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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