Ta có:
+) \({Q_{\left( {O,0} \right)}}\left( A \right) = A,\) \({Q_{\left( {O,0} \right)}}\left( B \right) = B,\) \({Q_{\left( {O,0} \right)}}\left( C \right) = C,\) \({Q_{\left( {O,0} \right)}}\left( D \right) = D\)
Do đó \({Q_{\left( {O,0} \right)}}\left( {ABCD} \right) = ABCD\).
+) \({Q_{\left( {O,\dfrac{\pi }{2}} \right)}}\left( A \right) = B,\) \({Q_{\left( {O,\dfrac{\pi }{2}} \right)}}\left( B \right) = C,\) \({Q_{\left( {O,\dfrac{\pi }{2}} \right)}}\left( C \right) = D,\) \({Q_{\left( {O,\dfrac{\pi }{2}} \right)}}\left( D \right) = A\)
Do đó \({Q_{\left( {O,\dfrac{\pi }{2}} \right)}}\left( {ABCD} \right) = BCDA\).
+) \({Q_{\left( {O,\pi } \right)}}\left( A \right) = C,\) \({Q_{\left( {O,\pi } \right)}}\left( B \right) = D,\) \({Q_{\left( {O,\pi } \right)}}\left( C \right) = A,\) \({Q_{\left( {O,\pi } \right)}}\left( D \right) = B\)
Do đó \({Q_{\left( {O,\dfrac{\pi }{2}} \right)}}\left( {ABCD} \right) = CDAB\).
+) \({Q_{\left( {O,\dfrac{{3\pi }}{2}} \right)}}\left( A \right) = D,\) \({Q_{\left( {O,\dfrac{{3\pi }}{2}} \right)}}\left( B \right) = A,\) \({Q_{\left( {O,\dfrac{{3\pi }}{2}} \right)}}\left( C \right) = B,\) \({Q_{\left( {O,\dfrac{{3\pi }}{2}} \right)}}\left( D \right) = C\)
Do đó \({Q_{\left( {O,\dfrac{\pi }{2}} \right)}}\left( {ABCD} \right) = DABC\).
Vậy có \(4\) phép quay cần tìm.
Chọn D.
