Answer:
The area of the triangle is 84.511 square units.
Step-by-step explanation:
First step, we need to calculate lengths of line segments AB, BC and AC by Pythagorean Theorem:
AB:
[tex]AB = \sqrt{(12-0)^{2}+(3-8)^{2}}[/tex]
[tex]AB = 13[/tex]
BC:
[tex]BC = \sqrt{(7-12)^{2}+(-9-3)^{2}}[/tex]
[tex]BC = 13[/tex]
AC:
[tex]AC = \sqrt{(7-0)^{2}+(-9-8)^{2}}[/tex]
[tex]AC \approx 18.385[/tex]
Now, we can determine the area of the triangle ([tex]A[/tex]) by Heron's formula:
[tex]A = \sqrt{s\cdot (s-AB)\cdot (s-BC)\cdot (s-AC)}[/tex] (1)
[tex]s = \frac{AB+BC+AC}{2}[/tex] (2)
Where [tex]s[/tex] is the semiperimeter of the triangle.
If we know that [tex]AB = 13[/tex], [tex]BC = 13[/tex] and [tex]AC \approx 18.385[/tex], then the area of the triangle is:
[tex]s = 22.193[/tex]
[tex]A = 84.511[/tex]
The area of the triangle is 84.511 square units.
