High School Mathematics Archives - Page 156 of 157

GeoGebra Tutorial 2 – Constructing an Equilateral Triangle

November 9, 2009 GB GeoGebra, High School Geometry, High School Mathematics

This is the second tutorial of the GeoGebra Intermediate Tutorial Series. If this is your first time to use GeoGebra, please read the GeoGebra Essentials Series. This tutorial, answers the following problem using GeoGebra.

Note: This tutorial has been updated for GeoGebra 4.2.

In the tutorial below, menu commands, located in the menu bar, are in brown bold text, and submenus are denoted by the > symbol. For example, Options>Labeling> New Points Only means, click the Options menu, choose Labeling from the list, then select New Points Only. The GeoGebra tools are denoted by orange texts. For example, New Point means the new point tool.

Problem: How will you draw an equilateral triangle without using the Regular Polygon tool? » Read more

20 comments constructing an equilateral triangle, geogebra 4.2 tutorial, GeoGebra Angle tool, GeoGebra circle tool, GeoGebra intersect two objects tool, GeoGebra polygon tool, GeoGebra Tutorial

GeoGebra Tutorial 1 – Quadrilaterals & Midpoints

November 4, 2009 GB GeoGebra, High School Geometry, High School Mathematics, Software Tutorials

If this is your first time to use GeoGebra, I suggest that you read Introduction to GeoGebra which contains the discussion about the basics of GeoGebra and parts of the GeoGebra window. The assumption in this tutorial is that you are already familiar with the parts of the GeoGebra window.

Problem: Investigate what happens if you connect the the consecutive midpoints of a quadrilateral.

In this tutorial, we use GeoGebra to explore the properties of the midpoints of a quadrilateral. In doing this tutorial, you will learn how to use the following tools: Move, Midpoint or Center, Segment between Two Points and Polygon. If you want to view the output of this tutorial, click here.

	1.) Open GeoGebra and select Algebra from the Perspective menu .
	2.) To construct the vertices of the quadrilateral, click the New Point tool, and then click four distinct places on the Graphics view.
	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.) Select the Segment tool and click point A and click point B two distinct points to construct segment AB. To construct BC, with the Segment tool still active, click point B and another location to create segment BC. Now, construct CD and AD to complete the quadrilateral. Your drawing should look like Figure 1. Figure 1 – Quadrilateral ABCD
	5.) Click the Move tool and move the vertices of the quadrilateral. What do you observe?
	6.) To determine the midpoint of each side of the quadrilateral, choose the Midpoint or Center and then click the segments (not the points) in the following order: AB, BC, CD, and AD.
	7.) Select the Move tool, right click the midpoints and click Show label from the context menu. After that step, your drawing should look like Figure 2. Figure 2 – Quadrilateral ABCD with Midpoints E, F, G and H.
	8.) Move the vertices of the quadrilateral. What do you observe?
	9.) To have a better view, connect the midpoints of the quadrilateral using the Polygon tool. To do this, click the Polygon tool and click the points in the following order: Point E, point F, point G, point H and then point E again to close the polygon.
	10. Move the vertices of the quadrilateral. What do you observe about the figure?
	11. What conjecture can you make based on your observation?

Updated: 11 January 2016 (GeoGebra 5.0)

9 comments geogebra 4.2 tutorial, GeoGebra midpoint tool, GeoGebra polygon tool, GeoGebra Tutorial, inner quadrilateral, midpoints of a quadrilateral

Sum of the interior angles of a polygon

October 21, 2009 GB High School Geometry, High School Mathematics, Questions and Quandaries

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.

twopentagons

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

pentagonanglesum

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.

table

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 $180(n-2)$ . $\blacksquare$