Question

In the below fig. ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60
cm. If X and Y are respectively, the mid-points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar (trap. DCYX) =`9/11`ar (trap. (XYBA))

Answer 1

(1) Join DY and produce it to meet AB produced at P
In Δ' s BYP  and CYDwe have

∠BYP =(∠CYD)           [Vertical opposite angles]

∠DCY ∠ (∠CYD)           [∴ DC || AP ]

And By = CY

So, by ASA congruence criterion, we have

  Δ BYP ≅ CYD

⇒ DY  =YP and DC = BP

⇒ y is the midpoint of DP
Also, x is the midpoint of AD

∴ XY || AP and XY = `1/2`AD 

⇒ `xy = 1/2 ( AB + BD)`

⇒ `xy = 1/2 (BA + DC )⇒ xy  = 1/2 ( 60 + 40 )`

(2) We have 

XY  || AP 

⇒ XY || AB and AB  || DC         [As proved above]

⇒  XY ||  DC

⇒ DCY is a trapezium

 (3)   Since x and y are the midpoint of DA and CB respectively
 ∴Trapezium DCXY and ABYX are of the same height say hm 

Now

ar (Trap DCXY) =`1/2`(DC + XY) × h 

 `= 1/2   (50 + 40) hcm^2  = 45  hcm^2`

⇒ `ar (trap  ABXY) = 1/2 (AB + XY) xx h = 1/2 (60 + 50) hm^3 `

⇒`ar (trap  ABXY) = 1/2 (AB + XY) xx h = 1/2 (60 + 50) hm^2 `

 = 55 cm²

` (ar trap (YX))/(ar trap (ABYX)) = (45h )/ (55h)= 9 /11  `  

⇒ ar (trap DCYX) =`9/11` ar (trap ABXY)