Question

Question 10
ABCD is a trapezium with parallel sides. The ratio of ar (ABFE) and ar (EFCD) is:


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

Answer 1

The answer is B.
Given, AB = a cm, DC = b cm and AB || DC.
Also, E and F are mid-points of AD and BC, respectively.
So, the distance between CD, EF and AB, EF will be same, say h.
Join BD which intersect EF at M.
Now, in $△$ ABD, E and F are the midpoints of AD and BC, respectively.

EM || AB
So, M is the mid-point of BD.
And $EM=\frac{1}{2}AB$ [by mid - point theorem] ….(i)
Similarly, in $\Delta$ CBD, ME = $\frac{1}{2}$ CD
On adding Eqs. (i) and (ii), we get,
$EM+MF=\frac{1}{2}AB+\frac{1}{2}CD$
$EF=\frac{1}{2}(AB+CD)=\frac{1}{2}(a+b)$
Now, area of trapezium ABFE = $\frac{1}{2}$ (sum of parallel sides) $\times$ (distance between parallel sides)
$=\frac{1}{2}(a+\frac{1}{2}(a+b\left)\right)\times h=\frac{1}{4}(3b+a)h$
Now, area of trapezium EFCD $=\frac{1}{2}\{b+\frac{1}{2}(a+b\left)\right]\times h=\frac{1}{4}(3b+a)h$
$\therefore$ Required ratio = $\frac{AreaofABEF}{AreaofEFCD}=\frac{\frac{1}{4}(3a+b)h}{\frac{1}{4}(3b+a)h}$
=$\frac{(3a+b)}{(a+3b)}$ or (3a+b) : (a+3b)