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 :

$DC$, if $MN=15$ cm and $AB=23$ cm.

• A. 7 cm
• B. 9 cm
• C. 2 cm
• D. 8 cm

Answer 1

The correct option is A 7 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, $2\times 15=23+CD$
$CD=30-23=7cm$