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).

Options

A. `a : b`
B. `(3a + b) : (a + 3b)`
C. `(a + 3b) :(3a + b)`
D. `(2a + b) : (3a + b)`

Answer

B

Explanation

Given, AB = a cm, DC = b cm and AB|| DC.
Also, E and F are mid-points of AD and BC, respectively.
So, distance between CD, EF and AB, EF will be same say h.
Join BD which intersect EF at M.

Now, in `DeltaABD, E` is the mid-point of AD and EM || AB
So, M is the mid-point of BD
and `" "EM = (1)/(2) AB" "` [by mid-point theorem] ...(i)
Similarly in `DeltaCBD" "`, `MF = (1)/(2)CD" "` ...(ii)
On adding, Eqs. (i) and (ii), we get
`EM + MF = (1)/(2)AB + (1)/(2)CD`
`rArr` `EF = (1)/(2)(AB + CD) = (1)/(2)(a + b)`
Now, area of trapezium `ABFE = (1)/(2) ("sum of parallel sides") xx ("distance between parallel sides")`
`=(1)/(2) (a + (1)/(2)(a + b)) xx h = (1)/(4)(3a + b)h`
Now, area of trapezium EFCD `=(1)/(2)[b + (1)/(2)(a + b)] xx h = (1)/(4)(3b + a)h`
`therefore` `"Required ratio" = ("Area of " ABFE)/("Area of " EFCD) = ((1)/(4)(3a + b)H)/((1)/(4)(3b + a)H)`
`= ((3a + b))/((a + 3b)) "or " (3a + b) : (a + 3b)`