^@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^@.


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Answer:

^@ 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 } ^@

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