A) 21 : 22
B) 4 : 9
C) 16 : 81
D) 11 : 37
Correct Answer: B
Solution :
Let ABCD be the given parallelogram. If\[\angle DAB\]equal to\[\theta \], then\[\angle DCB\]will also be equal to\[\theta \]. DM and DN are the two altitudes. Now, in right angled triangle AMD, \[DM=4\,\,\sin \theta \] ...(1) Similarly, in right angled triangle CND, \[DN=9\,\,\sin \theta \] ...(2), From equations (1) and (2), \[\frac{DM}{DN}=\frac{4\sin \theta }{9\sin \theta }=\frac{4}{9}.\]You need to login to perform this action.
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