To find the area of both triangles, we can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
1. Area of the larger triangle with vertices \((5, 3)\), \((5, 9)\), and \((2, 3)\):
Let:
- \( (x_1, y_1) = (5, 3) \)
- \( (x_2, y_2) = (5, 9) \)
- \( (x_3, y_3) = (2, 3) \)
Plugging into the area formula:
\[ \text{Area} = \frac{1}{2} \left| 5(9-3) + 5(3-3) + 2(3-9) \right| \] \[ = \frac{1}{2} \left| 5 \cdot 6 + 0 + 2(-6) \right| \] \[ = \frac{1}{2} \left| 30 - 12 \right| \] \[ = \frac{1}{2} \left| 18 \right| = 9 \]
2. Area of the smaller triangle with vertices \((12, 2)\), \((12, 6)\), and \((10, 2)\):
Let:
- \( (x_1, y_1) = (12, 2) \)
- \( (x_2, y_2) = (12, 6) \)
- \( (x_3, y_3) = (10, 2) \)
Plugging into the area formula:
\[ \text{Area} = \frac{1}{2} \left| 12(6-2) + 12(2-2) + 10(2-6) \right| \] \[ = \frac{1}{2} \left| 12 \cdot 4 + 0 + 10(-4) \right| \] \[ = \frac{1}{2} \left| 48 - 40 \right| \] \[ = \frac{1}{2} \left| 8 \right| = 4 \]
3. Ratio of the area of the smaller triangle to the area of the larger triangle:
Now that we have the areas:
- Area of the larger triangle = \(9\)
- Area of the smaller triangle = \(4\)
The ratio is:
\[ \text{Ratio} = \frac{\text{Area of smaller triangle}}{\text{Area of larger triangle}} = \frac{4}{9} \]
Thus, the ratio of the area of the smaller triangle in relation to the area of the larger triangle is \(\boxed{\frac{4}{9}}\).