



Solutions for Session 4, Part B
See solutions for Problems: B1  B2  B3  B4  B5  B6  B7  B8  B9  B10  B11  B12

Problem B1  
a.  Many polygons have acute angles. All but A, B, C, and H have at least one acute angle. 
b.  Polygons A, B, C, D, E, F, and L have at least one right angle. 
c.  Polygons G, H, J, K, M, and O have at least one obtuse angle. 
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Problem B2  
a.  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. 
b.  Polygons D, E, F, G, I, J, K, L, M, N, and O all have two or more acute angles. 
c.  Polygons G, H, K, M, and O all have two or more obtuse angles. 
d.  Polygons A, B, and C all have four right angles. 
e.  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." 
f.  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." 
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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.
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Problem B4  
a.  Polygons A, B, C, G, K, M, and O all have two angles that are supplementary. 
b.  Polygons D, E, F, and L all have two angles that are complementary. 
c.  All polygons except L have some congruent angles. In polygons A, B, C, H, I, and N, all angles are congruent! 
d.  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. 
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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.
a.  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. 
b.  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). 
c.  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.

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Problem B6  
a.  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.

b.  Every angle in every polygon is an interior, or vertex, angle.

c.  There are many possible answers. One is to connect polygons H and I so that one long side is formed. The 120degree angle in H is an exterior angle to polygon I, and the 60degree angle in I is an exterior angle to polygon H.

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Problem B7  


Polygon 

Angle 1 

Angle 2 

Angle 3 

Angle 4 

Angle 5 

Angle 6 

Name of Polygon 


A 

90° 

90° 

90° 

90° 

 

 

square 

B 
90° 
90° 
90° 
90° 
 
 
square 

C 
90° 
90° 
90° 
90° 
 
 
rectangle 

D 
45° 
45° 
90° 
 
 
 
triangle 

E 
45° 
45° 
90° 
 
 
 
triangle 

F 
45° 
45° 
90° 
 
 
 
triangle 

G 
60° 
120° 
60° 
120° 
 
 
parallelo gram 

H 
120° 
120° 
120° 
120° 
120° 
120° 
hexagon 

I 
60° 
60° 
60° 
 
 
 
triangle 

J 
30° 
30° 
120° 
 
 
 
triangle 

K 
60° 
60° 
120° 
120° 
 
 
trapezoid 

L 
30° 
60° 
90° 
 
 
 
triangle 

M 
60° 
120° 
60° 
120° 
 
 
parallelo gram 

N 
60° 
60° 
60° 
 
 
 
triangle 

O 
30° 
150° 
30° 
150° 
 
 
parallelo gram 


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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.
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Problem B9  
a.  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. 
b.  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. 
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Problem B10  
a.  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. 
b.  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.

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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.
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Problem B12  
a.  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°. 
b.  Four triangles are formed by drawing diagonals. 
c.  Since each triangle's angles sum to 180 degrees, and there are four triangles, the hexagon's angle measures sum to 180° • 4 = 720°. 
d.  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°. 
e.  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. 
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