10. Let $A B C D$ be a quadrilateral inscribed in a circle with diameter $A C$, and let $E$ be the foot of perpendicular from $D$ onto $A B$, as shown in the figure below. If $A D=$ $D C$ and the area of quadrilateral $A B C D$ is $24 \mathrm{~cm}^{2}$, find the length of $D E$ in $\mathrm{cm}$.
(A) $3 \sqrt{2}$
(B) $2 \sqrt{6}$
(C) $2 \sqrt{7}$
(D) $4 \sqrt{2}$
(E) 6
10. Answer: (B)

Let $r$ be the radius of the circle with centre $O$.
Since $A D=D C$ and $\angle A D C=90^{\circ}, \angle A C D=45^{\circ}$. Thus $\angle A B D=45^{\circ}$. As $\angle D E B=90^{\circ}$, this implies that $D E=B E$. Let $x=D E=B E$. Since $B C A E D$ and area of quadrilateral $A B C D=$ area of $\triangle A E D+$ area of $\triangle E B D+$ area of $\triangle B C D$, we have
$$
24=\frac{1}{2} \cdot(A B-x) x+\frac{1}{2} x^{2}+\frac{1}{2} \cdot B C \cdot x=\frac{1}{2}(A B+B C) x .
$$

On the other hand, as $O D$ is perpendicular to $A C$, and area of quadrilateral $A B C D=$ area of $\triangle A B C+$ area of $\triangle A C D$, we have
$$
24=\frac{1}{2} \cdot A B \cdot B C+\frac{1}{2} A C \cdot O D=\frac{1}{2} \cdot A B \cdot B C+r^{2} .
$$

Now equation (2) and $A B^{2}+B C^{2}=A C^{2}=4 r^{2}$ imply that
$$
(A B+B C)^{2}=4 r^{2}+2 \cdot A B \cdot B C=4 r^{2}+4\left(24-r^{2}\right)=96 .
$$

Therefore $A B+B C=4 \sqrt{6}$. Hence from equation (1), we obtain $x=2 \sqrt{6}$.