Q. 30 - Geometry

Question

In Exercises 27-32 find and then compare lengths of segments.

 The vertices of triangle KAT and triangle JES are K(3, -1), A(2, 6), T(5, 1), I(-4, 1), E(-3, -6), and S(-6, -1). What word best describes the relationship between triangle KAT and triangle IES?

Step-by-Step Solution

Answer

Triangle KAT and triangle IES are similar.

Step-1 – Given

The given vertices of the triangle are 

$K\left(3,-1\right),A\left(2,6\right),T\left(5,1\right)\text{ ,}I\left(-4,1\right)\text{ ,}E\left(-3,-6\right)\text{ and }S\left(-6,-1\right)$

Step-2 – To determine

We have to describe the relationship between triangle KAT and triangle IES.

Step-3 – Proof

We will find the lengths KA, KT and AT using the distance formula.

$\begin{array}{l}KA=\sqrt{{\left(2-3\right)}^2+{\left(6-\left(-1\right)\right)}^2}\left[\text{since }K\left(3,-1\right)\text{ and }A\left(2,6\right)\right]\\ KA=\sqrt{{\left(2-3\right)}^2+{\left(6+1\right)}^2}\\ KA=\sqrt{{\left(-1\right)}^2+{\left(7\right)}^2}\\ KA=\sqrt{1+49}\\ KA=\sqrt{50}\\ KA=5\sqrt{2}\end{array}$

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

$\begin{array}{l}AT=\sqrt{{\left(5-2\right)}^2+{\left(1-6\right)}^2}\left[\text{since }A\left(2,6\right)\text{ and }T\left(5,1\right)\right]\\ AT=\sqrt{{\left(3\right)}^2+{\left(-5\right)}^2}\\ AT=\sqrt{9+25}\\ AT=\sqrt{34}\end{array}$

Then we find IE, IS and ES using distance formula:

$\begin{array}{l}IE=\sqrt{{\left(-3-\left(-4\right)\right)}^2+{\left(-6-1\right)}^2}\left[\text{since }I\left(-4,1\right)\text{ and }E\left(-3,-6\right)\right]\\ IE=\sqrt{{\left(-3+4\right)}^2+{\left(-6-1\right)}^2}\\ IE=\sqrt{{\left(1\right)}^2+{\left(-7\right)}^2}\\ IE=\sqrt{1+49}\\ IE=\sqrt{50}\\ IE=5\sqrt{2}\end{array}$

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

$\begin{array}{l}ES=\sqrt{{\left(-6-\left(-3\right)\right)}^2+{\left(-1-\left(-6\right)\right)}^2}\left[\text{since }E\left(-3,-6\right)\text{ and }S\left(-6,-1\right)\right]\\ ES=\sqrt{{\left(-6+3\right)}^2+{\left(-1+6\right)}^2}\\ ES=\sqrt{{\left(-3\right)}^2+{\left(5\right)}^2}\\ ES=\sqrt{9+25}\\ ES=\sqrt{34}\end{array}$

Then we find the ratio of the corresponding sides:

$\begin{array}{l}\frac{KA}{IE}=\frac{5\sqrt{2}}{5\sqrt{2}}=1\\ \frac{KT}{IS}=\frac{2\sqrt{2}}{2\sqrt{2}}=1\\ \frac{AT}{ES}=\frac{\sqrt{34}}{\sqrt{34}}=1\end{array}$

So,

$\frac{KA}{IE}=\frac{KT}{IS}=\frac{AT}{ES}=1$

Hence, the relationship between triangle KAT and triangle IES is: they are similar.