Question

In a trapezium ABCD with AB = 20 cm, CD = 10 cm, AM parallel to CD, P and Q are the mid point of AC and BD then the length of PQ is :

Answer 1

Let E and F are midpoints of the diagonals AC and BD of trapezium ABCD respectively.

Draw DE and produce it to meet AB at G.

Consider DAEG and DCED 

∠AEG = ∠CED (vertically opposite angles)

AE = EC (E is midpoint of AC)

∠ECD = ∠EAG (alternate angles)

△AEG ≅ △CED

⇒ DE = EG ......(1)

And AG = CD  .....(2)

In △DGB

E is the midpoint of DG [From (1)]

F is the midpoint of BD

. ^.. EF is parallel to GB

⇒ EF is parallel to AB

⇒ EF is parallel to AB and CD

Also, EF = 1/2 GB

⇒ EF = 1/2(AB - AG)

⇒ EF = 1/2 (AB - CD) [From (2)]

So, here PQ = 1/2 (20 - 10)

= 5 CM