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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)
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Solution
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.
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