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ABCD is parallelogram, G is the point on AB such that AG = 2GB, E is point on DC such that CE = 2DE and F is the point on BC such that BF = 2FC. Then, match the following:
| Column-I | Column-II | 
| P. ar (ADEG) | (i)   ar (ABCD) | 
| Q. ar (∆EGB) | (ii) ar (∆EDG) | 
| R. ar (∆EFC) | (iii) ar (GBCE) | 
| S. ar (∆EGB) | (iv)   ar (∆EBF) | 
| P | Q | R | S | |
| A | (i) | (ii) | (iii) | (iv) | 
| B | (iii) | (i) | (iv) | (ii) | 
| C | (iii) | (ii) | (iv) | (i) | 
| D | (ii) | (i) | (iii) | (iv) | 
Why cant it be C
 ar (ABCD)
 ar (∆EBF)