1985 AMC 8 Problem 12

Attempt Problem 12 of the 1985 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 1985 AMC 8 solutions, or check the answer key.

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

12.

A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.26.2 cm, 8.38.3 cm, and 9.59.5 cm. The area of the square, in square centimeters, is
24 cm224 \text{ cm}^2
36 cm236 \text{ cm}^2
48 cm248 \text{ cm}^2
64 cm264 \text{ cm}^2
144 cm2144 \text{ cm}^2
Answer: B
Concepts:perimetersquare (geometry)
Small hint:
Add the triangle’s three sides to get the common perimeter
Big hint:
The square’s side is one fourth of its perimeter
Solution:
The triangle’s perimeter is 6.2+8.3+9.5=246.2 + 8.3 + 9.5 = 24 cm, so the square also has perimeter 2424 cm.
The square’s side is 244=6\dfrac{24}{4} = 6 cm, so its area is 62=36 cm2.6^2 = 36 \text{ cm}^2.
Thus, the correct answer is B .
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