Question

ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = 7/9 ar (XYBA)

Answer 1

According to the question,

We have,

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)

Since, alternate opposite angles of DC || AP and BC is the transversal

∠DCY = ∠PBY

Since, Y is the midpoint of BC,

BY = CY

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

So, DY = YP and DC = BP

⇒ Y is the midpoint of AD

∴ XY || AP and XY = ½ AP (by midpoint theorem)

⇒ XY = ½ AP

= ½ (AB + BP)

= ½ (AB + DC)

= ½ (50 + 30)

= ½ × 80 cm

= 40 cm

Since X is the midpoint of AD

And Y is the midpoint of BC

Hence, trapezium DCYX and ABYX are of same height, h, cm

Now

⇒ 9 ar(DCXY) = 7 ar(XYBA)

⇒ ar(DCXY) = 7/9 ar(XYBA)

Hence Proved.