question on LINES AND ANGLES.........

Asked by krithika.rrrr | 29th May, 2009, 11:12: AM

Expert Answer:

The correct option is (b)

Let us the denote the centre point by O.

AOJ = EOF    (vertically opposite angles)

Simlarly, JOI = DOE COD = IOH, HOG = COB, GOF= BOA  (vertically opposite angles)

And AOJ +JOI + IOH+ HOG+ GOF + EOF+ DOE+COD+ COB+ BOA =360o   

2 (JOI + HOG + EOF + COD + BOA ) = 360o

Therefore, JOI + HOG + EOF + COD + BOA =180o       ..............(eqn 1)

Now, (A+B)+(C+D)+(E+F)+(G+H)+(I+J)

         =(180o - BOA )+(180o - COD )+(180o - EOF )+(180o - HOG)+(180o - JOI)      

                                                                                                    [By using angle sum property]

         = 900o - (JOI + HOG + EOF + COD + BOA)

         = 900o - 180o = 720o

Answered by  | 29th May, 2009, 04:28: PM

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