Question

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

• A. $9.5$ cm
• B. $11.5$ cm
• C. $7.5$ cm
• D. None of these

Answer 1

Answer: (A)

$9.5$ 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