AMC 10 Daily Practice Round 1

Complete problem set with solutions and individual problem pages

Problem 12 Easy

In \triangle ABC, the incircle \odot O is tangent to sides BC, CA, and AB at points D, E, and F, respectively. Given that AB = 8, BC = 17, and CA = 15, what is the area of the shaded region (quadrilateral AEOF)?

  • A.

    3

  • B.

    36

  • C.

    12

  • D.

    9

  • E.

    6

Answer:D

Given that AB = 8, BC = 17, and CA = 15, we can apply the Pythagorean theorem: AB^2 + CA^2 = BC^2. Thus, \triangle ABC is a right triangle, with \angle A = 90^\circ.

Since the incircle \odot O is tangent to sides AB, AC, and BC at points F, E, and D, respectively, and since OF \perp AB and OE \perp AC, quadrilateral OFAE is a square.

Let OE = r, so AE = AF = r. The incircle \odot O is tangent to sides BC, CA, and AB at points D, E, and F. Thus, we have: BD = BF = 8 - r, \quad CD = CE = 15 - r. From the equation BD + CD = BC, we have: (8 - r) + (15 - r) = 17, which simplifies to: 23 - 2r = 17 \quad \Rightarrow \quad r = \frac{8 + 15 - 17}{2} = 3.

Therefore, the area of the shaded region (quadrilateral AEOF) is the area of the square, which is: r^2 = 3 \times 3 = 9.

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