Question

ABCD is a parallelogram.X and Y are mid-point of BC and CD respectively. Prove that area of triangle AXY is 3/8 area of ABCD

Answer 1

Answer:

Step-by-step explanation:

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)]