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2009 AMC 10A Problem 17

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2009 AMC 10A Problem 17AMC 10 Step by Step
Name:                             Date: September 13, 2026
Similar & Congruent Triangles · difficulty 4 of 5 · about 6 min · position P16-20

Rectangle ABCDABCD has AB=4AB=4 and BC=3BC=3 . Segment EFEF is constructed through BB so that EFEF is perpendicular to DBDB , and AA and CC lie on DEDE and DFDF , respectively. What is EFEF ?

  1. A)

    99

  2. B)

    1010

  3. C)

    12512\frac {125}{12}

  4. D)

    1039\frac {103}{9}

  5. E)

    1212

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