Okay, here we have this:
First we are going to find the distances CA, AB, and BC, let's do it:
CA segment:
CA=√((3-(-6))²+(5-(-5))²)=√((3+6)²+(5+5)²)=√((9)²+(10)²)=√(81+100)
CA=√181
AB segment:
AB=√((4-3)²+(-5-5)²)=√((1)²+(-10)²)=√(1+100)=√101
AB=√101
BC segment:
BC=√((4-(-6))²+(-5-(-5))²)=√((4+6)²+(-5+5)²)=√((10)²+(0)²)=√(100+0)=√100=10
BC=10
Now, let's use Heron's formula:
![\text{Area}=\sqrt[]{s(s-CA)(s-AB)(s-BC)}](https://img.qammunity.org/2023/formulas/mathematics/college/31vizpp7q506wpr69vuu6rndkxk17tqlfj.png)
And "s" is equal to=(CA+AB+BC)/2=(√181+√101+10)/2
So, the area is equal to:
![\begin{gathered} Area=\sqrt[]{(√(181)+√101+10)/(2)((√(181)+√101+10)/(2)-\sqrt[]{181})((√(181)+√101+10)/(2)-\surd101)((√(181)+√101+10)/(2)-10)} \\ =\sqrt[]{2500} \\ =\sqrt[]{50}^2 \\ =50 \end{gathered}]()
Finally we obtain that area of triangle ABC is equal to 50 square units.