ABCD is a trapezium in which AB is parallel to DC. DC=30 cm. and AB=50cm. if  X  and Y   are respectively the mid points of AD and BC, prove that     (Nine times the area of DCYX= seven times the area of XYBA.)

Dear Student!

Here is the answer to your query.

 

 

Given :  ABCD is a trapezium with AB || DC

Construction :  Join DY and produce it to meet AB produced at P.

 

In ∆BYP and ∆CYD

∠BYP  = ∠CYD    (vertically opposite angles)

 ∠DCY  = ∠PBY  (∵ Alternate opposite angles as DC || AP and BC is the transversal)

and BY = CY  (Y is the mid point of BC)

Thus ∆BYP ≅ ∆CYD (by ASA cogence criterion)

 

⇒ DY = YP and DC = BP

⇒ Y is the mid point of AD

∴ XY || AP and XY = AP  (mid point theorem)

⇒ XY = AP = (AB + BP) = (AB + DC) = (50 + 30) = × 80 cm = 40 cm

 

Since X and Y are the mid points of AD and BC respectively.

∴ trapezium DCYX and ABYX are of same height, say h cm

Now

⇒ 9 ar(DCXY) = 7 ar(XYBA)

 

Cheers!

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