Calculate the lengths of the sides of the triangle:
\(\overline{DE}=\sqrt{(8-7)^2+(1-3)^2}=\sqrt{1^2+(-2)^2}=\sqrt{1+4}=\sqrt{5} \\
\overline{DF}=\sqrt{(4-7)^2+(-1-3)^2}=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5 \\
\overline{EF}=\sqrt{(4-8)^2+(-1-1)^2}=\sqrt{(-4)^2+(-2)^2}=\sqrt{16+4}= \\
=\sqrt{4(4+1)}=\sqrt{4 \times 5}=2\sqrt{5}\)
The triangle has three different sides, so it's a scalene triangle.
Now if c is the longest side of the triangle, and a and b are the shorter sides, then:
- if \(c^2=a^2+b^2\), the triangle is right
- if \(c^2, the triangle is acute
- if \(c^2>a^2+b^2\), the triangle is obtuse
The longest side is 5.
\(c^2=5^2=25 \\
a^2+b^2=(\sqrt{5})^2+(2\sqrt{5})^2=5+4 \times 5=5+20=25 \\ \Downarrow \\
c^2=a^2+b^2\)
The square of the longest side is equal to the sum of the squares of the shorter sides, so it's a right triangle.
The triangle is a right scalene triangle.
