We are given a circle with center $A$. We have two chords, $BE$ and $CD$. We are given that $BE = 36$ and $CD = 32$. $AF$ is perpendicular to chord $CD$. We need to find the length of $AF$.

1. Problem Description

We are given a circle with center $A$. We have two chords, $BE$ and $CD$. We are given that $BE = 36$ and $CD = 32$. $AF$ is perpendicular to chord $CD$. We need to find the length of $AF$.

2. Solution Steps

Let the radius of the circle be $r$.

Since $BE = 36$, the radius $r = AE = AB = \frac{36}{2} = 18$.

Since $AF$ is perpendicular to $CD$, $F$ is the midpoint of $CD$.

Therefore, $CF = \frac{CD}{2} = \frac{32}{2} = 16$.

In the right triangle $ACF$, we have $AC = r = 18$ and $CF = 16$. By the Pythagorean theorem, we have

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

$18^2 = AF^2 + 16^2$

$324 = AF^2 + 256$

$AF^2 = 324 - 256 = 68$

$AF = \sqrt{68} = \sqrt{4 \cdot 17} = 2\sqrt{17}$

We need to round the answer to the nearest hundredth.

$AF = 2\sqrt{17} \approx 2(4.123) \approx 8.246$

Rounding to the nearest hundredth, we have $AF \approx 8.25$.

3. Final Answer

8.25