Given \( \sqrt{(224)_{r}}=(13)_{r} \) where \( r \) is the radix. The value of \( r \) is \( \qquad \) .
Write both numbers in base \(r\):
\((224)_r = 2r^2 + 2r + 4\) and \((13)_r = r + 3\).
Since \(\sqrt{(224)_r} = (13)_r\), we get \((r+3)^2 = 2r^2 + 2r + 4\).
\(r^2 + 6r + 9 = 2r^2 + 2r + 4 \Rightarrow r^2 - 4r - 5 = 0 \Rightarrow (r-5)(r+1) = 0\)
A radix must be positive, so \(r = 5\). (The digit 4 is also valid in base 5.)
Answer: (C) 5
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