Answer
$AB^2+AC^2=BC^2$
$41+41=82$
$82=82$
$Area = 20.5$
Work Step by Step
To show the following, first we have to find length of each side. So we will consider the longest as Hypotenuse and other two as legs.
Using the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$AB=\sqrt{(11-6)^2+(-3+7)^2}=\sqrt{25+16}=\sqrt{41}$
$BC=\sqrt{(2-11)^2+(-2+3)^2}=\sqrt{81+1}=\sqrt{82}$
$AC=\sqrt{(2-6)^2+(-2+7)^2}=\sqrt{16+25}=\sqrt{41}$
According to the Pythagorean Theorem: $AB^2+AC^2=BC^2$, where $BC$ is the biggest side (Hypotenuse) while $AB$ and $AC$ are smaller sides (legs).
$41+41=82$
$82=82$
The triangle is right-angle triangle.
$A=AB\times AC \times \frac{1}{2}=\sqrt{41} \times \sqrt{41} \times \frac{1}{2}= \frac{41}{2}=20.5$
