Answer:
Is an acute triangle
Step-by-step explanation:
we have
[tex]G(7, 3),H(9, 0),I(5, -1)[/tex]
so
The polygon is a triangle
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
Remember that
If applying the Pythagoras Theorem
[tex]c^{2}=a^{2}+b^{2}[/tex] -----> is a right triangle
[tex]c^{2}>a^{2}+b^{2}[/tex] -----> is an obtuse triangle
[tex]c^{2}<a^{2}+b^{2}[/tex] -----> is an acute triangle
where
c is the greater side
step 1
Find the distance GH
[tex]G(7, 3),H(9, 0),I(5, -1)[/tex]
substitute
[tex]d=\sqrt{(0-3)^{2}+(9-7)^{2}}[/tex]
[tex]d=\sqrt{(-3)^{2}+(2)^{2}}[/tex]
[tex]GH=\sqrt{13}\ units[/tex]
step 2
Find the distance HI
[tex]G(7, 3),H(9, 0),I(5, -1)[/tex]
substitute
[tex]d=\sqrt{(-1-0)^{2}+(5-9)^{2}}[/tex]
[tex]d=\sqrt{(-1)^{2}+(-4)^{2}}[/tex]
[tex]HI=\sqrt{17}\ units[/tex]
step 3
Find the distance GI
[tex]G(7, 3),H(9, 0),I(5, -1)[/tex]
substitute
[tex]d=\sqrt{(-1-3)^{2}+(5-7)^{2}}[/tex]
[tex]d=\sqrt{(-4)^{2}+(-2)^{2}}[/tex]
[tex]GI=\sqrt{20}\ units[/tex]
step 4
Let
[tex]c=GI=\sqrt{20}\ units[/tex]
[tex]a=HI=\sqrt{17}\ units[/tex]
[tex]b=GH=\sqrt{13}\ units[/tex]
Find [tex]c^{2}[/tex] ------> [tex]c^{2}=(\sqrt{20})^{2}=20[/tex]
Find [tex]a^{2}+b^{2}[/tex] ----> [tex](\sqrt{17})^{2}+(\sqrt{13})^{2}=30[/tex]
Compare
[tex]20 < 30[/tex]
therefore
Is an acute triangle
