Angle Measurement Part B: Angles in Polygons

Another way to classify angles is by their relationship to other angles. As you work on the types of classifications in Problems B4 and B5, think about the key relationships between angles.

The types of angles you will be looking at in Problems B4 and B5 are easily shown on two parallel lines cut by a transversal. You can use several copies of the same polygon and place them together to form parallel lines that are cut by a transversal.

Problem B4

Use two or more polygons to illustrate the angles below:

1. Supplementary angles (the sum of their measures equals 180 degrees)
2. Complementary angles (the sum of their measures equals 90 degrees)
3. Congruent angles (their angle measurement is the same)
4. Adjacent angles (they share a common vertex and side)

Problem B5

Use two or more polygons to illustrate the angles below, and explain how you would justify that some of the angles are congruent:

1. Vertical angles (the angles formed when two lines intersect; in the figure above, ad, cb, eh, and fg are pairs of vertical angles, and the angle measures in each pair are equal)
2. Corresponding angles (the angles formed when a transversal cuts two parallel lines; in the figure above, ae, bf, cg, and dh are pairs of corresponding angles, and the angle measures in each pair are equal)
3. Alternate interior angles (the angles formed when a transversal cuts two parallel lines; in the figure above, cf and de are pairs of alternate interior angles, and the angle measures in each pair are equal)

Problem B6

Find one or more polygons you can use to see examples of the following angles:

1. Central angles (for regular polygons, the central angle has its vertex at the center of the polygon, and its rays go through any two adjacent vertices)
2. Interior or vertex angles (an angle inside a polygon that lies between two sides)
3. Exterior angles (an angle outside a polygon that lies between one side and the extension of its adjacent side):
   

Problem B7

Angle measurement is recorded in degrees. Using only the polygons (no protractor! no formulas!) and logical reasoning, determine the measure of each of the angles in the polygons, and record your measures in the table. Be able to explain how you determined the angle size.

One approach would be to use the known measurement of one angle to determine the measures of other angles. For example, Polygon N is an equilateral triangle — notice that all the angles are equal. If we arrange six copies of the polygon around a center point, the angles completely fill up a circle. So each angle measure must be 60 degrees. Similarly, since two N blocks fit into the vertex angles of Polygon H, the measure of each of the angles in H must be 120 degrees.

Problem B8

Describe two different methods for finding the measure of an angle in these polygons.

One method would be to use multiple copies of the angle under investigation to form a circle around a point.

Problem B9

1. Based on the data in the table, what is the sum of the angles in a triangle? How might we prove this?
2. Trace around four different triangles and cut them out. Label 1, 2, and 3 in each triangle. Then tear off the angles and arrange them around a point on a straight line (i.e., each vertex point must meet at the point on the line), as shown below: Note 3

What do you notice? Is it true for all four of your triangles? Will it be the same for every triangle? Explain. Note 4

Take It Further

Problem B10

1. Now draw two parallel lines with a triangle between them so that the vertices of the triangle are on the two parallel lines. The angles of the triangle are A, B, and ACB:
   
   Earlier you reviewed the relationships between angles (e.g., alternate interior angles). Use some of these relationships to describe the diagram above.
2. Use what you know about angle relationships to prove that the sum of the angles in a triangle is 180 degrees.

Problem B11

Examine the polygons in the table in Problem B7 that are quadrilaterals. Calculate the sum of the measures of the angles in each quadrilateral. What do you notice? How can we explain this sum? Will the measures of the angles in an irregular quadrilateral sum to this amount? Explain.

We determined that the sum of the measures of the angles of a triangle is 180 degrees. Notice in this diagram that the diagonal from one vertex of a quadrilateral to the non-adjacent vertex divides the quadrilateral into two triangles:

The sum of the angle measures of these two triangles is 360 degrees, which is also the sum of the measures of the vertex angles of the quadrilateral. Note 5

Take It Further

Problem B12

1. Use this technique of drawing diagonals from a vertex to find the sum of the measures of the vertex angles in a regular pentagon (see below). What is the measure of each vertex angle in a regular pentagon?
   
2. How many triangles are formed by drawing diagonals from one vertex in a hexagon?
3. What is the sum of the measures of the vertex angles in a hexagon?
4. Find a rule that can be used to find the sum of the vertex angles in any polygon.
5. Can you use your rule to find the measure of a specific angle in any polygon? Why or why not?

Note 3

For this problem, be sure to tear, not cut, the three angles off the triangle. This way you will be sure which point is the angle under consideration. If you cut them, you will end up with three clean-cut points (angles), and it is easy to become confused about which are the actual angles of the triangle.

Note 4

This approach for justifying that the sum of the angles in a triangle is 180 degrees is an informal justification or proof. A more formal proof can be written using what you know about the angles formed when parallel lines are cut by a transversal, which will be explored next.

Note 5

You may want to draw different quadrilaterals to show visually how a quadrilateral can only be divided into two triangles (from any one vertex)

Problem B1

1. Many polygons have acute angles. All but A, B, C, and H have at least one acute angle.
2. Polygons A, B, C, D, E, F, and L have at least one right angle.
3. Polygons G, H, J, K, M, and O have at least one obtuse angle.

