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.
From Sum of the Interior Angles of an n-sided Polygon to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Apr 20, 24 05:39 PM
Apr 20, 24 05:29 PM
Apr 19, 24 04:01 PM
Apr 19, 24 01:50 PM
Apr 19, 24 01:22 PM