Question:
The parallel sides of a trapezium are a and b respectively. The line joining the mid-points of its non-parallel sides will be
(a) $\frac{1}{2}(a-b)$
(b) $\frac{1}{2}(a+b)$
(C) $\frac{2 a b}{(a+b)}$
(d) $\sqrt{a b}$
Solution:
(b) $\frac{1}{2}(a+b)$
Explanation:
Suppose ABCD is a trapezium.
Draw EF parallel to AB.
Join BD to cut EF at M.
Now, in ∆ DAB, E is the midpoint of AD and EM || AB.
$\therefore M$ is the mid point of $\mathrm{BD}$ and $\mathrm{EM}=\frac{1}{2}(a)$
Similarly, M is the mid point of BD and MF || DC.
i.e., $F$ is the midpoint of $\mathrm{BC}$ and $\mathrm{MF}=\frac{1}{2}(b)$
$\therefore E F=E M+M F=\frac{1}{2}(a+b)$