Q.We know the sum of the interior angles of a triangle is $180°$. Show that the sums of the interior angles of polygons with $3, 4, 5, 6, \ldots$ sides form an arithmetic progression. Find the sum of the interior angles for a $21$ sided polygon.

The key idea is that any $n$-sided polygon can be triangulated into $(n-2)$ non-overlapping triangles by drawing diagonals from a single vertex.

For polygons with $3, 4, 5, 6$ sides, the sums are:

• $n=3$: $S_3 = (3-2) \times 180^\circ = 1 \times 180^\circ = 180^\circ$
• $n=4$: $S_4 = (4-2) \times 180^\circ = 2 \times 180^\circ = 360^\circ$
• $n=5$: $S_5 = (5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ$
• $n=6$: $S_6 = (6-2) \times 180^\circ = 4 \times 180^\circ = 720^\circ$ …

The core idea behind finding the sum of interior angles of any polygon is to break down the complex shape into simpler ones whose angle sums we already know. The simplest polygon is a triangle, and we are given that the sum of its interior angles is $180^\circ$. We can use this fundamental fact to derive the sum for any polygon.

Imagine any convex polygon. If you pick one vertex and draw all possible diagonals from that vertex to the other non-adjacent vertices, you will divide the polygon into a set of non-overlapping triangles. The crucial insight is that the sum of the interior angles of the polygon is exactly equal to the sum of the interior angles of all these triangles. This is because the angles around the chosen vertex combine to form the polygon's angle at that vertex, and the other angles of the triangles simply correspond to the other interior angles of the polygon.

Let's apply this method step-by-step.

1. Triangle ($n=3$ sides):

   A triangle already is a single triangle.

   The number of triangles formed is $1$.

   The sum of its interior angles is $1 \times 180^\circ = 180^\circ$.

2. Quadrilateral ($n=4$ sides):

   Consider a quadrilateral (a 4-sided polygon). Pick one vertex, say A. You can draw one diagonal from A to the opposite vertex C. This diagonal divides the quadrilateral into two triangles (e.g., $\triangle ABC$ and $\triangle ADC$).

   The number of triangles formed is $2$.

   The sum of its interior angles is $2 \times 180^\circ = 360^\circ$.

3. Pentagon ($n=5$ sides):

   Consider a pentagon (a 5-sided polygon). Pick one vertex. You can draw two diagonals from this vertex. These diagonals divide the pentagon into three triangles.

   The number of triangles formed is $3$.

   The sum of its interior angles is $3 \times 180^\circ = 540^\circ$.

4. Hexagon ($n=6$ sides):

   Consider a hexagon (a 6-sided polygon). Pick one vertex. You can draw three diagonals from this vertex. These diagonals divide the hexagon into four triangles.

   The number of triangles formed is $4$.

   The sum of its interior angles is $4 \times 180^\circ = 720^\circ$.

5. Generalizing for an $n$-sided polygon:

   From the examples above, we can observe a pattern:

   • For $n=3$ sides, we get $1$ triangle ($3-2$).
   • For $n=4$ sides, we get $2$ triangles ($4-2$).
   • For $n=5$ sides, we get $3$ triangles ($5-2$).
   • For $n=6$ sides, we get $4$ triangles ($6-2$). In general, for an $n$-sided polygon, drawing diagonals from one vertex divides it into $(n-2)$ triangles.

   The sum of the interior angles of an $n$-sided polygon, denoted $S_n$, is given by:

   $S_n = (n-2) \times 180^\circ$

6. Showing the sums form an Arithmetic Progression (AP):

   Let's list the sums for polygons with $3, 4, 5, 6, \ldots$ sides:

   • $S_3 = (3-2) \times 180^\circ = 1 \times 180^\circ = 180^\circ$
   • $S_4 = (4-2) \times 180^\circ = 2 \times 180^\circ = 360^\circ$
   • $S_5 = (5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ$
   • $S_6 = (6-2) \times 180^\circ = 4 \times 180^\circ = 720^\circ$ The sequence of sums is $180^\circ, 360^\circ, 540^\circ, 720^\circ, \ldots$.

   To check if this is an arithmetic progression, we find the difference between consecutive terms: …