Question

In the given figure ABCD is a trapezium in which AB || DC such that AB = a cm and DC = b cm. If E and F are the midpoints of AD and BC respectively. Then, ar(ABFE) : ar(EFCD) = ?
(a) a : b
(b) (a + 3b) : (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)

Answer 1

(c) (3a +b) : (a +3b)

Clearly, EF = $\frac{1}{2}$ (a + b) [Mid point theorem]
Let d be the distance between AB and EF.
Then d is the distance between DC and EF.
$$\begin{gathered} \mathrm{Now},\mathrm{we}\mathrm{have}: \\ \mathrm{ar}(\mathrm{trap}ABEF)=\frac{1}{2}\left(a+\frac{a+b}{2}\right)d=\frac{\left(3a+b\right)d}{4} \\ \mathrm{ar}(\mathrm{trap}EFCD)=\frac{1}{2}\left(b+\frac{a+b}{2}\right)d=\frac{\left(a+3b\right)d}{4} \\ \therefore \mathrm{Required}\mathrm{ratio}=\frac{\left(3a+b\right)d}{4}:\frac{\left(a+3b\right)d}{4}=(3a+b):(a+3b) \end{gathered}$$