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# In a $\Delta$ABC, points M and N respectively lie on side AB and AC such that area of triangle ABC is double than the area of trapezium BMNC. The ratio AM : MB is

[ A ]    (2 - $\sqrt 2$) : 1
[ B ]    (2 + $\sqrt 2$) : 1
[ C ]    ($\sqrt 2$ + 1) : 1
[ D ]    ($\sqrt 2$ - 1) : 1
 Answer : Option C Explanation :