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the vertices of a triangle are located at P (0,0), Q(8,6), R(-3,4). what is the perimeter of this triangle

Trigonometry
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Vertices atr the corner points. plot them out, since this triangle is not sitting on the x plane to find the length of each side you are going to have to use the Pythagorean theorem (a²+b²=c²) or better in the form \[\sqrt{a ^{2}+b ^{2}}=c\] so to find the length of line QR we have\[\sqrt{(8-(-3))^2+(6-4)^2}=\sqrt{11^2+2^2}=\sqrt{121+4}=\sqrt{125}=5\sqrt{5}\] now you can do PQ and PR, once you have the length of all three sides then add them together to get the perimeter.
Line RP is 5 Line RQ² = 4 + 121 = 125 Line RQ = 11.1803398875 Line PQ² = 64 + 36 = 100 PQ = 10 Perimeter = 5 + 11.1803398875 + 100 = 116.1803398875
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