2024 AMC 10A Problem 16

Attempt Problem 16 of the 2024 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 2024 AMC 10A solutions, or check the answer key.

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16.

All of the rectangles in the figure below, which is drawn to scale, are similar to the enclosing rectangle. Each number represents the area of the rectangle. What is length AB?AB?

4+454 + 4\sqrt5

10210\sqrt2

5+555 + 5\sqrt5

108410\sqrt[4]{8}

2020

Answer: D
Concepts:similarityarea ratiorectangle
Difficulty rating: 1730
Solution:

The areas of the eleven pieces sum to 200.200. Let the enclosing rectangle have height hh and width w=AB.w=AB. Similar figures have corresponding side lengths in the square-root ratio of their areas. From the diagram, the area-3232 piece contributes its short side to the left edge, of length h32/200=25h.h\sqrt{32/200}=\tfrac25h. Above it, the area-3636 piece contributes its long side, of length w36/200=3210w.w\sqrt{36/200}=\tfrac{3\sqrt2}{10}w. These two segments make the full height, so h=25h+3210w,h=\tfrac25h+\tfrac{3\sqrt2}{10}w, which simplifies to h=w/2.h=w/\sqrt2. Since wh=200,wh=200, we get w2/2=200w^2/\sqrt2=200 and w=2002=1084.w=\sqrt{200\sqrt2}=10\sqrt[4]{8}. Therefore, the answer is D.

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