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Sagot :
Okay, here we have this:
Considering the provided vertices, we are going to calculate the requested area, so we obtain the following:
Then we will first calculate the measure of each side and later with Heron's formula we will find the area, then we have:
[tex]\begin{gathered} u=\sqrt{((-4-(-8))^2+(6-5)^2)} \\ u=\sqrt{4^2+1^2} \\ u=\sqrt{17} \end{gathered}[/tex][tex]\begin{gathered} w=\sqrt{(-6-(-4))^2+(2-6)^2} \\ w=\sqrt{2^2+(-4)^2} \\ w=\sqrt{20} \end{gathered}[/tex][tex]\begin{gathered} v=\sqrt{(-6-(-8))^2+(2-5)^2} \\ v=\sqrt{2^2+(-3)^2} \\ v=\sqrt{13} \end{gathered}[/tex]Then, the area is:
[tex]\begin{gathered} A=\sqrt{\frac{(\sqrt{13}+\sqrt{17}+\sqrt{20})}{2}(\frac{\sqrt{13}+\sqrt{17}+\sqrt{20}}{2}\sqrt{13})(\frac{\sqrt{13}+\sqrt{17}+\sqrt{20}}{2}\sqrt{17})(\frac{\sqrt{13}+\sqrt{17}+\sqrt{20}}{2}\sqrt{20})} \\ =\sqrt{49} \\ =7 \end{gathered}[/tex]Finally we obtain that the triangle's area is equal to 7 square units.
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