6. Ans: A
Observe that $\triangle A B D \cong \triangle E B D$. Thus, $B E=A B=4, A D=D E$. Hence
Area of $\triangle A C E=2 \times$ Area of $\triangle D C E$.
Since $B C=3, E C=E B-B C=4-3=1$. Thus,
$$
\text { Area of } \begin{aligned}
\triangle A B C & =3 \times \text { Area of } \triangle A C E . \\
& =6 \times \text { Area of } \triangle D C E .
\end{aligned}
$$

Thus,
$$
\text { Area of } \begin{aligned}
A B C D & =\text { Area of } \triangle A B C+\text { Area of } \triangle A C D \\
& =6 \times \text { Area of } \triangle D C E+\text { Area of } D C E \\
& =7 \times \text { Area of } \triangle D C E .
\end{aligned}
$$
6. In the following figure, $A B$ is the diameter of a circle with centre at $O$. It is given that $A B=4 \mathrm{~cm}, B C=3 \mathrm{~cm}, \angle A B D=\angle D B E$. Suppose the area of the quadrilateral $A B C D$ is $x \mathrm{~cm}^{2}$ and the area of $\triangle D C E$ is $y \mathrm{~cm}^{2}$. Find the value of the ratio $\frac{x}{y}$.
(A) 7
(B) 8
(C) 4
(D) 5
(E) 6