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

Answer

`(a+3b):(3a+b)`

Explanation

M and N are midpoints of AD and BC respectively
`rArr` MN||AB||DC and MN `=(1)/(2)(AB+DC)=((a+b)/(2))`.
Draw `DQ bot AB`. Let DQ cut MN at P.
Then, P is the midpoint of DQ,
i.e., DP = PQ = h (say).
Now, `ar(DCNM)=(1)/(2)xx(DC+MN)xxDP=(1)/(2)=(1)/(2)xx(b+(a+b)/(2))xxh=(h)/(4)xx(a+3b)`
and `ar(MNBA)=(1)/(2)xx(MN+AB)xxPQ = (1)/(2)xx((a+b)/(2)+a)xxh =(h)/(4)xx (3a+b)`.
`therefore ar(DCNM):ar(MNBA)=(a+3b):(3a+b)`.