Ans: A
Since $ABCD$ is a rectangle, $\angle ABC = \angle ADC = 90^\circ$.
Since $ABCD$ is a rectangle, $\angle ABC = \angle ADC = 90^\circ$.
By applying Pythagoras Theorem to $\Delta ABC$, we have
$\begin{array}{rcl}
AC^2 & = & AB^2 + BC^2 \\
AC^2 & = & 28^2 + 21^2 \\
AC & = & 35\text{ cm}
\end{array}$
Since $ABCD$ is a rectangle, $AD = BC = 12\text{ cm}$. In $\Delta ADC$,
$\begin{array}{rcl}
\cos CAD & = & \dfrac{AD}{AC} \\
\cos CAD & = & \dfrac{21}{35} \\
\cos CAD & = & \dfrac{3}{5}
\end{array}$
By applying the cosine law to $\Delta ADE$, we have
$\begin{array}{rcl}
DE^2 & = & AD^2 + AE^2 – 2 \times AD \times AE \times \cos \angle CAD \\
DE^2 & = & (21)^2 + (30)^2 – 2 \times 21 \times 30 \times \dfrac{3}{5} \\
DE^2 & = & 585 \\
DE & = & \sqrt{585} \\
DE & = & \sqrt{9 \times 65} \\
DE & = & 3\sqrt{65}
\end{array}$
