Question

11. On the set of axes below, triangle KLM has verticeswhose coordinates are K(-6, -2), L(6,-4), andM(2,6).What is the perimeter of triangle KLM? Round youranswer to the nearest tenth.

Answer 1

Step 1: Write out the formula for the distance between two points on the Cartesian plane

`d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}`
`\begin{gathered} \text{ Where} \\ (x_1,y_1)\text{ is a point on the Cartesian plane} \\ (x_2,y_2)\text{ is another point on the Cartesian plane} \\ d=\text{ the distance betw}een\text{ points (x1,y1) and (x2,y2)} \end{gathered}`

Step 2: Use the formula to find the distance KM between points K and M.

In this case,

`\begin{gathered} (x_1,y_1)=(-6,-2) \\ (x_2,y_2)=(2,6_{}) \\ \text{ Therefore,} \end{gathered}`
`\begin{gathered} d=\sqrt[]{(2-(-6))^2+(6-(-2))^2}=\sqrt[]{(2+6)^2+(6+2)^2} \\ d=\sqrt[]{8^2+8^2}=\sqrt[]{64+64}=\sqrt[]{128}\approx11.3 \end{gathered}`

Hence, KM = 11.3 units

Step 3: Use the formula to find the distance KL between points K and L.

In this case,

`\begin{gathered} (x_1,y_1)=(-6,-2) \\ (x_2,y_2)=(6,-4_{}) \\ \text{ Therefore,} \end{gathered}`
`\begin{gathered} d=\sqrt[]{(6-(-6))^2+(-4-(-2))^2}=\sqrt[]{(6+6)^2+(-4+2)^2} \\ d=\sqrt[]{12^2+(-2)^2}=\sqrt[]{144+4}=\sqrt[]{148}\approx12.2 \end{gathered}`

Hence, KL = 12.2 units

Step 4: Use the formula to find the distance LM between points L and M.

In this case,

`\begin{gathered} (x_1,y_1)=(6,-4) \\ (x_2,y_2)=(2,6_{}) \\ \text{ Therefore,} \end{gathered}`
`\begin{gathered} d=\sqrt[]{(2-6)^2+(6-(-4))^2}=\sqrt[]{(-4)^2+(6+4)^2} \\ d=\sqrt[]{16^{}+(10)^2}=\sqrt[]{16+100}=\sqrt[]{116}\approx10.8 \end{gathered}`

Hence, LM = 18.9 units

Step 4: Find the perimeter of the triangle KLM

The perimeter of the triangle is given by the sum of the sides of the triangle.

That is,

`\text{ the perimeter }=KL+KM+LM`

Therefore,

`\text{ the perimeter }=11.3+12.2+10.8=34.3\text{ units}`

Hence, the perimeter of triangle KLM is 34.3 units