Question

The parallel sides of a trapezium are ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ respectively. The line joining the midpoints of its nonparallel sides will be

(a) $\frac{1}{2}(a-b)$

(b) $\frac{1}{2}(a+b)$

(c) $\frac{2ab}{(a+b)}$

(d) $\sqrt{ab}$

Answer 1

The correct option is (b) $\frac{1}{2}(a+b)$

Given: The parallel sides ($AB$ and $CD$) of the given trapezium are of length ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$.

A line joining the midpoints of the non-parallel sides of a trapezium will always be parallel to the parallel sides of the trapezium

i.e. $EF\parallel AB\parallel CD$

In $△ADB,$ we have

$E$ is the midpoint of $AD,$ and $EG\parallel AB$ [$\because EF\parallel AB$]

By Converse of midpoint theorem, $G$ will midpoint of $BD.$ and $EG=\frac{1}{2}AB$

i.e. $EG=\frac{1}{2}b\dots \dots \left(i\right)$

In $△DBC,$ we have

$G$ is the midpoint of $BD$ and $F$ is the midpoint of $BC.$

By midpoint theorem, $GF\parallel CD,$ and $GF=\frac{1}{2}CD$

i.e. $GF=\frac{1}{2}a\dots \dots \left(ii\right)$

On adding eq.(i) and (ii), we get

$EG+GF=\frac{1}{2}b+\frac{1}{2}a$

$\therefore EF=\frac{1}{2}(a+b)$

Hence,the line joining the midpoints of its nonparallel sides is equal to $\frac{1}{2}(a+b)$.