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
• B. 1/3
• C. 2/5
• D. 1/6

Answer 1

The correct option is B 1/3


In the above question:
FE = AB = 6 cm
$\Delta ADF=\Delta BCE;soDF=ECLetDF=EC=x$
Solving through options; e.g., option (b) 1/3; x = 6
Then by Pythagoras triple AF = 8
Area of ABEF $=8\times 6=48c{m}^{2}Areaof\Delta AFD+\Delta BEC=2\times \frac{1}{2}\times 6\times 8\Rightarrow 48c{m}^2$
$\therefore$ Area of ABCD = 48 + 48 = 96 $c{m}^2$. 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 $c{m}^2$
Area of ABCD = $\frac{1}{2}\times (12+2x)\times h$
Given, 6*h = 1/2*$\frac{1}{2}\times (12+2x)\times h$
x = 6 cm
Therefore, required ratio = $\frac{6}{18}$ =1/3.