2014 AMC 12A Problem 10

Attempt Problem 10 of the 2014 AMC 12A 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 2014 AMC 12A solutions, or check the answer key.

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10.

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length 1.1. The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?

34\dfrac{\sqrt3}{4}

33\dfrac{\sqrt3}{3}

23\dfrac{2}{3}

22\dfrac{\sqrt2}{2}

32\dfrac{\sqrt3}{2}

Answer: B
Concepts:equilateral triangletriangle areaPythagorean Theorem
Difficulty rating: 1560
Solution:

The equilateral triangle has area 34.\dfrac{\sqrt3}{4}. Each isosceles triangle has base 11 and height h,h, so 312h=34,3\cdot\dfrac12 h=\dfrac{\sqrt3}{4}, giving h=36.h=\dfrac{\sqrt3}{6}.

A congruent side is the hypotenuse from the apex to a base endpoint: (12)2+(36)2=14+112=13=33. \begin{gathered} \sqrt{\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{6}\right)^2}\\ =\sqrt{\dfrac14+\dfrac{1}{12}}\\ =\sqrt{\dfrac13}=\dfrac{\sqrt3}{3}. \end{gathered}

Thus, the correct answer is B.

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