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
95 cm
B
85 cm
C
75 cm
D
65 cm

Solution

## The correct option is A $$9\cdot5$$ cmJoin 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 IICDIn $$\triangle$$, 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, $$\triangle BDC \sim \triangle BQN$$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 cmMathematics

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