Advertisement Remove all ads

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°. - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Answer in Brief

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°.

Advertisement Remove all ads

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.

Concept: Introduction to Lines and Angles
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 10 Lines and Angles
Exercise 10.2 | Q 4 | Page 14
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×