Question

To the nearest tenth, what is the perimeter of the triangle with vertices at (-2,3) , (3,6), and (2,-2)?

Answer 1

First, we need to draw the points on the plane to get a better idea of the problem, as shown below

We can obtain the distance between the points and then use the next formula

`\begin{gathered} A=\sqrt[]{s(s-a)(s-b)(s-c)} \\ \text{where} \\ s=(a+b+c)/(2) \end{gathered}`

where a, b and c are the sides of the triangle.

In our case,

`\begin{gathered} a=d(AB)=\sqrt[]{(-2-3)^2+(3-6)^2}=\sqrt[]{25+9}=\sqrt[]{34} \\ b=d(BC)=\sqrt[]{(3-2)^2+(6--2)^2}=\sqrt[]{1+64}=\sqrt[]{65} \\ c=d(CA)=\sqrt[]{(2--2)^2+(-2-3)^2}=\sqrt[]{16+25}=\sqrt[]{41} \end{gathered}`

Then, the area is

`\begin{gathered} \Rightarrow s=\frac{\sqrt[]{34}+\sqrt[]{65}+\sqrt[]{41}}{2} \\ \Rightarrow A=\sqrt[]{\frac{\sqrt[]{34}+\sqrt[]{65}+\sqrt[]{41}}{2}(\frac{-\sqrt[]{34}+\sqrt[]{65}+\sqrt[]{41}}{2})(\frac{\sqrt[]{34}-\sqrt[]{65}+\sqrt[]{41}}{2})(\frac{\sqrt[]{34}+\sqrt[]{65}-\sqrt[]{41}}{2})} \end{gathered}`

Simplifying the expression,

`\Rightarrow A=\sqrt[]{((1)/(16))(72+2\sqrt[]{2665})(2\sqrt[]{2665}-72)}`

Therefore,

`\Rightarrow A=\sqrt[]{(1)/(16)(4\cdot2665-72^2)}=\sqrt[]{((1)/(16))5476}=\sqrt[]{342.25}`

Finally,

`\Rightarrow A=18.5`

The area of the triangle is 18.5 square units, and the perimeter is

`P=\sqrt[]{34}+\sqrt[]{65}+\sqrt[]{41}=20.296333\ldots\approx20.3`

The perimeter is 20.3 units