Question

9) Find the area of triangle ABC with vertices A(3,5), B(4,-5), and C(-6,-5).

Answer 1

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:

$\text{Area}=\sqrt[]{s(s-CA)(s-AB)(s-BC)}$

And "s" is equal to=(CA+AB+BC)/2=(√181+√101+10)/2

So, the area is equal to:

$\begin{gathered} Area=\sqrt[]{\frac{\sqrt{181}+√101+10}{2}(\frac{\sqrt{181}+√101+10}{2}-\sqrt[]{181})(\frac{\sqrt{181}+√101+10}{2}-\surd101)(\frac{\sqrt{181}+√101+10}{2}-10)} \\ =\sqrt[]{2500} \\ =\sqrt[]{50}^2 \\ =50 \end{gathered}$

Finally we obtain that area of triangle ABC is equal to 50 square units.