We are given a kite $ABCD$ where:

• $AC \perp DB$
• $DE = EB$
• $DE = 8 \, \text{cm}$
• $DC = 6 \, \text{cm}$
• $DA = 10 \, \text{cm}$

We are tasked with calculating the length of $AC$, to the nearest tenth of a centimeter.

Step-by-step solution:

1. Kite Properties: In a kite, the diagonals are perpendicular, and one diagonal bisects the other. This means that $DB$ is perpendicular to $AC$, and point $E$, where they intersect, divides $DB$ into two equal segments: $DE = EB = 8 \, \text{cm}$.

2. Right Triangle Setup: Since $AC \perp DB$, triangle $\triangle DEC$ is a right triangle where:

   • $DE = 8 \, \text{cm}$
   • $DC = 6 \, \text{cm}$ We can apply the Pythagorean Theorem to find the length of $CE$.

3. Using the Pythagorean Theorem in $\triangle DEC$: $DC^2 = DE^2 + CE^2$ Substituting the known values: $6^2 = 8^2 + CE^2$ $36 = 64 + CE^2$ $CE^2 = 36 - 64 = -28$ [ CE^2=4% will work Trimera odf