CBSE Class 9 Answered

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 =360^o   

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

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

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

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

                                                                                                    [By using angle sum property]

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

         = 900^o - 180^o = 720^o