Asked by Ronnie
To the nearest tenth, find the perimeter of ABC with vertices A(-2-2), B(0,5) and C(3,-1).
Answers
Answered by
Elena
AB=sqrt{(x₂-x₁)²+(y₂-y₁)²} =
=sqrt{(0-(-2))²+(5-(-2)) ²} =
=sqrt(4+49)=7.3
BC = sqrt{(x₃-x₂ )²+(y₃-y₂)²} =
=sqrt{(3-0)²+(-1-5) ²} =
=sqrt(9+36)=6.7
CA = sqrt{(x₁-x₃ )²+(y₁-y₃)²} =
=sqrt{(-2-3)²+(-2+1) ²} =
=sqrt(25+1)=5.1
P=AB+BC+CA = 7.3+6.7+5.1 =14.1
=sqrt{(0-(-2))²+(5-(-2)) ²} =
=sqrt(4+49)=7.3
BC = sqrt{(x₃-x₂ )²+(y₃-y₂)²} =
=sqrt{(3-0)²+(-1-5) ²} =
=sqrt(9+36)=6.7
CA = sqrt{(x₁-x₃ )²+(y₁-y₃)²} =
=sqrt{(-2-3)²+(-2+1) ²} =
=sqrt(25+1)=5.1
P=AB+BC+CA = 7.3+6.7+5.1 =14.1
Answered by
Abdullah
19.1 units
AB= 7.3
BC=6.8
CA=5
AB= 7.3
BC=6.8
CA=5
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