Given: OA, OB, OC, OD and OE have common end point O.To prove: ∠ AOB + ∠ BOC+ ∠ COD + ∠ DOE + ∠ EOA = 360&#x00b0.Construction: Draw the ray opposite to OA.Proof: Adjacents, ∠COF and ∠COA form a linear pair.&#x21d2 ∠COF + ∠COA = 180&#x00b0 (Linear Pair Axiom) --------------(1)Similarly, adjacents, &#x00d0FOD and ∠DOA form a linear pair.&#x21d2 ∠FOD + ∠DOA = 180&#x00b0 --------------- (2) Adding (1) and (2) ∠COF + ∠COA + ∠FOD + ∠DOA = 180&#x00b0 + 180&#x00b0(∠COF + ∠FOD) + ∠COB + ∠BOA + (∠EOA + ∠DOE) = 360&#x00b0 ∠ AOB + ∠ BOC+ ∠ COD + ∠ DOE + ∠ EOA = 360&#x00b0.

Q. 205.0( 2 Votes )

Rays OA, OB, OC, OD

Class 9th

Answer :

Given: OA, OB, OC, OD and OE have common end point O.
To prove: ∠ AOB + ∠ BOC+ ∠ COD + ∠ DOE + ∠ EOA = 360°.
Construction: Draw the ray opposite to OA.
Proof: Adjacents, ∠COF and ∠COA form a linear pair.
⇒ ∠COF + ∠COA = 180°     (Linear Pair Axiom) --------------(1)
Similarly, adjacents, ഏOD and ∠DOA form a linear pair.
⇒ ∠FOD + ∠DOA = 180°                              --------------- (2)
Adding (1) and (2)
                            ∠COF + ∠COA + ∠FOD + ∠DOA = 180° + 180°
(∠COF + ∠FOD) + ∠COB + ∠BOA + (∠EOA + ∠DOE) = 360°
            ∠ AOB + ∠ BOC+ ∠ COD + ∠ DOE + ∠ EOA = 360°.

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