prove that sum of the measurement of all the interior angles of a polygon of sides n is 2(n-2) right angle
Answers
Answer:
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.
Sum of the Interior Angles of a Polygon
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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)