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2000 AMC 10 Problem 19

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2000 AMC 10 Problem 19AMC 10 Step by Step
Name:                             Date: September 13, 2026
Similar & Congruent Triangles · difficulty 3 of 5 · about 4 min · position P16-20

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is mm times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

  1. A)

    12m+1\frac{1}{2m+1}

  2. B)

    mm

  3. C)

    1−m1-m

  4. D)

    14m\frac{1}{4m}

  5. E)

    18m2\frac{1}{8m^2}

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