Problem: In trapezoid ABCDABCDABCD, leg BC‾\overline{BC}BC is perpendicular to bases AB‾\overline{AB}AB and CD‾\overline{CD}CD, and diagonals AC‾\overline{AC}AC and BD‾\overline{BD}BD are perpendicular. Given that AB=11AB=\sqrt{11}AB=11 and AD=1001AD=\sqrt{1001}AD=1001, find BC2BC^2BC2.
Solution:
The problems on this page are the property of the MAA's American Mathematics Competitions