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INTERIOR AND EXTERIOR ANGLES IN POLYGONS
Jump right to the exercises!
We'll start by talking about triangles, and then extend the ideas to any polygon.
It's easy to see that the sum of the angles in any triangle is 180°, as follows:
First, start with a right triangle (i.e., a triangle that has a 90° angle).
Make it into a rectangle, as shown below, to see that
α+β=
90° .
Thus, in any right triangle, the remaining two angles sum to 90° .
If you have trouble seeing the rectangle that is formed then study the following,
paying attention to angle α :
- Start with triangle (1).
- Slide it three units to the right, to get (2).
- Reflect about line h, to get (3).
- Reflect about line v, to get (4).
- Slide down two units, to get (5).
Now, consider any triangle.
Drop a perpendicular, as shown below, bringing two
right triangles into the picture.
Thus, we have
α+γ1
=90°
and
β+γ2
=90° .
Together,
α+γ1
+γ2+β
=180° , and so we have:
THEOREM: SUM OF ANGLES IN A TRIANGLE
The sum of the angles in any triangle is 180° .
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A quick note about terminology: you can't really SUM angles.
Angles are geometric figures, and we don't "add" geometric figures.
When we talk about adding angles, it is understood that we're adding the measures of the angles.
Not only can we explore the interior (inside) angles of polygons,
but we can also explore the exterior angles:
DEFINITION: EXTERIOR ANGLE IN A POLYGON
An exterior angle of a polygon is an angle between one side of a polygon,
and the extension of an adjacent side.
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Although two (equal) exterior angles can be drawn at each vertex of a polygon,
when we talk about summing the exterior angles of a polygon,
we agree to include only one exterior angle at each vertex.
For a triangle with interior angles
α ,
β , and
γ ,
we have the following:
(α+
αexterior)
+
(β+β
exterior)
+ (γ
+γexterior
)=3(180°
)
Regrouping gives:
(α+β
+γ)+(
αexterior
+βexterior
+γ
exterior)=3
(180°)
Since the sum of the (interior) angles of the triangle is 180°, it follows immediately
that
αexterior
+β
exterior+γ
exterior=
2(180°)=360°
Thus we have the following:
THEOREM: SUM OF EXTERIOR ANGLES IN A TRIANGLE
The sum of the exterior angles in any triangle is 360° .
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Indeed, you can think about the sum of the exterior angles in the following way.
Walk around the perimeter of the triangle above in a counter-clockwise direction.
At each vertex, you are turning an amount that is equal to an exterior angle.
When you reach your starting point, you are facing in exactly the same direction that you started,
so you have turned through a total of 360°—one complete revolution!
This is a nice way to think of why the exterior angles should sum to 360° .
(Try applying this thought process to the exterior angles in a more general polygon!)
SUM OF INTERIOR ANGLES IN ARBITRARY POLYGONS
The interior angles of any polygon always sum to a constant value, which depends only on the number of sides.
This is easy to see in many cases—take a polygon with
n sides, and divide it into
n-2 triangles,
as shown below:
Thus, the sum of the angles is (n-2)
(180°) .
If the polygon "caves in" then this method may break down, as illustrated below.
The result is still true–it's just harder to prove!
THEOREM: SUM OF INTERIOR ANGLES IN ARBITRARY POLYGONS
The sum of the interior angles in any n-gon is
(n-2)
(180°) .
Thus, in a regular polygon with n sides, each angle has measure
(n-2
)(180°)n
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You can explore interior and exterior angles of general polygons at these excellent links.
(Be patient; they may take a few minutes to load.)
exterior angle exploration
interior angle exploration
On this exercise, you will not key in your answers.
However, you can check to see if your answer is correct.
