Math, asked by kyabhai634, 5 months ago

Solve this queation .​

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Answers

Answered by BrainlicaLDoll
109

Check the attachment :-

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Answered by TheRise
883

Given :

  • Verties of right angled isoceles triangle

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To prove :

  • Prove the points
  • (1, 10), (2, 1) and (-7, 0) ?

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\large\star As we know that,

\large\dag Formula Used :

  • \boxed{\sf{\sqrt{\bigg(x_{2}\:x_{1}\bigg)^{2}\:+\:\bigg(y_{2}\:y_{1}\bigg)^{2}}}}\large\dag

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Solution :

{\sf{AB\:=\:\sqrt{(2\:-\:1)^{2}\:+\:(1\:-\:10)^{2}}}}

\:\:\:\:\:\:\:\:{\sf{=\:\sqrt{(1)^{2}\:+\:(-9)^{2}}}}

\:\:\:\:\:\;\:\:{\sf{=\:\sqrt{1\:+\:18}}}

\:\:\:\:\:\:\:\:{\sf{=\:\sqrt{82}}}

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{\sf{BC\:=\:\sqrt{(-7\:-\:-2)^{2}\:+\:(0\:-\:-1)^{2}}}}

\:\:\:\:\:\:\:\:{\sf{=\:\sqrt{(-9)^{2}\:+\:(-1)^{2}}}}

\:\:\:\:\:\;\:\:{\sf{=\:\sqrt{81\:+\:1}}}

\:\:\:\:\:\:\:\:{\sf{=\:\sqrt{82}}}

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{\sf{AC\:=\:\sqrt{(-7\:-\:-1)^{2}\:+\:(0\:-\:10)^{2}}}}

\:\:\:\:\:\:\:\:{\sf{=\:\sqrt{(-8)^{2}\:+\:(10)^{2}}}}

\:\:\:\:\:\;\:\:{\sf{=\:\sqrt{64\:+\:100}}}

\:\:\:\:\:\:\:\:{\sf{=\:\sqrt{164}}}

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\therefore{\sf{AB\:=\:BC\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sf{\large\star\bigg(This\: means}\:\bf{ABC}\:\sf{is\:an}\:\bf{isoceles\:triangle\bigg)\large\star}}}

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Also, we have,

{\sf{AB^2\:+\:BC^2\:=\:82\:+\:82\:=\:164}}

{\sf{AC^2\:=\:164}}

\therefore{\sf{AB^2\:+\:BC^2\:=\:AC^2\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\large\star\bigg(Pythagoras\: theorem\bigg)\large\star}}

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