ABCD is a trapezium with AB||DC . x and y are mid points of sides AD and BC respectively .If CD =24cm and AB=16cm , show that area of ABXY = 9/11 area of XYCD
Answers
Answer:
Area of ABYX = 9 *(Area of XYCD)/11
Step-by-step explanation:
See the attached diagram for the problem description.
Join AY and extend it to meet DC at point P.
From the picture we can see
∠BYP = ∠BYA (Vertically opposite angles)
∠ABY = ∠PCY [Since AB || DP and CB is line cutting both parallel lines]
and BY = CY (Since Y is the mid-point of BC).
So, according to Angle - Side - Angle congruence criterion, we get
ΔCYP ≅ ΔAYB
⇒ AY = YP and AB = CP
In ΔADP, X is the midpoint of AD and Y is midpoint of AP
∴ XY || AP and XY = ½DP
⇒ XY = ½ (CD + CP)
⇒ XY = ½ (CD + AB)
⇒ XY = ½ (24 + 16)
= ½ × 40 = 20 cm.
Also, We have
XY || DP
⇒ XY || DC and AB || DC
⇒ XY || DC
⇒ DCYX is a trapezium.
Since X and Y are the mid-points of AD and BC respectively.
Therefore trapezium DCYX and ABYX are of same height and assuming it as h cm.
Area of Trapezium DCYX = (DC + XY) × h/2
= (24 + 20) h/2
= 22h cm2
Area of Trapezium ABYX) = (AB + XY) × h/2
= (16 + 20)h/2
= 18h cm2
So, Area of ABYX) / Area of DCYX = 18h / 22h = 9/ 11
⇒ Area of ABYX = 9 *(Area of XYCD)/11
