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

Given: In a trapezium ABCD, AB || DC, DC = 30 cm and AB = 50 cm.

Also, X and Y are respectively the mid-points of AD and BC.

To prove: `ar (DCYX) = 7/9 ar (XYBA)`

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

Proof: In ΔDCY and ΔPBY,

CY = BY  ...[Since, Y is the mid-point of BC]

∠DCY = ∠PBY  ...[Alternate interior angles]

And ∠2 = ∠3  ...[Vertically opposite angles]

∴ ΔDCY ≅ ΔPBY  ...[By ASA congruence rule]

Then, DC = BP  ...[By CPCT]

But DC = 30 cm  ...[Given]

∴ DC = BP = 30 cm

Now, AP = AB + BP

= 50 + 30

= 80 cm

In ΔADP, by mid-point theorem,

`XY = 1/2 AP` 

= `1/2 xx 80`

= 40 cm

Let distance between AB, XY and XY, DC is h cm.

Now, area of trapezium `DCYX = 1/2 h (30 + 40)`  ...[∵ Area of trapezium = `1/2` sum of parallel sides × distance between them]

= `1/2 h (70)`  

= 35 h cm² 

Similarly, area of trapezium XYBA

= `1/2 h (40 + 50)`

= `1/2 h xx 90`

= 45 h cm²

∴ `(ar (DCYX))/(ar (XYBA)) = (35h)/(45h) = 7/9`

⇒ `ar (DCYX) = 7/9 ar (XYBA)`

Hence proved.