In a trapezium ABCD,AB||DC,AB=a cm,and DC=b cm.If M and N are the midpoints of the nonparallel lines,AD and BC respectively then find the ratio of ar(DCNM) and ar(MNBA).

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Solution

Construction: Draw a perpendicular from point D to the opposite side AB, meeting AB at Q and MN at P.
Let length DQ = h
Given, M and N are the midpoints of AD and BC respectively.
So, MN || AB || DC and MN = 1/2AB+DC=a+b/2
M is the mid point of AD and MP || AB so by converse of mid point theorem,
MP || AQ and P will be the mid point of DQ.
arDCNM=1/2×DPDC+MN

ar(DCNM) : ar(MNBA) = (a + 3 b) : (3 a + b)

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