# 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:

 Statement Reason 1. As the polygon has n sides, n triangles are formed, namely, ∆OPQ, ∆QR, ...., ∆OZP. 1. On each side of the polygon one triangle has been drawn. 2. The sum of all the angles of the n triangles is 2n right angles. 2. The sum of the angles of each triangle is 2 right angles. 3. ∠P + ∠Q + ∠R + ..... + ∠Z + (sum of all angles formed at O) = 2n right angles. 3. From statement 2. 4. ∠P + ∠Q + ∠R + ..... + ∠Z + 4 right angles = 2n right angles. 4. Sum of angles around the point O is 4 right angles. 5. ∠P + ∠Q + ∠R + ..... + ∠Z            = 2n right angles - 4 right angles           = (2n – 4) right angles            = (2n – 4) 90°.        (Proved) 5. From statement 4.

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.

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