Problem B2

1. Answers will vary. Polygons G, H, K, M, and O have more than one obtuse angle, while only J has exactly one obtuse angle. Other angles in polygons with obtuse angles may be acute, right, or obtuse.
2. Polygons D, E, F, G, I, J, K, L, M, N, and O all have two or more acute angles.
3. Polygons G, H, K, M, and O all have two or more obtuse angles.
4. Polygons A, B, and C all have four right angles.
5. No. When you draw a side of a triangle and then an obtuse angle at each vertex, the resulting lines point away from each other. They will never meet for the triangle to “close up.”
6. No. When you draw a side of a triangle and then one right angle at each vertex, the resulting lines are parallel (perpendicular to the same line). They will never meet, so the triangle will never “close up.”

Problem B3

Equilateral triangle polygons are I and N. They are also isosceles. Equilateral triangles have three angles that are equal in measure.

Isosceles triangle polygons are D, E, F, and J. All these triangles have two equal side lengths, opposite the equal angles. Isosceles triangles have at least two equal angles.

The only scalene triangle polygon is L. All of its sides have different lengths. Scalene triangles have no equal angles.

Problem B4

1. Polygons A, B, C, G, K, M, and O all have two angles that are supplementary.
2. Polygons D, E, F, and L all have two angles that are complementary.
3. All polygons except L have some congruent angles. In polygons A, B, C, H, I, and N, all angles are congruent!
4. If you take any two polygons and line them up so that one side from each meets at a common vertex, you’ve created a pair of adjacent angles.

Problem B5

One way to review these relationships is to use either the G, M, or O polygon. Place multiple copies of one of these polygons together as illustrated. Next, identify segments that are parallel and segments that are transversals.

1. The vertical angles are congruent (the pairs of vertical angles are a and d, c and b, e and h, and f and g). We know that a + b sum to 180 degrees because a and b form a straight line. Similarly, a + c sum to 180 degrees, as a and c also form a straight line. We can conclude from this that b = c; thus, angles b and c are congruent. This works for any pair of vertical angles.
2. Angles a and e, and angles b and f, are congruent. Angles on the same side of the transversal and in the same position relative to the parallel lines are called corresponding angles and are congruent. Since the transversal cuts the parallel lines at the same angle, the angles formed by the parallel lines and the transversal are identical in measure (e.g., d and h, and c and g).
3. Angles a and d are vertical angles, so they have an equal measure. Angles a and e are corresponding angles, so they have an equal measure. Therefore, angles d and e must have an equal measure, since both are congruent to angle a.Other ways of determining the above conjectures are also possible.

Problem B6

1. Many answers are possible, but regular polygons (A, B, H, I, and N) work best. Tiling (covering an area with no gaps or overlaps) with many of these produces a central angle at any vertex. The rectangular tile C can also be used in this way.
   
2. Every angle in every polygon is an interior, or vertex, angle.
   
3. There are many possible answers. One is to connect polygons H and I so that one long side is formed. The 120-degree angle in H is an exterior angle to polygon I, and the 60-degree angle in I is an exterior angle to polygon H.
   

Problem B7

Problem B8

One method is to tile the polygon until a complete circle is formed, then divide 360 degrees by the number of polygons required to complete the circle. A second method is to figure out angles for the regular polygons, then lay other polygons on top to compare the angles.

Problem B9

1. All the triangles made up from the polygons in the table have a total angle measure of 180 degrees. Later in this part, we’ll explore how to prove that all triangles have angle measures that sum to 180 degrees.
2. This works for all four triangles. Since the three angles form a straight line, their sum should be 180 degrees. It appears from the drawing that this should work for any triangle.

Problem B10 

1. Angle 1 is congruent to ∠A, and ∠3 is congruent to ∠B. Both of these are a pair of alternate interior angles, since DE and AB are parallel.
2. Since the sum of ∠1, ∠2, and∠3 is 180 degrees, the sum of the angles inside the triangle is also 180 degrees (∠1 =∠A,∠2 =∠ACB, and∠3 =∠B;
   if∠1 +∠2 +∠3 = 180°, then∠A +∠B +∠ACB = 180°).Notice that there’s nothing special about the triangle we drew. Starting with any triangle, you can pick a vertex and draw a line parallel to the opposite side. The angle relationships you found here will hold, and so will the fact that the sum of the angles is equal to 180 degrees.

Problem B11

The sum of the four angles in any quadrilateral is 360 degrees. One way to explain this is to draw an interior diagonal in the quadrilateral (a line connecting opposite vertices). This divides the quadrilateral into two triangles. Since we know each triangle’s angles add up to 180 degrees, the two triangles’ angles must add up to 360 degrees.

Problem B12

1. Since there are now three triangles, the angle sum is 180° • 3 = 540°. In a regular pentagon, where all five angles have the same measure, each measures 540 5 = 108°.
2. Four triangles are formed by drawing diagonals.
3. Since each triangle’s angles sum to 180 degrees, and there are four triangles, the hexagon’s angle measures sum to 180° • 4 = 720°.
4. If there are n sides in the polygon, there are (n – 2) triangles formed. Each triangle’s angles sum to 180 degrees, so the angle sum for the polygon is (n – 2) • 180°.
5. For regular polygons, by definition the angles all have the same measure, so we can divide the angle sum by n (the number of angles) to find the measure of a specific angle. But for an irregular polygon, this won’t work. It takes a little work to show, but even oddly shaped polygons with lots of sides can always be divided into (n – 2) triangles, where n is the number of sides, and where each triangle has all of its angles on the polygon. So the angle sum will still be (n – 2) • 180°. If, however, the angles are not all equal, there’s no way to use the angle sum to find the measure of a particular angle.