GeoGebra Tutorial 7 – Sliders and Rotation

This is the seventh tutorial of the GeoGebra Intermediate Tutorial Series. If this is your first time to use GeoGebra, please read the GeoGebra Essentials Series.

In the Graphs and Sliders posts (click here and here ), we have discussed how to use number sliders.  In this tutorial, we use the Angle slider to rotate a triangle in order to show that its angle sum is 180 degrees. This is the same GeoGebra worksheet shown in my Parallel Lines and Transversals post, but we will change some of the labels. Although this tutorial is the seventh of the GeoGebra Tutorial Series.

Figure 1 – Rotated triangles using sliders.

Construction Overview

The construction will start by drawing line AB and constructing triangle ABC using the Polygon tool. Afterwards, we reveal the interior angle measures of the triangle and create two angle sliders namely $\alpha$  and $\beta$. Next, we rotate the triangle 180 clockwise about the midpoint of BC producing triangle A’B’C’ (see Figure 1-B). We then repeat the process, and rotate triangle A’B’C’ 180 degrees clockwise about the midpoint of A’C’ to produce A’’B’’C’’.

Part I – Constructing Triangle ABC

 1.) Open GeoGebra and select Geometry from the Perspectives menu. 2.) Click the Line through Two Points tool, and click two distinct locations on the Graphics view to construct line AB. 3.) If the labels of the points are not displayed, click the Move tool, right click each point and click Show label from the context menu. 4.) Click the New Point tool and construct a point C not on line AB. 5.) Display the name of the third point. GeoGebra would automatically name it C, otherwise right click and rename it C. 6.) Click the Polygon tool and click the points in the following order: point A, point B, point C, and click again on point A to close the polygon. Your drawing should look like Figure 1. Figure 2 – Triangle ABC on line AB. 7.) Move the vertices of the polygon. What do you observe? 8.) Now we construct two angle sliders $\alpha$ and $\beta$. To do this, click the Slider tool, and click on the Graphics view. 9.) In the Slider dialog box (see Figure 3), choose the Angle radio button, and then leave the name angle as $\alpha$.  In the Interval tab, choose 0° as minimum, 180° as maximum and 1°, and then click the Apply button when finished. Figure 3 – The Slider dialog box 10.) Using steps 8-9, create another slider with the same specifications shown in Figure 3 and name it $\beta$. You can find the Greek letters by pressing the $\alpha$ button located at the right of the text box. 11. ) We reveal the angle measures of the interior angles of the triangle, the change the colors of the angle symbols (green sectors). To do this, click the Angle tool and then click the interior of triangle ABC. 12.)  We now hide the measures of the angles. To do this, right click each angle symbol and uncheck Show label from the context menu. 13.  We set angle colors: angle A red, angle B blue and angle C green. To change the color of the angle symbol of angle A, right click the angle symbol (not point A) and click Object Properties from the context menu to display the Preferences window. 14.  In the Preferences window, click the Color tab and choose the color you want from the color palette then click the Close button. 15.  Change the color of angle B to blue and leave angle C as is.  Your drawing should look line Figure 1-A after step 15.

Part II – Rotating the Triangle

We already have the sliders ready. The next thing that we will do is to rotate the triangle. The idea is to create a rotation point. Our choice would be the midpoint of BC. That is because if we rotate ABC by 180 degrees producing A’B’C’, angle A’C’B’ will be adjacent to angle ABC (see Figure 1-B). This is also the idea when we rotate A’B’C producing A’’B’’C’’.

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 1.) To construct D, the midpoint of BC, click the Midpoint or Center tool, and click side BC (the segment, not the points). 2.) Note that we want ABC to rotate around D $\alpha$ degrees clockwise. To do this, choose Rotate around a Point by Angle tool, click the interior of the triangle and click point D to reveal the Rotate Object dialog box. 3.) In the Rotate Object around Point by Angle dialog box, change the measure of the angle to $\alpha$, choose the clockwise radio button, and then click the OK button. Figure 4 – The Rotate Dialog Box 4.) Now move slider $\alpha$. What do you observe? 5.) Adjust slider $\alpha$ to 90 degrees, and show the labels of the vertices of the rotated triangle. (Refer to Part I – Step 3). 6.) While the triangle is still rotated 90 degrees, click the Angle tool and click the interior of triangle A’B’C’. Hide the labels of the angles symbols. 7.) Change the colors of the angle measures. Refer to Part I – Steps 13 through 15. Be sure that angle A and A’ have the same color, B and B’ have the same color, and C and C’ have the same color. Your drawing should look like the drawing in Figure 6. Figure 5 – The Rotated Triangle

Part III – Creating the Third Triangle

The idea of creating the third triangle is basically the same as that of creating the second triangle, so I will just enumerate the steps and left the construction as an exercise.

1. Get the midpoint of A’C’. (Refer to Part II Step 1)
2. Rotate triangle A’B’C’ $\beta$ degrees clockwise around the midpoint of A’C’.  (Refer to Part II – Steps 2 – 3).
3. Reveal the labels of the vertices of the third triangle which is A’’B’’C’’.  (Refer to Part II – Step 5 and Part I – Step 3).
4. Reveal the angle symbols of triangle A’’B’’C’’.  (Refer to Part I – Step 11)
5. Hide the labels of the angle symbols, and change the colors of the angle symbols of triangle A’’B’’C’’. (Refer to Part I – Step 13-15)

The explanation of the theory behind this construction is in my Parallel Lines and Transversal post.

Why is the angle sum of the interior angles of a triangle 180 degrees?

We were taught that the sum of the measures of interior angles of a triangle is 180 degrees. But how come? Is it true no matter what the shape or size of the triangle?

