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Perimeter of Triangle

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ABC A(1,3) B(6,10) C(11,18)?

asked Dec 8, 2014 in ALGEBRA 2 by anonymous

1 Answer

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Vertices of the triangle are  A(1,3)  ,  B(6,10)  ,  C(11,18)

Formula : Distance between two points = √[(x2-x1)²+(y2-y1)²]

 

A = (x1, y1) = (1,3) and B = (x2, y2) = (6,10)

AB = √[6-1)²+(10-3)²] = √[(5)²+(7)²] = √[ 25+49 ] = √74

 

B = (x1, y1) = (6,10) and C = (x2, y2) = (11,18)

BC =  = √[11-6)²+(18-10)²] = √[(5)²+(8)²] = √[ 25+64 ] = √89

 

C = (x1, y1) = (11,18)  and A = (x2, y2) = (1,3)

CA =  = √[1-11)²+(3-18)²] = √[(-10)²+(-15)²] = √[ 100+225] = √325 = 5√13

 

Perimeter = AB + BC + CA = √74 + √89 + 5√13

=  8.61 + 9.43 + 18.03 = 36.07.

Solution : Perimeter of the triangle is 36.07 units.

answered Dec 8, 2014 by Shalom Scholar

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