1995 AMC 8 Problem 6

Attempt Problem 6 of the 1995 AMC 8 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 1995 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

6.

Figures I,I, IIII and IIIIII are squares. The perimeter of II is 1212 and the perimeter of IIII is 24.24. The perimeter of IIIIII is
99
1818
3636
7272
8181
Answer: C
Concepts:perimetersquare (geometry)
Small hint:
The perimeter of a square is 44 times its side, so find the side of II and of IIII
Big hint:
From the figure, the side of IIIIII equals the side of II plus the side of IIII
Solution:
Square II has side 12÷4=3,12 \div 4 = 3, and square IIII has side 24÷4=6.24 \div 4 = 6.
From the figure, the side of IIIIII is 3+6=9,3 + 6 = 9, so its perimeter is 4×9=36.4 \times 9 = 36.
Thus, the correct answer is C .
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