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Illustrative
Mathematics
A-CED Sum of angles in a
polygon
Alignments to Content Standards: A-CED.A.1
Task
a. Below is a quadrilateral
Show, by dividing
ABCD:
ABCD into triangles, that the sum of the interior angles is 360∘ :
m(∠A) + m(∠B) + m(∠C) + m(∠D) = 360.
b. Below is a pentagon
ABCDE:
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Show that the sum of the interior angles is
540∘ :
m(∠A) + m(∠B) + m(∠C) + m(∠D) + m(∠E) = 540.
c. Suppose P is a polygon with n ≥ 3 sides and assume that all interior angles of P
measure less than 180 degrees. Show that the sum of the measures of the interior
angles of P is
(n − 2) × 180 degrees.
Check that this formula gives the correct value for equilateral triangles and squares.
IM Commentary
This problem provides students with an opportunity to discover algebraic structure in a
geometric context. More specifically, the student will need to divide up the given
polygons into triangles and then use the fact that the sum of the angles in each
triangle is 180∘ . The algebraic formula which occurs in step (c) can be seen as a
generalization of parts (a) and (b) but an argument establishing this formula requires a
delicate argument which successively reduces the number of sides of the polygon in
question. Alternatively, step (c) can be accomplished by providing an explicit division of
the given polygon into triangles. Both solutions are provided and it is important to note
that the logic is rather different: for the first solution, the case of a polygon with n sides
is reduced to the case of a polygon with one fewer, or n − 1 sides. This argument
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requires subscripts and the teacher may wish to provide guidance here. The second
argument is a direct generalization, a careful subdivision into triangles for parts (a) and
(b) and requires much less work with subscripts.
When the polygon has angles measuring more than 180 degrees the result is still true
but more care is needed to explain why. For example, in the picture below the
quadrilateral has one reflex angle (that is, one angle measuring greater than 180
degrees):
The sum of the four angles is still equal to 360 degrees, however, as can be seen by
drawing the auxiliary line indicated which divides the quadrilateral into two triangles.
The division needs to be done carefully, however, if there are several angles measuring
over 180 degrees which may occur for polygons with more sides. This could make for a
challenging and interesting classroom activity. The pentagon in part (b) was chosen
with one angle measuring greater than 180 degrees in order to give students a chance
to think about how to deal with this situation.
This task illustrates three of the mathematical practices. The first two parts require a
careful combination of geometric and algebraic reasoning (MP2) and, in order to
provide a convincing argument (MP3), the algebra and geometry need to be carefully
linked. This link is accomplished through a picture which the students must provide.
Finally, in order to complete part (c), students must identify a pattern from parts (a)
and (b) and then explain why this works for a polygon with any number of sides: this is
a good example of MP8.
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Solutions
Edit this solution
Solution: 1
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a. We can divide the quadrilateral ABCD into two triangles by drawing segment AC as
below:
The sum of the angles in triangle ABC is 180 degrees and the sum of the angles in
triangle ACD is 180 degrees:
m(∠CAB) + m(∠B) + m(∠BCA) = 180
m(∠CAD) + m(∠D) + m(∠DCA) = 180.
We have m(∠CAB) + m(∠CAD) = m(∠A) and m(∠BCA) + m(∠DCA)
adding the above two equations gives
= m(∠C) so
m(∠A) + m(∠B) + m(∠C) + m(∠D) = 180 + 180 = 360.
⎯⎯⎯⎯⎯⎯⎯⎯
⎯⎯⎯⎯⎯⎯⎯⎯
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⎯⎯⎯⎯⎯⎯⎯⎯
⎯⎯⎯⎯⎯⎯⎯⎯
The same argument applies if segment BD is added instead of segment AC .
b. As in part (a) we divide the pentagon into triangles by connecting some of its
vertices. We have to be careful here to make sure that the segments joining vertices do
not go outside of the pentagon and we also do not want these segments to cross inside
of the pentagon:
We have divided pentagon ABCDE into triangles ABC, CDE, and
the angles in each of these triangles is 180 degrees:
ACE. The sum of
m(∠CAB) + m(∠B) + m(∠BCA) = 180
m(∠ECD) + m(∠D) + m(∠DEC) = 180
m(∠EAC) + m(∠ACE) + m(∠CEA) = 180.
As in part (a) we can use the facts that m(∠BCA) + m(∠ECD) + m(∠ACE) = m(∠C),
m(∠DEC) + m(∠CEA) = m(∠E), and m(∠CAB) + m(∠EAC) = m(∠A) to conclude,
adding these three equations, that
m(∠A) + m(∠B) + m(∠C) + m(∠D) + m(∠E) = 540.
There is a second way to show that the sum of the angles in pentagon ABCDE is 540
degrees which helps build an intuition for part (c) below. We know from part (a) that
the sum of the angles in quadrilateral ABCE is 360 degrees:
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m(∠A) + m(∠B) + m(∠BCE) + m(∠AEC) = 360.
We also know that the sum of the angles in triangle CDE is 180 degrees:
m(∠D) + m(∠CED) + m(∠ECD) = 180.
