Question

ABCD is a trapezium in which AB || CD, AB = 16 cm and DC = 24 cm. If E and F are respectively the midpoints of AD and BC, prove that ar(ABFE) = $\frac{9}{11}$ ar(EFCD).

Answer 1

Construction: Draw a perpendicular from point D to the opposite side CD, meeting CD at Q and EF at P.
Let length AQ = h
Given, E and F are the midpoints of AD and BC respectively.
So, EF || AB || DC and EF = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)=\left(\frac{a+b}{2}\right)$
E is the mid point of AD and EP || AB so by converse of mid point theorem,
EP || DQ and P will be the mid point of AQ.
$$\begin{gathered} \mathrm{ar}\left(\mathrm{ABFE}\right)=\frac{1}{2}\times \mathrm{AP}\left(\mathrm{AB}+\mathrm{EF}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right) \\ \mathrm{ar}\left(\mathrm{EFCD}\right)=\frac{1}{2}\times \mathrm{PQ}\left(\mathrm{CD}+\mathrm{EF}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right) \end{gathered}$$
ar(ABEF) : ar(EFCD) = (a + 3b) : (3a + b)
Here a = 24 cm and b = 16 cm
So, $\frac{\mathrm{ar}\left(\mathrm{ABEF}\right)}{\mathrm{ar}\left(\mathrm{EFCD}\right)}=\frac{24+3\times 16}{16+3\times 24}=\frac{9}{11}$