Sum of the Interior Angles of an n-sided Polygon

Here we will discuss the theorem of sum of the interior angles of an n-sided polygon and some related example problems.

The sum of the interior angles of a polygon of n sides is equal to (2n - 4) right angles.

Given: Let PQRS .... Z be a polygon of n sides.

To prove: ∠P + ∠Q + ∠R + ∠S + ..... + ∠Z = (2n – 4) 90°.

Construction: Take any point O inside the polygon. Join OP, OQ, OR, OS, ....., OZ.

Proof:

Note:

1. In a regular polygon of n sides, all angles are equal.

Therefore, each interior angle = \(\frac{(2n - 4) × 90°}{n}\).

2. A quadrilateral is a polygon for which n = 4.

Therefore, the sum of interior angles of a quadrilateral = (2 × 4 – 4) × 90° = 360°

Solved examples on finding the sum of the interior angles of an n-sided polygon:

1. Find the sum of the interior angles of a polygon of seven sides.

Solution:

Here, n = 7.

Sum of the interior angles = (2n – 4) × 90°

                                      = (2 × 7 - 4) × 90°

                                      = 900°

Therefore, the sum of the interior angles of a polygon is 900°.

2. Sum of the interior angles of a polygon is 540°. Find the number of sides of the polygon.

Solution:

Let the number of sides = n.

Therefore, (2n – 4) × 90° = 540°

⟹ 2n - 4 = \(\frac{540°}{90°}\)

⟹ 2n - 4 = 6

⟹ 2n = 6 + 4

⟹ 2n = 10

⟹ n = \(\frac{10}{2}\)

⟹ n = 5

Therefore, the number of sides of the polygon is 5.

3. Find the measure of each interior angle of a regular octagon.

Solution:

Here, n = 8.

The measure of each interior angle = \(\frac{(2n – 4) × 90°}{n}\)

                                                   = \(\frac{(2 × 8 – 4) × 90°}{8}\)

                                                   = \(\frac{(16 – 4) × 90°}{8}\)

                                                   = \(\frac{12 × 90°}{8}\)

                                                   = 135°

Therefore, the measure of each interior angle of a regular octagon is 135°.

4. The ratio of the number of sides of two regular polygons is 3:4, and the ratio of the sum of their interior angles is 2:3. Find the number of sides of each polygon.

Solution:

Let the number of sides of the two regular polygons be n\(_{1}\) and n\(_{2}\).

According to the problem,

\(\frac{n_{1}}{n_{2}}\) = \(\frac{3}{4}\)

⟹ n\(_{1}\) = \(\frac{3n_{2}}{4}\) ........... (i)

Again, \(\frac{2(n_{1} – 2) × 90°}{2(n_{2} – 2) × 90°}\) = \(\frac{2}{3}\)

⟹ 3(n\(_{1}\) – 2) = 2(n\(_{2}\) – 2)

⟹ 3n\(_{1}\) = 2n\(_{2}\) + 2

⟹ 3 × \(\frac{3n_{2}}{4}\) = 2n\(_{2}\) + 2

⟹ 9n\(_{2}\) = 8n\(_{2}\) + 8

Therefore, n\(_{2}\) = 8.

Substituting the value of n\(_{2}\) = 8 in (i) we get,

n\(_{1}\) = \(\frac{3}{4}\) × 8

⟹ n\(_{1}\) = 6.

Therefore, the number of sides of the two regular polygons be 6 and 8.

9th Grade Math

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