A triangle has vertices at $(-3,2),(6,-2),(3,5)$. How many square units are in the area of the triangle? Express your answer as a decimal to the nearest tenth.
[asy]
import graph;
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
dotfactor=4;
real x = 7;
pair A=(-3,2), B=(6,-2), C=(3,5);
fill(A--C--(-3,5)--cycle,gray(0.6));
fill(B--C--(6,5)--cycle,gray(0.6));
fill(A--B--(-3,-2)--cycle,gray(0.6));
pair[] dots = {A,B,C};
dot(dots);
xaxis(Ticks(" ",1.0,begin=false,end=false,NoZero,Size=3),Arrows(4),above=true);
yaxis(Ticks(" ",1.0,begin=false,end=false,NoZero,Size=3),Arrows(4),above=true);
draw(A--B--C--cycle);
label("$(-3,2)$",A,W);
label("$(6,-2)$",B,SE);
label("$(3,5)$",C,N);
draw((-3,5)--(6,5)--(6,-2)--(-3,-2)--cycle,dotted);[/asy] We find the area of the given triangle by subtracting the sum of the areas of the three shaded triangles in the figure from the area of the rectangle formed by all four triangles.

The area of the rectangle is $9(7)=63$ square units, and the sum of the areas of the shaded triangles is $$\frac{1}{2}(6)(3)+\frac{1}{2}(3)(7)+\frac{1}{2}(4)(9)=37.5$$ square units.  The area of the fourth triangle is $63-37.5=\boxed{25.5}$ square units.
25.5
Level 4
Geometry
Problem: A triangle has vertices at $(-3,2),(6,-2),(3,5)$. How many square units are in the area of the triangle? Express your answer as a decimal to the nearest tenth.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the area of a triangle given its vertices, we can use the formula for the area based on the coordinates of the vertices. The vertices of the triangle are given as A(-3, 2), B(6, -2), and C(3, 5). 

The area \( A \) of the triangle can be calculated using the formula:

\[
A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]

Substituting the coordinates of the vertices into the formula, we identify \( (x_1, y_1) = (-3, 2) \), \( (x_2, y_2) = (6, -2) \), and \( (x_3, y_3) = (3, 5) \).

Calculating each term step-by-step:

1. Calculate \( y_2 - y_3 \):
   \[
   -2 - 5 = -7
   \]

2. Calculate \( y_3 - y_1 \):
   \[
   5 - 2 = 3
   \]

3. Calculate \( y_1 - y_2 \):
   \[
   2 - (-2) = 4
   \]

Now substitute these values into the area formula:

\[
A = \frac{1}{2} \left| -3(-7) + 6(3) + 3(4) \right|
\]

Calculating each term:

- The first term: 
  \[
  -3 \times -7 = 21
  \]

- The second term: 
  \[
  6 \times 3 = 18
  \]

- The third term: 
  \[
  3 \times 4 = 12
  \]

Now sum these results:

\[
21 + 18 + 12 = 51
\]

Now, calculate the area:

\[
A = \frac{1}{2} \left| 51 \right| = \frac{51}{2} = 25.5
\]

Thus, the area of the triangle is \( 25.5 \) square units.<|END_OF_LATENT|> The final answer is $25.5$.