InterMath | Workshop Support | Write Up Template
Title
Sum of Angles in a Polygon
Problem Statement
What is the sum of the angles of a triangle? of a quadrilateral? of a pentagon? of a
hexagon? What is the sum of the angles in convex polygons in terms of the number
of sides? Move the mouse pointer over the picture for a hint.
Problem setup
In this problem, I want determine the sum of angles in different polygons and then find a
common formula in determining the sum of angles in any convex polygon with n sides.
Plans to Solve/Investigate the Problem
Prediction: For a triangle, the angles will add up to equal 180º. I am not sure about the
other polygons.
To begin the problem:

Construct a triangle, quadrilateral, pentagon, and hexagon

Divide the polygons into triangles since you know the sum of the angles in a
triangle is equal to 180º

Then measure the angles of each polygon and find their sums

Make a chart showing the relationships of the sums of angles

Create a formula for finding a polygon with n number of sides

Show another example using a different polygon, proving formula works

Do extension
Investigation/Exploration of the Problem
1. First I plan to construct a triangle in GSP (see figure below). Then using the
measure tool, I will measure the angles in the triangle and find the sum.
a. To construct my triangle, I created line segment AB and then connected it
to line segment BC. Then I made a convex polygon by completing the
triangle with line segment AC.
mABC = 98.59
mBCA = 37.90
B
mCAB = 43.51
A
mCAB+mBCA+mABC = 180.00
C
2. Second, I plan to construct a quadrilateral in GSP (see figure below). Then using
the measure tool, I will measure the angles in the quadrilateral and find the sum of
the angles.
a. To construct my quadrilateral, I created four random points and connected
the points with line segments to create quadrilateral EFGD.
mDEF = 93.63
F
E
mEFG = 96.55
mFGD = 81.21
mGDE = 88.61
mDEF+mEFG+mFGD+mGDE = 360.00
D
G
3. Now I wanted to divide the quadrilateral in half to form two triangles. This helps
me because I know that the sum of the angles of a triangle is 180º, so therefore if I
divide my quadrilateral into two triangles my sum of the angles should equal
360º.
K
J
mKIH = 86.88
mKJH = 91.86
mIHK = 35.32
mJHK = 57.39
mHKI = 57.80
mHKJ = 30.75
mKIH+mIHK+mHKI = 180.00
I
H
mKJH+mJHK+mHKJ = 180.00
mKIH+mIHK+mHKI+mKJH+mJHK+mHKJ = 360.00
4. Next I plan to construct a pentagon in GSP (see figure below). Then using the
measure tool, I will measure the angles in the pentagon and find the sum of the
angles.
a. To construct my pentagon, I created five random points and connected the
points with line segments to create a pentagon.
L
M
mLMN = 139.85
mMNO = 69.37
N
mNOP = 147.94
mLMN+mMNO+mNOP+mOPL+mPLM = 540.00
mOPL = 88.78
P
O
mPLM = 94.06
5. Now I wanted to divide the pentagon into three triangles. This helps me because I
know that the sum of the angles of a triangle is 180º, so therefore if I divide the
pentagon into three triangles my sum of the angles should equal 540º.
A
B
E
C
D
mACD = 88.78
mADB = 57.27
mDBE = 66.75
mCAD = 44.43
mDBA = 73.10
mBED = 69.37
mADC = 46.79
mBAD = 49.63
mEDB = 43.88
mDBE+mBED+mEDB = 180.00
mACD+mCAD+mADC = 180.00 mADB+mBAD+mDBA = 180.00
mDBE+mBED+mEDB+mADB+mBAD+mDBA+mACD+mCAD+mADC = 540.00
6. Next I plan to construct a hexagon in GSP (see figure below). Then using the
measure tool, I will measure the angles in the hexagon and find the sum of the
angles.
a. To construct my hexagon, I created six random points and connected the
points with line segments to create a hexagon.
Q
mQRS = 144.38
R
mRST = 142.10
mST U = 71.92
S
mTUV = 141.25
mUVQ = 131.56
V
mVQR = 88.79
U
T
mQRS+mRST+mST U+mTUV+mUVQ+mVQR = 720.00
7. Now I wanted to divide the hexagon into four triangles. This helps me because I
know that the sum of the angles of a triangle is 180º, so therefore if I divide the
hexagon into four triangles my sum of the angles should equal 720º.
mBEC = 40.57
A
B
mAFB = 47.11
mECB = 75.30
mFBA = 44.63
mCBE = 64.13
mBAF = 88.25
mBEC+mECB+mCBE = 180.00
mFBA+mBAF+mAFB = 180.00
C
F
E
D
mFEB = 60.54
mCDE = 75.50
mEBF = 35.62
mDEC = 37.71
mBFE = 83.85
mECD = 66.79
mFEB+mEBF+mBFE = 180.00
