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couple days ago in my math high school lessons I learned that sum of interior angles in convex polygon is:

$Z$ = sum of the angles, $n$ = number of sides in polygon

$Z=(n-2)\cdot 180$

Can someone help me understanding this formula, and why is it like this?

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    $\begingroup$ Hint chop the shape into triangles. $\endgroup$
    – Karl
    Feb 26, 2017 at 15:50
  • $\begingroup$ BTW, this is not restricted to the convex case $\endgroup$ Feb 26, 2017 at 15:53
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    $\begingroup$ The polygon can be "concave," but the theorem does not apply to _all_polygons: if you allow self-intersecting polygons, you can get $180n,$ $180(n+2),$ or a larger sum. $\endgroup$
    – David K
    Feb 26, 2017 at 16:24

6 Answers 6

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Here is a useful diagram:

enter image description here

The sum of all the angles in all the triangles equals the sum of the interior angles of the polygon. Notice that the shape has $7$ sides, and we are able to fit $5$ triangles inside it each of whose angles sum to $180$ degrees. This generalises to a $n$ sided polygon being able to fit $n-2$ triangles inside it as shown above. Then the sum of the angles must = $(n-2)*180$ degrees

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    $\begingroup$ If true for n=3, then true for all n>3. Induction on steroids. $\endgroup$
    – PatrickT
    Jun 2, 2020 at 20:16
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One way to understand this is to look at the exterior angles in a traverse one way around the polygon - the exterior angle being the angle turned in passing each vertex. (For example, in a regular decagon, the exterior angle at each vertex is $36°$.) Clearly the sum of such angles $\{\epsilon_i\}$ is $360°$, since you arrive back on the starting edge having turned fully around the polygon. The interior angles $\{\theta_i\}$ are each $\theta_k=180°-\epsilon_k$, so you have $$\sum^n \theta_i = \sum^n (180°-\epsilon_i) = n\cdot180° - \sum^n \epsilon_i = n \cdot 180°-360° = (n-2)180°$$


diagram for clarity

enter image description here

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  • $\begingroup$ I've seen this before but struggle to understand. Would you care to explain. Why should it be $360$, can't you say we also traverse around the inside, ending up in the same place. So the interior angles add to $360$? $\endgroup$ Feb 26, 2017 at 16:05
  • $\begingroup$ The exterior angle represent how much you have deviated from going straight on at the vertex. So we're not "inside" or "outside" the polygon, we're just using different labels to describe different aspects of the angle at the vertex. $\endgroup$
    – Joffan
    Feb 26, 2017 at 16:07
  • $\begingroup$ Also, if you imagine traversing around the inside of the polygon, take note of the fact that you don't make any turns that are fully through any of the interior angles. $\endgroup$
    – Dancrumb
    Feb 26, 2017 at 16:54
  • $\begingroup$ Here's another way to think about it $\endgroup$ Feb 28, 2017 at 9:52
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Hint Pick a point in the interior of the polygon. Then, drawing line segments from the point to each vertex divides the polygon into triangles, but you already know that the sum of the angles of a triangle is $180^{\circ}$.

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See the case, when $n=3$ i.e. the polygon is triangle. Now sum of all interior angles of a triangle is $180^0$ which matches with formula.
For $n>3$, we can divide the polygon into $n-2$ triangles. So, the sum of interior angles will be $(n-2)180$.

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Start with a triangle.

Choose an edge on the triangle and mark a point on it.

Now imagine pulling the point outwardly away from the edge.

What happens is the original triangle gains 2 more edges and there is an extra $180^{\circ} $ from the triangle formed by the original edge and the two new ones.

This has the advantage of explaining why there are $n-2$ less triangles as each triangle is paired with the $2$ parts of a broken edge.

If I get chance I'll add an animation or if someone wants to edit feel free. This can be extended to make the polygon grow.

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Hold two opposite corners of a polygon and start pulling the two away from each other. After a while, the interior angles of two corners you are holding would become zero and remaining corners will become straight line so total is (n-2) *180.

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