AMC 10 Daily Practice Round 4

Complete problem set with solutions and individual problem pages

Problem 10 Medium

Let ABCD be a convex quadrilateral such that \angle DAB=\angle DBC=90{}^\circandDA=AB=6. Let E be the intersection of the diagonals AC and BD. If BE=2\sqrt{2}, find the area of ABCD.

  • A.

    30

  • B.

    36

  • C.

    48

  • D.

    54

  • E.

    60

Answer:D

Construct AH\bot BD.

∵ \angle DAB=90{}^\circ, DA=AB=6,

∴ BD=\sqrt{D{{A}^{2}}+A{{B}^{2}}}=6\sqrt{2}.

∵ AH\bot BD,

∴ AH=BH=DH=\frac{1}{2}BD=3\sqrt{2}, \angle AHE=90{}^\circ,

∵ BE=2\sqrt{2},

∴ EH=BH-BE=3\sqrt{2}-2\sqrt{2}=\sqrt{2},

∵ \angle AHE=\angle CBE=\angle DBC=90{}^\circ, \angle AEH=\angle CEB,

∴ \triangle AHE \sim \triangle CEB,

∴ \frac{AH}{BC}=\frac{EH}{BE}, or \frac{3\sqrt{2}}{BC}=\frac{\sqrt{2}}{2\sqrt{2}}, BC=6\sqrt{2},

∴ {{A}_{\triangle BCD}}=\frac{1}{2}BD\times BC=\frac{1}{2}\times 6\sqrt{2}\times 6\sqrt{2}=36,

{{A}_{\triangle ABD}}=\frac{1}{2}AD\times AB=\frac{1}{2}\times 6\times 6=18,

∴ {{A}_{ABCD}}={{A}_{\triangle BCD}}+{{A}_{\triangle ABD}}=36+18=54.

Therefore, A_{ABCD}=54.

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