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Q.

If M and N are the mid points of the sides BC and CD respectively of a parallelogram ABCD, then $\bar{AM} + \bar{AN} =$

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a

$\frac{5}{3} \bar{AC}$

b

$\frac{4}{3} \bar{AC}$

c

$\frac{3}{2} \bar{AC}$

d

$\frac{6}{5} \bar{AC}$

answer is C.

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Detailed Solution

Question Image

$OM = \frac{a + a + b}{2} ON = \frac{a + b + b}{2} AM + AN = OM + ON = \frac{3 a + 3 b}{2} = \frac{3}{2} (AC)$

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