Question

The diagonals of a parallelogram ABCD intersect at O. A line through O meets AB at X and CD at Y. Show that the
area of quadrilateral AXYD is half the area of parallelogram ABCD

Answer 1

hey these is the answer of your easy question .

Answer 2

Area of Quadrilateral AXYD   = (1/2)  Area of parallelogram ABCD

Step-by-step explanation:

Diagonal of a parallelogram Bisect each other

=> AO = CO

ΔAOX  & ΔCOY

∠XAO = ∠YCO   ( as opposite sides of parallelogram are parallel)

∠AOX = ∠COY  ( oppoiste angles)

AO = CO

=> ΔAOX  ≅ ΔCOY

=> AX = CY

Now Area of Quadrilateral AXYD

= Area of ΔAXY + Area of ΔADY

Area of ΔAXY = (1/2) AX  * (Distance between AB & CD ) =  (1/2) CY  * (Distance between AB & CD )

Area of ΔADY =  (1/2) DY  * (Distance between AB & CD)

=> Area of Quadrilateral AXYD  = 1/2) CY  * (Distance between AB & CD ) + (1/2) DY  * (Distance between AB & CD)

= (1/2) (CY + DY) *  (Distance between AB & CD)

= (1/2) CD * (Distance between AB & CD)

CD * (Distance between AB & CD) = Area of parallelogram ABCD

=> Area of Quadrilateral AXYD   = (1/2)  Area of parallelogram ABCD

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