Question

and bd 9f parallelogram abcd prove that ar(cyx)=1/4ar(dbc)

Answer 1

Answer:

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Step-by-step explanation:

Construction :- Join points B and D. Since X and Y are the mid points of sides BC and CD respectively, In ∆BCD, XY|| BD and XY = BD. ⇒ ar (∆CYX) = ar (∆DBC) ⇒ ar (∆CYX) = ar (||gm ABCD) [Area of parallelogram is twice the area of triangle made by the diagonal] Since parallelogram ABCD and ∆ABX are between the same parallel lines AD and BC and BX = BC. ar (∆ABX) = ar (||gm ABCD) Similarly, ar (∆AYD) = ar (||gm ABCD) Now, ar (∆AXY) = ar (||gm ABCD) – [ar (∆ABX) + ar (∆AYD) + ar (∆CYX)] ar (||gm ABCD) + ar (||gm ABCD) + ar (||gm ABCD)]

Answer 2

Step-by-step explanation:

1

Since X and Y are the mid points of sides BC and CD respectively, In ∆BCD, XY|| BD and XY = 1/2 BD.

⇒ ar (∆CYX) = 1/4 ar (∆DBC)

⇒ ar (∆CYX) = 1/8 ar (||gm ABCD)

SINCE, Area of Parallelogram is twice the Area of triangle made by the diagonal.

And parallelogram ABCD and ∆ABX are between the same parallel lines AD and BC and BX = 1/2BC.

=> ar( ABX ) = 1/4 ar (llgm ABCD)

Now, ar (∆AXY) = ar (||gm ABCD) – [ar (∆ABX) + ar (∆AYD) + ar (∆CYX)]

=> Ar(ABCD) - 1/4 ar (||gm ABCD) + 1/4 ar (||gm ABCD) + 1/8ar (||gm ABCD)]

=> Ar(ABCD) - 5/8 (llgm ABCD)

=> 3/8 ar( IIgm ABCD).

✔✔✔

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Comments Report

lava88

lava88 Ace

Given that ABCD ia a //gm. X and Y are the mid points of BC and CD

Construction: Join BD

Since X and Y are the mid points of sides BC and CD respectively, therefore in triangle BCD, XY//BD and XY=1/2 BD

implies area of triangle CYX= 1/4 area of triangle DBC

{ In triangle BCD, if X is the mid point of BC and Y is the mid pt of CD then area triangle CYX=1/4 area triangle DBC}

IMPLIES AREA TRIANGLE CYX= 1/8 area //gm ABCD

[Area of//gm is twice the area of triangle made by the diagonal]

Since //gm ABCD and triangle ABX are between same // lines AB and BC and BX=1/2BC

Therefore, area triangle ABX= 1/4 area //gm ABCD

Similarly, area triangle AYD= 1/4 area //gm ABCD

Now, area triangle AXY= area//gm ABCD- {ar triangleABX + ar AYD + ar CYX}

= ar//gmABCD - {1/4 + 1/4 + 1/8} area //gmABCD

=area//gmABCD- 5/8 area //gmABCD

=3/8 area //gm ABCD.