Triangle in Three Dimensions

Calculate the area of triangle ABC with the following vertices: A(1,3,0), B(0,2,5), and C(\textrm{-}1,0,2).
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
Apply the distance formula in three dimensions
AB=\sqrt{(1-0)^2+(3-2)^2+(0-5)^2}
=\sqrt{1+1+25}
=\sqrt{27}
AC=\sqrt{(1+1)^2+(3-0)^2+(0-2)^2}
=\sqrt{4+9+4}
=\sqrt{17}
BC=\sqrt{(0+1)^2+(2-0)^2+(5-2)}^2
=\sqrt{1+4+9}
=\sqrt{14}
Apply Heron’s formula for the area of a triangle ABC with side lengths a,b,c
area =\sqrt{s(s-a)(s-b)(s-c)} where s=1/2(a+b+c)
s=1/2(\sqrt{27}+\sqrt{17}+\sqrt{14})
=6.53
area =\sqrt{6.53(6.53-\sqrt{27})(6.53-\sqrt{17})(6.53-\sqrt{14})}
=\sqrt{6.53(1.33)(2.41)(2.79)}
=7.64

Answer: 7.64 square units

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About mvtrinh

Retired high school math teacher.
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