mCDE+mDEC+mECD = 180.00
mFBA+mBAF+mAFB +mFEB+mEBF+mBFE +mBEC+mECB+mCBE+mCDE+mDEC+mECD = 720.00
8. Finally I created a chart showing the relationships I found with the sum of the
interior angles of polygons. Therefore, I came to the conclusion that if a polygon
has n number of sides then to find the sum of the interior angles, you would use
the formula we derived: (n-2) 180º, where n is the number of sides of a polygon.
This formula was found by creating triangles within polygons. So, if a polygon
has 3 sides (triangle) then (3-2) =1 x 180º = 180º. You can also view the formula,
for example, in a triangle there is only one triangle but in a quadrilateral there are
two triangles and so forth.
# of sides: n
3
4
5
6
N
Regular polygon :
measure of 1 interior angle
= sum of interior angles/n
60
90
108
120
((n-2)180)/ n
Sum of interior angles
180 = 1 (180)
360 = 2 (180)
540 = 3 (180)
720 = 4 (180)
= (n-2) 180
So, using the formula I can complete the chart and find the sum of interior angles for
polygons with any number of sides. In my chart however, I just went up to polygons
consisting of twelve sides.
# of sides: n
3
4
5
6
7
8
9
10
11
12
Regular polygon :
measure of 1 interior angle
= sum of interior angles/n
60
90
108
120
((n-2)180)/ n
128.57
135
140
144
147.27
150
Sum of interior angles
180 = 1 (180)
360 = 2 (180)
540 = 3 (180)
720 = 4 (180)
= (n-2) 180
900 = 5 (180)
1080 = 6 (180)
1260 = 7 (180)
1440 = 8 (180)
1620 = 9 (180)
1800 = 10 (180)
9. Just to prove this works, I am going to construct another polygon (10 sided figure
called a decagon) in GSP. Then using the measure tool, I will measure the angles
in the decagon and find the sum of the angles.
a. To construct my decagon, I created ten random points and connected the
points with line segments to create the decagon
Y
X
Z
mWXY = 156.32
mXYZ = 147.27
E1mYZE
W
1
= 146.40
mZE1F1 = 134.86
F1mE F G
1 1
G1
J1
H1
I1
1
= 159.33
mF1G1H1 = 151.02
mG1H1I 1 = 159.24
mH1I 1J1 = 124.28
mI 1J1W = 137.23
mJ1WX = 124.04
mWXY+mXYZ+mYZE1+mZE1F1+mE1F1G1+mF1G1H1+mG1H1I 1+mH1I 1J1+mI 1J1W+mJ1WX = 1440.00
10. Now I wanted to divide the decagon into eight triangles. This helps me because I
know that the sum of the angles of a triangle is 180º, so therefore if I divide the
decagon into eight triangles my sum of the angles should equal 1440º.
C1
A1
J
D1
mD1A1E = 38.00
mB1GH = 64.72
mA1ED1 = 23.18
mGHB1 = 88.10
mED1A1 = 118.82
mD1A1E+mA1ED1+mED1A1 = 180.00
I
H
E
G
F
mHB1G = 27.18
mB1GH+mGHB1+mHB1G = 180.00
mB1HI = 55.93
mA1EB1 = 17.51
mEB1A1 = 32.95
mHIB1 = 109.28
mIB1H = 14.80
mB1A1E = 129.55
mA1EB1+mEB1A1+mB1A1E = 180.00
mEFB1 = 65.49
mFB1E = 17.97
mB1EF = 96.54
mEFB1+mFB1E+mB1EF = 180.00
mB1HI+mHIB1+mIB1H = 180.00
mB1IJ = 50.06
mIJB1 = 113.32
mJB1I = 16.62
mB1IJ+mIJB1+mJB1I = 180.00
mFGB1 = 110.07
mC1B1J = 12.06
mGB1F = 19.70
mB1JC1 = 21.54
mB1FG = 50.23
mJC1B1 = 146.40
mFGB1+mGB1F+mB1FG = 180.00
mC1B1J+mB1JC1+mJC1B1 = 180.00
mD1A1E+mA1ED1+mED1A1+mA1EB1+mEB1A1+mB1A1E+mEFB1+mFB1E+mB1EF+mFGB1+mGB1F+mB1FG  = 720.00
mB1GH+mGHB1+mHB1G+mB1HI+mHIB1+mIB1H+mB1IJ+mIJB1+mJB1I +mC1B1J+mB1JC1+mJC1B1 = 720.00
FB1+mFB1E+mB1EF+mFGB1+mGB1F+mB1FG +mB1GH+mGHB1+mHB1G+mB1HI+mHIB1+mIB1H+mB1IJ+mIJB1+mJB1I +mC1B1J+mB1JC1+mJC1B1 = 1440.00
Extensions of the Problem
Does the relationship for convex polygons also hold true for nonconvex polygons?
Explain.
Yes, the relationship holds true even for nonconvex polygons. I am able to justify my
answer by once again dividing the polygon into triangles. Without the triangles, I am
unable to measure some of the angles like angle BAE. So, when I divide the nonconvex
polygon into three different triangle, I can use the formula (n-2) 180º = (5-2) 180º = 540º.
mABC = 58.61
E
mAED+mEAD+mEDA = 180.00
mAED = 35.05
mEAD = 99.33
B
mCAB = 79.03
A = 62.14
mCAD
mCAD+mADC+mDCA = 180.00
mCAB+mACB+mABC = 180.00
mAED+mEAD+mEDA+mCAD+mADC+mDCA+mCAB+mACB+mABC = 540.00
mEDA = 45.62
mACB = 42.36
D
mDCA = 52.76 mADC = 65.09
C
mAED+mEDA+mADC+mDCA+mACB+mABC = 299.50
Author & Contact
Lauren Johnson, Middle Grades Cohort at Georgia College and State University
Lauren_johnson@ecats.gcsu.edu
Link(s) to resources, references, lesson plans, and/or other materials
Link 1
Link 2