Question

A triangle has a verticles at (1,3), (2,-3) and (-1,-1) what is the approximate perimeter of the triangle

Answer 1

Perimeter of any figure is the sum of all sides

then to calculate sides of the triangle , we need the distance between each point, the distance between two points will be the length of a side

then to find the distance between two points we use the formula

`d=\sqrt[]{(x2-x1)^2+(y2-y1)^2}^{}`

(1,3) and (2,-3)

replacing

`\begin{gathered} d_1=\sqrt[]{(1-2)^2+(3-(-3))^2} \\ d_1=\sqrt[]{(-1)^2+(6)^2} \\ d_1=\sqrt[]{1+36} \\ d_1=\sqrt[]{37}\approx6.08 \end{gathered}`

first distance side is 6.08 units

(1,3) and (-1,-1)

replacing

`\begin{gathered} d_2=\sqrt[]{(1-(-1))^2+(3-(-1))^2} \\ d_2=\sqrt[]{(2)^2+(4)^2} \\ d_2=\sqrt[]{4+16} \\ d_2=\sqrt[]{20}\approx4.47 \end{gathered}`

second distance is 4.47 units

(2,-3) and (-1,-1)

replacing

`\begin{gathered} d_3=\sqrt[]{(2-(-1))^2+(-3-(-1))^2} \\ d_3=\sqrt[]{(3)^2+(-2)^2} \\ d_3=\sqrt[]{9+4} \\ d_3=\sqrt[]{13}\approx3.6 \end{gathered}`

Third distance is 3.6 units

Perimeter

sum all distances

`\begin{gathered} P=\sqrt[]{37}+\sqrt[]{20}+\sqrt[]{13} \\ P\approx14.16 \end{gathered}`

Perimeter of the triangle is 14.16 units