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

Problem 1 – Investigation of 45^\circ-45^\circ-90^\circ 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.

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

Construct line segments BD and CD (\pi > Segment) and then measure the segments (GRAPH > Measure > D. & Length).

BD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && CD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}

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

Drag point C 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.

\angle{BAD} = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && \angle{CAD} = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}

Will they always be equal? _____________________________

Problem 2 – Investigation of 30^\circ-60^\circ-90^\circ Triangles

Open the file EQUIL. Note that all three angles are 60^\circ angles.

Construct the perpendicular from A to side BC. Label the point of intersection as D.

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

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

This completes the construction of two 30^\circ-60^\circ-90^\circ triangles. We will work only with the triangle BAD.

Measure the three sides of triangle BAD.

AB = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && BD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && AD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}

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

BD:AB = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && AD:AB = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && AD:BD = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}

Drag point C to another location. What do you notice about the three ratios?

Problem 3 – Investigation of 45^\circ-45^\circ-90^\circ Triangles

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

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

Use the compass tool with center A and radius AB. The circle will intersect the perpendicular line at C.

Hide the circle and construct segments AC and BC.

Explain why AB = AC and why angle ACB = \text{angle} \ ABC?

Why are these two angles 45^\circ each?

Measure the sides of the triangle.

AC = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && BC = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} && AB = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}

Use the Calculate tool to find the ratio of AC:BC and AC:AB. Once again, these ratios will be important when you study trigonometry.

Drag point B and observe what happens to the sides and ratios.

Why do the ratios remain constant while the sides change?

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Date Created:

Feb 23, 2012

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

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