Question

in trapezium ABCD with AB=20CM, CD=10CM, AB//CD, P and Q are mid points of AC and BD then length of PQ is:-
a) 10cm
b) 5cm
c) 2cm
d) none
And give explanation ​

Answer 1

Step-by-step explanation:

take one more mid point S on PQ.

then in two triangles ABD and ACD.

Using similar triangles concept :

PS/CD=QS/AB=PS/SQ=CD/AB=10/20.

add the values u will get your answer.

Answer 2

Length of PQ is b) 5cm

Let midpoints of the diagonals AC and BD of trapezium ABCD be = E and F

In ΔDAG and ΔDCE

∠AEG = ∠CED (vertically opposite angles)

AE = EC (E is midpoint of AC)

∠ECD = ∠EAG (alternate angles)

Therefore, △AEG ≅ △CED

DE = EG --- eq 1

AG = CD  --- eq 2

In △DGB E is the midpoint of DG ( 1 )

F is the midpoint of BD

Therefore, EF ║ GB  and EF ║ AB = EF is ║ AB and CD

Also, EF = 1/2 GB

⇒ EF = 1/2(AB - AG)

⇒ EF = 1/2 (AB - CD) (2)

Thus, PQ = 1/2 (20 - 10)

= 1/2 ( 10)

= 5