To find the area of triangle \( C \), \( A \), and \( B \) with the coordinates given:
- \( C(3, 2) \)
- \( A(-5, -4) \)
- \( B(3, -3) \)
We can use the formula for the area of a triangle determined by three points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates into this formula:
Let
- \((x_1, y_1) = (3, 2) , (C)\)
- \((x_2, y_2) = (-5, -4) , (A)\)
- \((x_3, y_3) = (3, -3) , (B)\)
Now, substitute these values into the area formula:
\[ \text{Area} = \frac{1}{2} \left| 3(-4 - (-3)) + (-5)(-3 - 2) + 3(2 - (-4)) \right| \]
Calculating each term inside the absolute value:
- \( 3(-4 + 3) = 3(-1) = -3 \)
- \( -5(-3 - 2) = -5(-5) = 25 \)
- \( 3(2 + 4) = 3 \times 6 = 18 \)
Now substitute these results back into the formula:
\[ \text{Area} = \frac{1}{2} \left| -3 + 25 + 18 \right| \]
Combine the values inside the absolute value:
\[ -3 + 25 + 18 = 40 \]
Then we compute the area:
\[ \text{Area} = \frac{1}{2} \left| 40 \right| = \frac{1}{2} \times 40 = 20 \]
Thus, the area of Leonardo’s triangle is
\[ \boxed{20} \text{ square units.} \]