The parallel sides of a trapezium are a and b respectively.

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 $\mathrm{MF}=\frac{1}{2}(b)$

$\therefore E F=E M+M F=\frac{1}{2}(a+b)$

 

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