Math, asked by prokarthik, 9 months ago

Find the perimeter of a triangle whose vertices have the (3,10) (5,2) (4,12)

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Answered by vedangISRO
1

Answer:

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Answered by Anonymous
7

 \sf \fcolorbox{red}{pink}{ \huge{Solution :)}}

Let , the vertices of triangle be A(3,10) , B(5,2) and C(4,12)

The distance between point A(3,10) and B (5,2) is

\sf \mapsto AB =  \sqrt{ {(5- 3)}^{2} +  {(2 - 10)}^{2}  }  \\  \\\sf \mapsto AB =  \sqrt{ {(2)}^{2} +  {( - 8)}^{2}  }   \\  \\ \sf \mapsto AB =  \sqrt{4 + 64}  \\  \\\sf \mapsto  AB =  \sqrt{68}  \:  \: units

The distance between the point B(5,2) and C(4,12) is

\sf \mapsto BC =  \sqrt{ {(4 - 5)}^{2}  +  {(12 - 2)}^{2} }  \\  \\ \sf \mapsto BC =  \sqrt{ {(1)}^{2} +  {(10)}^{2}  } \\  \\ \sf \mapsto BC =  \sqrt{1 + 100}   \\  \\ \sf \mapsto BC =  \sqrt{101}  \:  \: units

The distance between A(3,10) and C(4,12) is

 \sf \mapsto AC =  \sqrt{ {(4 - 3)}^{2}  +  {(12 - 10)}^{2} }  \\  \\\sf \mapsto AC =  \sqrt{ {(1)}^{2} +  {(2)}^{2}  }  \\  \\  \sf \mapsto AC = \sqrt{1 + 4}  \\  \\ \sf \mapsto AC =  \sqrt{5}  \:  \: units

Therefore , the perimeter of triangle whose vertices are A(3,10) , B(5,2) and C(4,12) is

 \sf \hookrightarrow Perimeter = \sqrt{68}  +  \sqrt{101}  +  \sqrt{5}  \\  \\  \sf \hookrightarrow Perimeter = 20.53  \:  \: units

Hence , the perimeter of triangle is 20.53 units

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