The problem asks to find the area of the kite with vertices at $(-3, -2)$, $(-6, -5)$, $(-3, -8)$, and $(4, -5)$.

1. Problem Description

The problem asks to find the area of the kite with vertices at $(-3, -2)$, $(-6, -5)$, $(-3, -8)$, and $(4, -5)$.

2. Solution Steps

The area of a kite is given by half the product of the lengths of its diagonals.

Let the vertices be $A(-3, -2)$, $B(-6, -5)$, $C(-3, -8)$, and $D(4, -5)$.

The length of the diagonal $AC$ is the distance between the points $A(-3, -2)$ and $C(-3, -8)$.

Since the x-coordinates are the same, the length is the absolute difference in the y-coordinates:

$|AC| = |-2 - (-8)| = |-2 + 8| = |6| = 6$.

The length of the diagonal $BD$ is the distance between the points $B(-6, -5)$ and $D(4, -5)$.

Since the y-coordinates are the same, the length is the absolute difference in the x-coordinates:

$|BD| = |-6 - 4| = |-10| = 10$.

The area of the kite is given by

$Area = \frac{1}{2} |AC| |BD| = \frac{1}{2} (6)(10) = \frac{1}{2}(60) = 30$.

3. Final Answer

30