2026 AMC 8 Problem 13

Below is the video solution and professionally curated solution for Problem 13 of the 2026 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2026 AMC 8 solutions, or check the answer key.

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Concepts:Pythagorean Theoremsquare (geometry)

Difficulty rating: 1330

13.

The figure below shows a tiling of 1×11 \times 1 unit squares. Each row of unit squares is shifted horizontally by half a unit relative to the row above it. A shaded square is drawn on top of the tiling. Each vertex of the shaded square is a vertex of one of the unit squares. In square units, what is the area of the shaded square?

1010

212\dfrac{21}{2}

323\dfrac{32}{3}

1111

343\dfrac{34}{3}

Video solution:
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Written solution:

One side of the shaded square goes 33 unit widths horizontally and 11 unit vertically in the grid. Its length squared is therefore 32+12=103^2+1^2=10. The area of a square equals its side length squared, so the shaded area is 1010.

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