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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.

ID: 7896

Time required: 45 minutes

Activity Overview

In this activity, students will investigate the properties of an isosceles triangle. Then students will construct a \begin{align*}30^\circ - 60^\circ-90^\circ\end{align*} triangle and a \begin{align*}45^\circ-45^\circ - 90^\circ\end{align*} 'triangle to explore the ratios of the lengths of the sides.

Topic: Right Triangles & Trigonometric Ratios

• Calculate the trigonometric ratios for \begin{align*}45^\circ-45^\circ - 90^\circ, 60^\circ - 60^\circ - 60^\circ\end{align*} and \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangles.

Teacher Preparation and Notes

This activity is designed to be used in a high school or middle school geometry classroom.

• If needed, review or introduce the term median of a triangle. Any median of an equilateral triangle is also an altitude, angle bisector, and perpendicular bisector.
• This activity is designed to be student-centered with the teacher acting as a facilitator while students work cooperatively.
• The worksheet guides students through the main ideas of the activity and provides a place for students to record their work. You may wish to have the class record their answers on separate sheets of paper, or just use the questions posed to engage a class discussion.

Associated Materials

Problem 1 – Investigation of 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 \begin{align*}AB = AC\end{align*}.

Using the Perpendicular tool (ZOOM > Perp.), construct a perpendicular from point \begin{align*}A\end{align*} to side \begin{align*}BC\end{align*}. Label the point of intersection of this line with \begin{align*}BC\end{align*} as \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 \begin{align*}x^{-1}\end{align*} ENTER for the letter \begin{align*}D\end{align*}.

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

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

Drag point \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 and CAD.

Will they always be equal?

Problem 2 – Investigation of 30-60-90 Triangles

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

Construct the perpendicular from \begin{align*}A\end{align*} to side \begin{align*}BC\end{align*}. Label the point of intersection as \begin{align*}D\end{align*}, like in Problem 1.

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

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

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

You may choose to have the students hide the segments \begin{align*}AC\end{align*} and \begin{align*}CD\end{align*}. To do this, construct segments \begin{align*}BD\end{align*} and \begin{align*}AB\end{align*} on top of the larger triangle. Then hide the original triangle. Keep the point \begin{align*}C\end{align*}. We will need that point later to resize the triangle.

Measure the three sides of the triangle.

Press GRAPH and select the Calculate tool. Click on the length of \begin{align*}BD\end{align*}, then on the length of \begin{align*}AB\end{align*}. Press the \begin{align*}\div\end{align*} key. Students will see the result 0.5. Move it to the upper corner. Repeat this step to find the ratio of \begin{align*}AD:AB\end{align*} and \begin{align*}AD:BD\end{align*}. These ratios will become important when you start working with trigonometry.

Drag point \begin{align*}C\end{align*} to another location.

What do you notice about the three ratios?

Problem 3 – Investigation of 45-45-90 Triangles

Press the \begin{align*}Y=\end{align*} button and select New to open a new document.

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

Can you explain why \begin{align*}AB = AC\end{align*} and why angle \begin{align*}ACB = \end{align*} angle \begin{align*}ABC\end{align*}?

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

Students should notice that the two angles must be equal, and angle \begin{align*}A\end{align*} is \begin{align*}90^\circ\end{align*}. Therefore, because the sum of the angles in a triangle is \begin{align*}180^\circ\end{align*}, the two angles must be \begin{align*}45^\circ\end{align*} each.

Measure the sides of the triangle. This verifies that \begin{align*}AB = AC\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?

Students should notice that \begin{align*}AC\end{align*} and \begin{align*}AB\end{align*} are equal, so the ratios will always remain the same.

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