Question

4. In trapezium ABCD, M and N are mid-points
of AC and BD, and AB II CD. Find
(1) MN if AB=7.5 cm, CD=5.5 cm
(11) AB if CD = 16 cm, MN = 10 cm
(in) CD if AB = 14 cm, MN = 16.5 cm

Answer 1

Answer:

ANSWER

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,

∠B=∠B (Common)

∠BDC=∠BQN (Corresponding angles of parallel lines)

∠BCD=∠BNQ (Corresponding angles of parallel lines)

thus, △BDC∼△BQN

Thus,

QB

BD

=

QN

DC

2=

QN

DC

(Q is the mid point of BD)

QN=

2

1

DC

Similarly, QM=

2

1

AB

Hence, QM+QN=

2

1

(AB+DC)

MN=

2

1

(AB+CD)

Hence, MN=

2

1

Answer 2

Answer:

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