Chapter 13.1, Problem 29WE

To determine

To show that the triangle JAN is similar to triangle RFK

Explanation of Solution

Given information:

The triangle has vertices is

  $J\left(-2,-2\right),A\left(4,-2\right),N\left(2,2\right)\text{ ,}R\left(8,1\right)\text{ ,}F\left(8,4\right)\text{ and }K\left(6,3\right)$

Proof:

The triangle JAN is similar to triangle RFK can be show as

Use distance formula to show that triangle JAN is similar to triangle RFK

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

First find JA, JN and AN

Therefore,

  $\begin{array}{l}JA=\sqrt{{\left(4-\left(-2\right)\right)}^2+{\left(-2-\left(-2\right)\right)}^2}\left[\text{since }J\left(-2,-2\right)\text{ and }A\left(4,-2\right)\right]\\ JA=\sqrt{{\left(4+2\right)}^2+{\left(-2+2\right)}^2}\\ JA=\sqrt{{\left(6\right)}^2+{\left(0\right)}^2}\\ JA=\sqrt{36}\\ JA=6\end{array}$

  $\begin{array}{l}JN=\sqrt{{\left(2-\left(-2\right)\right)}^2+{\left(2-\left(-2\right)\right)}^2}\left[\text{since }J\left(-2,-2\right)\text{ and }N\left(2,2\right)\right]\\ JN=\sqrt{{\left(2+2\right)}^2+{\left(2+2\right)}^2}\\ JN=\sqrt{{\left(4\right)}^2+{\left(4\right)}^2}\\ JN=\sqrt{16+16}\\ JN=\sqrt{32}\\ JN=4\sqrt{2}\end{array}$

  $\begin{array}{l}AN=\sqrt{{\left(2-4\right)}^2+{\left(2-\left(-2\right)\right)}^2}\left[\text{since }A\left(4,-2\right)\text{ and }N\left(2,2\right)\right]\\ AN=\sqrt{{\left(2-4\right)}^2+{\left(2+2\right)}^2}\\ AN=\sqrt{{\left(-2\right)}^2+{\left(4\right)}^2}\\ AN=\sqrt{4+16}\\ AN=\sqrt{20}\\ AN=2\sqrt{5}\end{array}$

Again, find RF , RK and FK

Therefore,

  $\begin{array}{l}RF=\sqrt{{\left(8-8\right)}^2+{\left(4-1\right)}^2}\left[\text{since }R\left(8,1\right)\text{ and }F\left(8,4\right)\right]\\ RF=\sqrt{{\left(0\right)}^2+{\left(3\right)}^2}\\ RF=\sqrt{0+9}\\ RF=\sqrt{9}\\ RF=3\end{array}$

  $\begin{array}{l}RK=\sqrt{{\left(6-8\right)}^2+{\left(3-1\right)}^2}\left[\text{since }R\left(8,1\right)\text{ and }K\left(6,3\right)\right]\\ RK=\sqrt{{\left(-2\right)}^2+{\left(2\right)}^2}\\ RK=\sqrt{4+4}\\ RK=\sqrt{8}\\ RK=2\sqrt{2}\end{array}$

  $\begin{array}{l}FK=\sqrt{{\left(6-8\right)}^2+{\left(3-4\right)}^2}\left[\text{since }F\left(8,4\right)\text{ and }K\left(6,3\right)\right]\\ FK=\sqrt{{\left(-2\right)}^2+{\left(-1\right)}^2}\\ FK=\sqrt{4+1}\\ FK=\sqrt{5}\end{array}$

  $\begin{array}{l}\frac{JA}{RF}=\frac{6}{3}=2\\ \frac{JN}{RK}=\frac{4\sqrt{2}}{2\sqrt{2}}=2\\ \frac{AN}{FK}=\frac{2\sqrt{5}}{\sqrt{5}}=2\end{array}$

  $\frac{JA}{RF}=\frac{JN}{RK}=\frac{AN}{FK}=2$

Hence,

the triangle JAN is similar to triangle RFK