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# 9.2: Investigating Special Triangles

Difficulty Level: At Grade Created by: CK-12

This activity is intended to supplement Geometry, Chapter 8, Lesson 4.

## Problem 1 – Investigation of 45∘−45∘−90∘\begin{align*}45^\circ-45^\circ-90^\circ\end{align*} Triangles

First, turn on your TI-84 and press APPS. Arrow down until you see Cabri Jr and press ENTER. Open the file ISOSC. This file has a triangle with an isosceles triangle with AB=AC\begin{align*}AB = AC\end{align*}.

Using the Perpendicular tool (ZOOM > Perp.), construct a perpendicular from point A\begin{align*}A\end{align*} to side BC\begin{align*}BC\end{align*}. Label the point of intersection of this line with BC\begin{align*}BC\end{align*} as D\begin{align*}D\end{align*}. To name the point, they need to select the Alph-Num tool (GRAPH > Alph-Num), select the point, and press x1\begin{align*}x^{-1}\end{align*} ENTER for the letter D\begin{align*}D\end{align*}.

Construct line segments BD\begin{align*}BD\end{align*} and CD\begin{align*}CD\end{align*} (π\begin{align*}\pi\end{align*} > Segment) and then measure the segments (GRAPH > Measure > D. & Length).

BD=CD=\begin{align*}BD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && CD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

Would you have expected these segments to be equal in length?

Drag point C\begin{align*}C\end{align*} to see the effect on the lengths of the line segments. It appears that the perpendicular from the vertex always bisects the opposite side. Measure the angles BAD\begin{align*}BAD\end{align*} and CAD\begin{align*}CAD\end{align*}.

BAD=CAD=\begin{align*}\angle{BAD} = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && \angle{CAD} = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

Will they always be equal? _____________________________

## Problem 2 – Investigation of 30∘−60∘−90∘\begin{align*}30^\circ-60^\circ-90^\circ\end{align*} Triangles

Open the file EQUIL. Note that all three angles are 60\begin{align*}60^\circ\end{align*} angles.

Construct the perpendicular from A\begin{align*}A\end{align*} to side BC\begin{align*}BC\end{align*}. Label the point of intersection as D\begin{align*}D\end{align*}.

From the construction above, we know that D\begin{align*}D\end{align*} bisects BC\begin{align*}BC\end{align*} and that mBAD=30\begin{align*}m\angle{BAD} = 30^\circ\end{align*}.

Construct segment BD\begin{align*}BD\end{align*}. We now have triangle BAD\begin{align*}BAD\end{align*} where mD=90\begin{align*}m\angle{D} = 90^\circ\end{align*}, mB=60\begin{align*}m\angle{B} = 60^\circ\end{align*} and mA=30\begin{align*}m\angle{A} = 30^\circ\end{align*}. We also have triangle ACD\begin{align*}ACD\end{align*} where mA=30\begin{align*}m\angle{A} = 30^\circ\end{align*}, mC=60\begin{align*}m\angle{C} = 60^\circ\end{align*} and mD=90\begin{align*}m\angle{D} = 90^\circ\end{align*}.

This completes the construction of two 306090\begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangles. We will work only with the triangle BAD\begin{align*}BAD\end{align*}.

Measure the three sides of triangle BAD\begin{align*}BAD\end{align*}.

AB=BD=AD=\begin{align*}AB = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && BD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && AD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

Press σ\begin{align*}\sigma\end{align*} and select the Calculate tool. Click on the length of BD\begin{align*}BD\end{align*}, then on the length of AB\begin{align*}AB\end{align*}. Press the \begin{align*}\infty\end{align*} key. Move it to the upper corner. Repeat this step to find the ratio of AD:AB\begin{align*}AD:AB\end{align*} and AD:BD\begin{align*}AD:BD\end{align*}. These ratios will become important when you start working with trigonometry.

BD:AB=AD:AB=AD:BD=\begin{align*}BD:AB = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && AD:AB = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && AD:BD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

Drag point C\begin{align*}C\end{align*} to another location. What do you notice about the three ratios?

## Problem 3 – Investigation of 45∘−45∘−90∘\begin{align*}45^\circ-45^\circ-90^\circ\end{align*} Triangles

Press the o button and select New to open a new document.

To begin the construction of the 454590\begin{align*}45^\circ-45^\circ-90^\circ\end{align*} triangle, construct line segment AB\begin{align*}AB\end{align*} and a perpendicular to AB\begin{align*}AB\end{align*} at A\begin{align*}A\end{align*}.

Use the compass tool with center \begin{align*}A\end{align*} and radius \begin{align*}AB\end{align*}. The circle will intersect the perpendicular line at \begin{align*}C\end{align*}.

Hide the circle and construct segments \begin{align*}AC\end{align*} and \begin{align*}BC\end{align*}.

Explain why \begin{align*}AB = AC\end{align*} and why angle \begin{align*}ACB = \text{angle} \ ABC\end{align*}?

Why are these two angles \begin{align*}45^\circ\end{align*} each?

Measure the sides of the triangle.

\begin{align*}AC = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && BC = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && AB = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

Use the Calculate tool to find the ratio of \begin{align*}AC:BC\end{align*} and \begin{align*}AC:AB\end{align*}. Once again, these ratios will be important when you study trigonometry.

Drag point \begin{align*}B\end{align*} and observe what happens to the sides and ratios.

Why do the ratios remain constant while the sides change?

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Feb 23, 2012
Nov 03, 2014
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TI.MAT.ENG.SE.1.Geometry.9.2