Ans: A
By Heron’s formula, we have
By Heron’s formula, we have
$\begin{array}{rcl}
s & = & \dfrac{10+4+2k}{2} \\
& = & 7+k \text{ cm}
\end{array}$
Hence, the area of $\Delta ABC$
$\begin{array}{cl}
= & \sqrt{(7+k)(7+k-10)(7+k-4)(7+k-2k)} \\
= & \sqrt{ (7+k)(-3+k)(3+k)(7-k)} \\
= & \sqrt{k^2-9)(49-k^2)} \text{ cm}^2
\end{array}$
