Question

In trapezium $ABCD,AB//DC.M$ is mid-point of $AD$ and $N$ is mid-point of $BC$, then $DC=5.2$ cm. ( Enter 1 if true or 0 otherwise)

Answer 1

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, $QM+QN=\frac{1}{2}(AB+DC)$
$MN=\frac{1}{2}(AB+DC)$
Hence, $5.7=\frac{1}{2}(6.2+DC)$
$2\times 5.7=6.2+DC$
$DC=5.2$ cm