Question

Classify ABC by its sides. Then determine whether it is a right triangle.

Answer 1

Answer:

∴Given Δ ABC is not a right-angle triangle

a= AB = √45 = 3√5

b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

            $AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

            AB = $\sqrt{(9-3)^{2}+(6-3)^{2} } = \sqrt{6^{2}+3^{2} } =\sqrt{45}$

Given vertices are  B(6,9) and C( 6,-3)

       $B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

             =  $\sqrt{(-3-9)^{2}+(6-6)^{2} } =\sqrt{12^{2} } = 12$

    BC = 12

Given vertices are  A(3,3) and C( 6,-3)

 $AC = \sqrt{(6-3)^{2}+(-3-3)^{2} } = \sqrt{9+36} = \sqrt{45}$

AC² = AB²+BC²

45  = 45+144

 45  ≠ 189

∴Given Δ ABC is not a right angle triangle