Question

ABCD is a parallelogram X and Y are the midpoints of BC amd CD respectively. prove that area ∆AXY=3/8area∆BQC​

Answer 1

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.

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