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Find the perimeter of the triangle with vertices at A(0,0), B(-4,-3), and C(-9,0)simplify the answer using radicals as needed

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In order to find the perimeter of the given figure, use the following formula to find the distance in between the given points:


d=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}

where (x1,y1) and (x2,y2) are the coordinates of two points.

calculate the distance AB, BC, CA, which are the sides of the triangle:


\begin{gathered} AB=\sqrt[]{(-3-0)^2+(-4-0)^2}=\sqrt[]{9+16}=\sqrt[]{25}=5 \\ BC=\sqrt[]{(0-(-3))^2+(-9-(-4))^2}=\sqrt[]{9+25}=\sqrt[]{34} \\ CA=\sqrt[]{(0-0)^2+(0-(-9))^2}=\sqrt[]{81}=9 \end{gathered}

The perimeter of the triangle is:

P = AB + BC + CA = 5 + √34 + 9 = 14 + √34 ≈ 19.8

User Dscyre Scotti
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