. We want to find $AF$. Using the Pythagorean theorem, we have:

$AC^2 = AF^2 + CF^2$

$17^2 = AF^2 + 13^2$

$289 = AF^2 + 169$

$AF^2 = 289 - 169$

$AF^2 = 120$

$AF = \sqrt{120}$

Now, we can approximate the value of $\sqrt{120}$ to the nearest hundredth:

$AF = \sqrt{120} \approx 10.95445 \approx 10.95$

3. Final Answer

$AF = 10.95$