Figure 1 – Triangles with interior angles a, b and c

Recall that an angle is the amount of rotation of a ray.  In Figure 2, the ray was rotated from A to C, and the amount of rotation is 60 degrees. We can say that the measure of angle ABC is 60 degrees.

Figure 2 – Ray containing A is rotated to to C making a 60-degree angle rotation

The rotation from A to D forms a straight line and measures 180 degrees.  Therefore, straight angle ABD measures 180 degrees.  It follows that a 180-degree rotation is a half-circle. Therefore, a complete rotation is 360 degrees.

We can verify if our question about the sum of the interior angles of a triangle by drawing a triangle on a paper, cutting the corners, meeting the corners (vertices) at one point  such that the sides coincide with no gaps and overlaps (see Figure 3). Notice that no matter what the size or shape of a triangle, as long as the previous conditions are met, the two of its sides will be collinear as shown in the Figure below.

Figure 4 – The corners of a triangle put together

However, this is not the proof. To discuss the proof, we are going to use Euclid’s fifth postulate. Euclid’s fifth postulate tells us that if a parallel line is cut by a transversal, their corresponding angles are congruent. In the diagram below, lines p and q are parallel lines and angles shown with the same letters are corresponding angles, and hence congruent.

Figure 5 – Parallel lines p and q cut by a transversal.

Now how do we connect Parallel Postulate to the angle sum of a triangle. Consider a triangle with interior angles a, b, and c. Using the fifth postulate, we use the side containing angles a and c as base of the triangle and extend it to both sides. Next, we draw a line to the base and passing thorugh the vertex not on the base.

Figure 6 – Triangle with three points on two parallel lines

Now, the sum of angles a, b and c is 180 degrees (why?). But a, b and c are also the interior angles of a polygon. Therefore, the sum of the interior angles of a triangle is 180 degrees.

You may also want to read Sum of the Interior Angles of a Polygon.

Sum of the interior angles of a polygon

We were taught that if we let $m$ be the angle sum (the total measure of the interior angles) and $n$ be the number of vertices (corners)  of a polygon, then $m = 180(n-2)$.  For example, a quadrilateral has $4$ vertices, so its angle sum is $180(4-2) = 360$ degrees.  Similarly, the angle sum of a hexagon (a polygon with $6$ sides) is $180(6-2) = 720$ degrees.

But where did this formula come from?  Does this formula work for all polygons?  (Note that in this discussion, when we say polygon, we only refer to convex polygons).

Before we answer these questions, let us first have a brief review of some elementary concepts.

Polygons and Interior Angles

A polygon is a closed figure with finite number of sides. In the figures below, $ABCDE$ is a polygon with $5$ sides and ($5$ vertices).  It is clear that the number of sides of a polygon is always equal to the number of its vertices.

A polygon has interior angles.  In the first figure below, angle $B$ measuring $91$ degrees is an interior angle of polygon $ABCDE$. The angle sum $m$ of $ABCDE$ (not drawn to scale) is given by the equation

$m = 126 + 91 + 113 + 102 + 108 = 540$ degrees.

In the second figure, if we let $a_1, a_2$ and $a_3$ be the measure of the interior angles of triangle $ABE$, then the angle sum m of triangle $ABE$ is given by the equation $m = a_1 + a_2 + a_3$.

Angle Sum

To generalize our calculation of  angle sum, we use the fact that the angle sum of a triangle is $180$ degrees. Notice that any polygon maybe divided into triangles by drawing diagonals from one vertex to all of the non-adjacent vertices.  In the second figure above, the pentagon was divided into three triangles by drawing diagonals from vertex $E$ to the non-adjacent vertices $B$ and $C$ forming $BE$ and $CE$.  Now let $a_k, b_k$ and $c_k$, where $k = 1, 2, 3$ be measures of the interior angles of the three triangles as shown on the second figure.

Calculating the angle sum of pentagon $ABCDE$ we have

Notice that the angle measures in the first line of our equation is just a rearrangement of the measures of the interior angles of the three triangles. Hence, the angle sum of the pentagon is equal to the angle sum of the three triangles. Therefore, we can conclude that the sum of the interior angles of a polygon is equal to the angle sum of the number of triangles that can be formed by dividing it using the method described above. Using this conclusion, we will now relate the number of sides of a polygon, the number of triangles that can be formed by drawing diagonals and the polygon’s angle sum.

From the table above, we observe that the number of triangles formed is $2$ less than the number of sides of the polygon.  This is true, because $n - 2$ triangles can be formed by drawing diagonals from one of the vertices to $n - 3$ non-adjacent vertices. Therefore, there the angle sum $m$ of a polygon with $n$  sides is given by the formula

$m = 180(n - 2)$

A More Formal Proof

Theorem: The sum of the interior angles of a polygon with $n$ sides is $180(n-2)$ degrees.

Proof:

Assume a polygon has $n$ sides. Choose an arbitrary vertex, say vertex $V$.  Then there are $n - 3$ non-adjacent vertices to vertex $V$.  If diagonals are drawn from vertex $V$ to all non-adjacent vertices, then $n - 2$ triangles will be formed.  The sum the interior angles of $n -2$ triangles is $180(n - 2)$. Since the angle sum of the polygon with $n$ sides is equal to the sum the interior angles of $n - 2$ triangles, the angle sum of a polygon with $n$ sides is $180(n-2)$. $\blacksquare$

Exercises:

1.)    Find the number the angle sum of a dodecagon ($12$-sided polygon).

2.)    The angle sum of a polygon is $3240$ degrees. What is the number of its sides?

3.)    The measure of one of the angles of a regular polygon is $150$. Find its number of sides.

4.)    From this, prove that the sum of the interior angles of a polygon is $360$ degrees.