$ABCD$ is a kite. $AC$ is perpendicular to $DB$, and $DE = EB$. $AD = 9$ cm, $DC = 10$ cm, and $AE = 8$ cm. Calculate the length of $AC$ to the nearest tenth of a centimeter.

1. Problem Description

$ABCD$ is a kite. $AC$ is perpendicular to $DB$, and $DE = EB$. $AD = 9$ cm, $DC = 10$ cm, and $AE = 8$ cm. Calculate the length of $AC$ to the nearest tenth of a centimeter.

2. Solution Steps

Since $ABCD$ is a kite and $AC \perp DB$, $DB$ is bisected by $AC$ at point $E$. Thus, $DE = EB$.

In $\triangle ADE$, we have $AD = 9$ cm and $AE = 8$ cm. Since $\triangle ADE$ is a right triangle, we can use the Pythagorean theorem to find $DE$:

$AD^2 = AE^2 + DE^2$

$9^2 = 8^2 + DE^2$

$81 = 64 + DE^2$

$DE^2 = 81 - 64 = 17$

$DE = \sqrt{17}$

Since $DE = EB$, $EB = \sqrt{17}$.

Thus, $DB = DE + EB = \sqrt{17} + \sqrt{17} = 2\sqrt{17}$.

In $\triangle CEB$, we have $EB = \sqrt{17}$ and $\angle CEB = 90^\circ$. In $\triangle CDE$, we have $DE = \sqrt{17}$ and $CD = 10$.

Consider $\triangle CDE$, where $CD = 10$ and $DE = \sqrt{17}$. Since $\triangle CDE$ is a right triangle, we can use the Pythagorean theorem to find $CE$:

$CD^2 = DE^2 + CE^2$

$10^2 = (\sqrt{17})^2 + CE^2$

$100 = 17 + CE^2$

$CE^2 = 100 - 17 = 83$

$CE = \sqrt{83}$

Now we can find the length of $AC$, which is $AE + CE = 8 + \sqrt{83}$.

$AC = 8 + \sqrt{83} \approx 8 + 9.11 = 17.11$

Rounding to the nearest tenth of a centimeter, $AC \approx 17.1$ cm.

3. Final Answer

17.1 cm