Chapter 9.5, Problem 2PSA

To determine

To calculate:Theperimeter of a triangleABC .

Answer to Problem 2PSA

The perimeter of a triangleABCis 18.1.

Explanation of Solution

Given information:

Points are $A=(2,6),B=(5,10)$ and $C=(0,13)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

D $=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

  $\begin{array}{l}AB=\sqrt{{\left(5-2\right)}^2+{\left(10-6\right)}^2}\\ =\sqrt{9+16}\\ =\sqrt{25}\\ =5\end{array}$

  $\begin{array}{l}BC=\sqrt{{\left(5-0\right)}^2+{\left(10-13\right)}^2}\\ =\sqrt{25+9}\\ =\sqrt{34}\end{array}$

  $\begin{array}{l}CA=\sqrt{{\left(2-0\right)}^2+{\left(6-13\right)}^2}\\ =\sqrt{4+49}\\ =\sqrt{53}\end{array}$

  $P(\Delta ABC)=AB+BC+CA=5+\sqrt{34}+\sqrt{53}\approx 18.1$