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))

#### Solution

(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 cm^{2}

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

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