$ABCD$ is a rectangle and point $P$ is such that $PA = 5 \text { cm }, PB = 4 \sqrt { 2 } \text { cm, } \text { and } PD = 3 \text { cm. }$ Find the length of $PC$.

$4 \text{ cm }$

Step by Step Explanation:
1. We are given that $ABCD$ is a rectangle with a point $P$ in its interior such that $PA = 5 \text { cm }, PB = 4 \sqrt { 2 } \text { cm, } \text { and } PD = 3 \text { cm. }$
2. Let us assume that $AD = x \text{ cm }$ and $CD = y \text{ cm. }$
Now, let us plot the perpendiculars at the points $S, T, U,$ and $V$ on the sides of the rectangle from the point $P.$
Let, $PS = m \text{ cm }$ and $PU = n \text{ cm }$
Now, $$PT = (x-n) \text{ cm } \\ PV = (y-m) \text{ cm }$$
3. Now, using pythagoras theorem in $\triangle PAS,$ we have $$PA^2 = PS^2 + AS^2 \ldots (1)$$ Similarly, using pythagoras theorem in $\triangle PCV,$ $$PC^2 = PV^2 + CV^2 \ldots (2)$$
4. Adding $(1)$ and $(2),$ we have, \begin{align} PA^2 + PC^2 &= PS^2 + AS^2 + PV^2 + CV^2 \\ &= m^2 + n^2 + (y-m)^2 + (x-n)^2 \ldots (3) \end{align}
5. Now, using pythagoras theorem in $\triangle PBV,$ we have $$PB^2 = PV^2 + BV^2 \ldots (4)$$ Similarly, using pythagoras theorem in $\triangle PDS,$ $$PD^2 = PS^2 + SD^2 \ldots (5)$$
6. Adding $(4)$ and $(5),$ we have, \begin{align} PB^2 + PD^2 &= PV^2 + BV^2 + PS^2 + SD^2 \\ &= (y-m)^2 + n^2 + m^2 + (x-n)^2 \\ &= m^2 + n^2 + (y-m)^2 + (x-n)^2 \ldots (6) \end{align}
7. On comparing, equation $(3)$ and $(6),$ we have \begin{align} & PA^2 + PC^2 = PB^2 + PD^2 \\ \implies & PC^2 = PB^2 + PD^2 - PA^2 \\ \implies & PC^2 = { (4 \sqrt{ 2 }) } ^2 + 3^2 - 5^2 \\ \implies & PC^2 = 32 + 9 - 25 \\ \implies & PC^2 = 41 - 25 \\ \implies & PC^2 = 16 \\ \implies & PC = \sqrt{ 16 } \\ \implies & PC = 4 \end{align}
8. Thus, the length of $PC = 4 \text{ cm }$