2020-II-38

Ans: A
By applying Pythagoras Theorem to $\Delta ABP$, we have

$\begin{array}{rcl}
BP^2 & = & AP^2 + AB^2 \\
BP^2 & = & 9^2 + 12^2 \\
BP^2 & = & 225 \\
BP & = & 15\text{ cm}
\end{array}$

By applying Pythagoras Theorem to $\Delta DEP$, we have

$\begin{array}{rcl}
DP^2 & = & EP^2 + DE^2 \\
DP^2 & = & 5^2 + 12^2 \\
DP^2 & = & 169 \\
DP & = & 13\text{ cm}
\end{array}$

Consider $\Delta BDP$.

$\begin{array}{rcl}
s & = & \dfrac{BD + BP + DP}{2} \\
s & = & \dfrac{13 + 15 + 2k}{2} \\
s & = & 14+k \text{ cm}
\end{array}$

Hence, the area of $\Delta BDP$

$\begin{array}{cl}
= & \sqrt{s(s-BD)(s-BP)(s-DP)} \\
= & \sqrt{ (14+k)(14+k – 15)(14+k – 13)(14+k – 2k)} \\
= & \sqrt{(14+k)(k-1)(k+1)(14-k)} \\
= & \sqrt{(k^2-1)(196-k^2)}\text{ cm}^2
\end{array}$