2013 AMC 10A Problem 16

Attempt Problem 16 of the 2013 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 10A solutions, or check the answer key.

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16.

A triangle with vertices (6,5),(6, 5), (8,3),(8, -3), and (9,1)(9, 1) is reflected about the line x=8x = 8 to create a second triangle. What is the area of the union of the two triangles?

99

283\dfrac{28}{3}

1010

313\dfrac{31}{3}

323\dfrac{32}{3}

Answer: E
Concepts:coordinate geometryreflection (geometry)triangle areasymmetry
Difficulty rating: 2060
Solution:

The reflected triangle has vertices (7,1)(7,1), (8,3)(8,-3), and (10,5)(10,5).

The line through (6,5)(6,5) and (9,1)(9,1) is y=43x+13y=-\frac43x+13, so it meets x=8x=8 at y=73y=\frac73. By symmetry, the union is two congruent triangles with vertical base from (8,3)(8,-3) to (8,73)(8,\frac73).

That base has length 73+3=163\frac73+3=\frac{16}{3}, and each triangle has horizontal height 22. Hence the union area is 2(121632)=3232\left(\frac12\cdot\frac{16}{3}\cdot2\right)=\frac{32}{3}.

Thus, E is the correct answer.

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