A) \[A{{Q}^{2}}+C{{P}^{2}}=2{{(A{{C}^{2}}+PQ)}^{2}}\]
B) \[2(A{{Q}^{2}}+C{{P}^{2}})=A{{C}^{2}}+P{{Q}^{2}}\]
C) \[A{{Q}^{2}}+C{{P}^{2}}=A{{C}^{2}}+P{{Q}^{2}}\]
D) \[AQ+CP=\frac{1}{2}(AC+PQ)\]
Correct Answer: C
Solution :
[c] In right angled \[\Delta ABQ\] and \[\Delta CPB,\] |
\[C{{P}^{2}}=C{{B}^{2}}+B{{P}^{2}}\] |
and \[A{{Q}^{2}}=A{{B}^{2}}+B{{Q}^{2}}\] |
\[C{{P}^{2}}+A{{Q}^{2}}=C{{B}^{2}}+B{{P}^{2}}+A{{B}^{2}}+B{{Q}^{2}}\] |
\[=C{{B}^{2}}+A{{B}^{2}}+B{{P}^{2}}+B{{Q}^{2}}\] |
\[=A{{C}^{2}}+P{{Q}^{2}}\] |
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