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avanturin [10]
3 years ago
12

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

Mathematics
1 answer:
m_a_m_a [10]3 years ago
5 0

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

 

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but am very much unsure if this is correct or not.



Perhaps you meant 

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<span>
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