Using the distance(d) formula to obtain the length AB,BC,CA.
The distance formula is,
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Given
[tex]\begin{gathered} A\rightarrow(5,-6) \\ B\rightarrow(-5,-3) \\ C\rightarrow(5,6) \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} AB=\sqrt{(-5-5)^2+(-3--6)^2}=\sqrt{(-10)^2+(-3+6)^2}=\sqrt{100+3^2}=\sqrt{109} \\ BC=\sqrt{(5--5)^2+(6--3)^2}=\sqrt{10^2+9^2}=\sqrt{100+81}=\sqrt{181} \\ CA=\sqrt{(5-5)^2+(6--6)^2}=\sqrt{0^2+12^2}=\sqrt{144}=12 \end{gathered}[/tex]
Therefore,
[tex]\begin{gathered} AB=\sqrt{109}=10.44030\approx10.4 \\ BC=\sqrt{181}=13.45362\approx13.5 \\ CA=12 \end{gathered}[/tex]
Using Heron's formula to solve for the area
[tex]\begin{gathered} Area=\sqrt{s(s-a)(s-b)(s-c)} \\ s=\frac{a+b+c}{2} \end{gathered}[/tex]
where,
[tex]\begin{gathered} a=10.4 \\ b=13.5 \\ c=12 \\ \\ s=\frac{10.4+13.5+12}{2}=17.95 \end{gathered}[/tex]
Therefore, the area is
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