Question

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 1

Construction: Draw a perpendicular from point D to the opposite side AB, meeting AB at Q and MN at P.
Let length DQ = h
Given, M and N are the midpoints of AD and BC respectively.
So, MN || AB || DC and MN = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)=\left(\frac{a+b}{2}\right)$
M is the mid point of AD and MP || AB so by converse of mid point theorem,
MP || AQ and P will be the mid point of DQ.
$$\begin{gathered} \mathrm{ar}\left(\mathrm{DCNM}\right)=\frac{1}{2}\times \mathrm{DP}\left(\mathrm{DC}+\mathrm{MN}\right)=\frac{1}{2}h\left(b+\frac{a+b}{2}\right)=\frac{h}{4}\left(a+3b\right) \\ \mathrm{ar}\left(\mathrm{MNBA}\right)=\frac{1}{2}\times \mathrm{PQ}\left(\mathrm{AB}+\mathrm{MN}\right)=\frac{1}{2}h\left(a+\frac{a+b}{2}\right)=\frac{h}{4}\left(b+3a\right) \end{gathered}$$
ar(DCNM) : ar(MNBA) = (a + 3b) : (3a + b)