2002 AMC 8 Problem 11

Below is the professionally curated solution for Problem 11 of the 2002 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 8 solutions, or check the answer key.

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Concepts:perfect square pattern recognition

Difficulty rating: 1060

11.

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?

$11$

$12$

$13$

$14$

$15$

Solution:

The sixth and seventh squares have side lengths $6$ and $7$ tile lengths.

They use $6^2=36$ and $7^2=49$ tiles, respectively, so the seventh square requires $49-36=13$ more tiles.

Thus, C is the correct answer.

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