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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.

178848.JPG


A
95 cm
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B
85 cm
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C
75 cm
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D
65 cm
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Solution

The correct option is A $$ 9\cdot5 $$ 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 $$\triangle $$, 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 cm

Mathematics

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