Triangle STU, with vertices \( \mathrm{S}(-5,-8), \mathrm{T}(-3,-6) \), and \( \mathrm{U}(-6,-5) \), is drawn inside a rectangle, as shown below.
What is the area, in square units, of triangle STU?

Step 1 ：\( \mathrm{ST} = \sqrt{(-3 - (-5))^2 + (-6 - (-8))^2} = \sqrt{4 + 4} = \sqrt{8} \)

Step 2 ：\( \mathrm{SU} = \sqrt{(-6 - (-5))^2 + (-5 - (-8))^2} = \sqrt{1 + 9} = \sqrt{10} \)

Step 3 ：\( \mathrm{TU} = \sqrt{(-3 - (-6))^2 + (-6 - (-5))^2} = \sqrt{9 + 1} = \sqrt{10} \)

Step 4 ：\( \mathrm{s} = \frac{\mathrm{ST} + \mathrm{SU} + \mathrm{TU}}{2} = \frac{\sqrt{8} + 2\sqrt{10}}{2} \)

Step 5 ：\( \mathrm{Area} = \sqrt{\mathrm{s}(\mathrm{s}- \mathrm{ST})(\mathrm{s}- \mathrm{SU})(\mathrm{s} - \mathrm{TU})} = \sqrt{(\frac{\sqrt{8} + 2\sqrt{10}}{2})(\frac{\sqrt{8} + 2\sqrt{10}}{2} - \sqrt{8})(\frac{\sqrt{8} + 2\sqrt{10}}{2} - \sqrt{10})(\frac{\sqrt{8} + 2\sqrt{10}}{2} - \sqrt{10})} \)

Step 6 ：\( \mathrm{Area} = 6.0000 \)