    Question

# ABCD is a trapezium in which AB is parallel to DC, AD = BC, AB = 6 cm, AB = EF and DF = EC. If the area of ABEF is half of ABCD, then find DF/CD. A
1/4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1/3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2/5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1/6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 1/3 In the above question: FE = AB = 6 cm ΔADF=ΔBCE;so DF=ECLet DF=EC=x Solving through options; e.g., option (b) 1/3; x = 6 Then by Pythagoras triple AF = 8 Area of ABEF =8×6=48 cm2Area of ΔAFD+ΔBEC=2×12×6×8⇒48 cm2 ∴ Area of ABCD = 48 + 48 = 96 cm2. Hence the condition is proved. Alternatively, Let the height of the trapezium ABCD be h cm and DF = EC = x cm Now, Area of quadrilateral ABEF = 6*h cm2 Area of ABCD = 12× (12+2x)× h Given, 6*h = 1/2*12× (12+2x)× h x = 6 cm Therefore, required ratio = 618 =1/3.  Suggest Corrections  1      Similar questions  Explore more