CBSE Class 9CBSE
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

In the Given Below Fig, Rays Oa, Ob, Oc, Op and 0e Have the Common End Point O. Show that ∠Aob + ∠Boc + ∠Cod + ∠Doe + ∠Eoa = 360°. - CBSE Class 9 - Mathematics

Login
Create free account


      Forgot password?

Question

In the given below fig, rays OA, OB, OC, OP and 0E have the common end point O. Show
that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.

Solution 1

Given that

Rays OA, OB, OD and OE have the common end point O.

A ray of opposite to OA is drawn

Since `∠`AOB, `∠`BOF are linear pairs

`∠`AOB + `∠`BOF = 180°

`∠`AOB + `∠`BOC + `∠`COF = 180°

Also

`∠`AOE, `∠`EOF are linear pairs

`∠`AOE + `∠`EOF = 180°

`∠`AOE + `∠`DOF + `∠`DOE = 180°

By adding (1) and (2) quations we get                       

`∠`AOB + `∠`BOC + `∠`COF + `∠`AOE + `∠`DOF + `∠`DOE = 360°

`∠`AOB + `∠`BOC + `∠`COD + `∠`DOE + `∠`EOA = 360°

Hence proved.

Solution 2

Let us draw AOXa straight line.

∠AOE,∠DOE and ∠DOXform a linear pair. Thus, their sum should be equal to180°.

Or, we can say that:

 ∠AOE +∠DOE +∠DOX  = 180°   (I)

Similarly,, ∠AOB,∠BOC and ∠COXform a linear pair. Thus, their sum should be equal to180°.

Or, we can say that:

 ∠AOB +∠BOC+ ∠COX =  180°       (II)

On adding (I) and (II), we get:

∠AOB +∠BOC + ∠COX +∠DOX +∠AOE +∠DOE = 180°+180°

∠AOB +∠BOC + ∠COD +∠AOE +∠DOE = 360°

Hence proved.

  Is there an error in this question or solution?

APPEARS IN

Solution In the Given Below Fig, Rays Oa, Ob, Oc, Op and 0e Have the Common End Point O. Show that ∠Aob + ∠Boc + ∠Cod + ∠Doe + ∠Eoa = 360°. Concept: Concept to Lines and Angles.
S
View in app×