Question

Determine, to the nearest tenth, the perimeter of the triangle shown in the accompanying diagram.A. 23.3B. 24.9C. 29.7D. 28.5Image Link-https://www.founderseducation.net/assets/images/f0a910c5c9993586ab25f6badfb6f65cae458348.gif

Answer 1

Given:

The vertices of triangle A(-9, 6), B(-3, 10) and C(-2, 2).

Required:

Determine the perimeter of the triangle.

Explanation:

The concept required:

Perimeter is the distance around the edge of a shape.

Distance formula:

$\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ Where, \\ d(distance) \\ (x_1,y_1),(x_2,y_2)\text{ are points.} \end{gathered}$

Now,

$\begin{gathered} AB=\sqrt{(-3+9)^2+(10-6)^2} \\ AB=\sqrt{6^2+4^2} \\ AB=\sqrt{36+16} \\ AB=\sqrt{52} \end{gathered}$$\begin{gathered} BC=\sqrt{(-2+3)^2+(2-10)^2} \\ BC=\sqrt{1^2+8^2} \\ BC=\sqrt{1+64} \\ BC=\sqrt{65} \end{gathered}$$\begin{gathered} CA=\sqrt{(-9+2)^2+(6-2)^2} \\ CA=\sqrt{7^2+4^2} \\ CA=\sqrt{49+16} \\ CA=\sqrt{65} \end{gathered}$

Now, the perimeter of triangle

$\begin{gathered} Perimeter(P)=AB+BC+CA \\ P=\sqrt{52}+\sqrt{65}+\sqrt{65} \\ P=\sqrt{52}+2\sqrt{65} \\ P=23.3 \end{gathered}$

Answer:

The perimeter of the triangle equal 23.3