Question

the verticals of a triangle are located at (-3, -1), (2, 3), and (5, 2). Record the triangles perimeter to the nearest whole number.

Answer 1

To find the perimeter we have to find the distance between each pair of points and add them. Therefore, we get that:

`\begin{gathered} d_(12)=\sqrt[]{(-3-2)^2+(-1-3)^2}=\sqrt[]{25+16}=\sqrt[]{41}\approx6.4 \\ d_(23)=\sqrt[]{(2-5)^2+(3-2)^2}=\sqrt[]{9+1}=\sqrt[]{10}\approx3.2 \\ d_(13)=\sqrt[]{(-3-5)^2+(-1-2)^2}=\sqrt[]{64+9}=\sqrt[]{73}\approx8.5 \\ P=d_(12)+d_(23)+d_(13)=6.4+3.2+8.5=18.1\approx18 \end{gathered}`

so the perimeter is 18