Question

The figure, given below, shows a trapezium $ABCD$. $M$ and $N$ are the mid-points of the non-parallel sides $AD$ and $BC$ respectively. Find :

$MN$, if $AB=11$ cm and $DC=8$ cm.

• A. $9\cdot 5$ cm
• B. $8\cdot 5$ cm
• C. $7\cdot 5$ cm
• D. $6\cdot 5$ cm

Answer 1

The correct option is A $9\cdot 5$ cm

Join BD, let BD and MN meet at Q. Since, M is the mid point of AD and N is the mid point of BC. So by mid point theorem, AB II MN IICD
In $△$, BDC and BQN,
$\angle B=\angle B$ (Common)
$\angle BDC=\angle BQN$ (Corresponding angles of parallel lines)
$\angle BCD=\angle BNQ$ (Corresponding angles of parallel lines)
thus, $△BDC\sim △BQN$
Thus, $\frac{BD}{QB}=\frac{DC}{QN}$
$2=\frac{DC}{QN}$ (Q is the mid point of BD)
$QN=\frac{1}{2}DC$
Similarly, $QM=\frac{1}{2}AB$
Hence, $QM+QN=\frac{1}{2}(AB+DC)$
$MN=\frac{1}{2}(AB+CD)$
Hence, $MN=\frac{1}{2}(11+8)$ = 9.5 cm