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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
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B
1/3
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C
2/5
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D
1/6
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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=2×12×6×848 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.

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