Since m(∠BCE) + m(∠ECD) = m(∠C) and m(∠AEC) + m(∠CED)
add these two equations to conclude again that
= m(∠E) we can
m(∠A) + m(∠B) + m(∠C) + m(∠D) + m(∠E) = 540.
c. We may assume that the polygon in question has at least n ≥ 6 sides because
quadrilaterals and pentagons have been treated in parts (a) and (b) above. We will call
the original polygon with n sides Pn . We begin by identifying three successive vertices
A, B, C on the polygon and drawing segment AC as pictured below:
⎯⎯⎯⎯⎯⎯⎯⎯
Since none of the angles in the polygon are greater than 180 degrees, segment AC
divides the polygon into triangle ABC and a new polygon with n − 1 sides which omits
vertex B. We call this new polygon Pn−1 . The sum of the angles of Pn is the sum of the
angles of Pn−1 together with the angles of triangle ABC:
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Sum of angles of Pn = 180 + Sum of angles of Pn−1 .
We can now apply the same argument to Pn−1 , dividing it into a triangle and a polygon
with n − 2 sides which we call Pn−2 :
Sum of angles of Pn−1 = 180 + Sum of angles of Pn−2 .
Combining these two formulas gives
Sum of angles of Pn = 360 + Sum of angles of Pn−2 .
We can continue this process until we reach a triangle, P3 , and we will find
Sum of angles of Pn = (n − 3) × 180 + Sum of angles of Pn−(n−3).
The sum of the angles of the triangle P3
= Pn−(n−3) is 180 so we find
Sum of angles of Pn = (n − 2) × 180 degrees.
Plugging in n = 4 and n = 5 to the formula gives 360∘ for the sum of the angles of a
quadrilateral and 540∘ for the sum of the angles of a pentagon.
Edit this solution
Solution: 2. Alternate approach
a. Instead of dividing quadrilateral ABCD into two triangles, another method uses an
interior point of the quadrilateral to make four triangles as pictured below:
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Since the sum of the angles in a triangle is 180∘ we have
m(∠OAB) + m(∠OBA) + m(∠AOB) = 180
m(∠OBC) + m(∠OCB) + m(∠BOC) = 180
m(∠OCD) + m(∠ODC) + m(∠COD) = 180
m(∠ODA) + m(∠OAD) + m(∠DOA) = 180.
So adding all twelve of these angles gives 4 ×
180∘ = 720∘ . We know that
m(∠OAB) + m(∠OAD) = m(∠A)
m(∠OBC) + m(∠OBA) = m(∠B)
m(∠OCD) + m(∠OCB) = m(∠C)
m(∠ODA) + m(∠ODC) = m(∠D).
This leaves the four angles which are interior to the quadrilateral: ∠AOB, ∠BOC,
∠COD, and ∠DOA. These four angles make a full circle which means that
m(∠AOB) + m(∠BOC) + m(∠COD) + m(∠DOA) = 360.
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Putting all of this information together we conclude that
m(∠A) + m(∠B) + m(∠C) + m(∠D) = 720 − 360 = 360.
b. The same method applies to the pentagon ABCDE provided we carefully choose an
interior point, as below:
Since the sum of the angles in a triangle is 180∘ we have
m(∠OAB) + m(∠OBA) + m(∠AOB) = 180
m(∠OBC) + m(∠OCB) + m(∠BOC) = 180
m(∠OCD) + m(∠ODC) + m(∠COD) = 180
m(∠ODE) + m(∠OED) + m(∠EOD) = 180
m(∠OEA) + m(∠OAE) + m(∠EOA) = 180.
So adding all fifteen of these angles gives 5 ×
180∘ = 900∘ . We know that
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m(∠OAB) + m(∠OAE) = m(∠A)
m(∠OBC) + m(∠OBA) = m(∠B)
m(∠OCD) + m(∠OCB) = m(∠C)
m(∠ODE) + m(∠ODC) = m(∠D)
m(∠OEA) + m(∠OED) = m(∠E).
This leaves the five angles which are interior to the quadrilateral: ∠AOB, ∠BOC,
∠COD, ∠DOE, and ∠EOA. These five angles make a full circle which means that
m(∠AOB) + m(∠BOC) + m(∠COD) + m(∠DOE) + m(∠EOA) = 360.
Putting all of this information together we conclude that
m(∠A) + m(∠B) + m(∠C) + m(∠D) + m(∠E) = 900 − 360 = 540.
c. This method also works for part (c) but we cannot draw a complete picture since we
do not know how many sides the polygon Pn has. The assumption that Pn is convex
guarantees that any point O on the interior will divide Pn into n triangles by drawing
the line segments connecting O to each of the n vertices of Pn . The sum of the angles
in these n triangles is n × 180∘ . This gives the sum of the angles of Pn together with
the 360∘ of the circle at point O. So the sum of the angles of Pn is (n − 2) × 180∘ .
Note that if Pn were not convex then there is no guarantee that there is a point O
inside of Pn which can be joined to each vertex of Pn without going outside of the
polygon.
A-CED Sum of angles in a polygon
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