1958 AMC 12 Problem 31

Attempt Problem 31 of the 1958 AMC 12 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 1958 AMC 12 solutions, or check the answer key.

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

31.

The altitude drawn to the base of an isosceles triangle is 8,8, and the perimeter is 32.32. The area of the triangle is:
5656
4848
4040
3232
2424
Answer: B
Concepts:isosceles trianglePythagorean Theoremtriangle area
Small hint:
Let each equal side be aa and half the base be bb
Big hint:
Use a+b=16a+b=16 and a2−b2=82a^2-b^2=8^2
Solution:
Let each equal side be aa and half the base be b.b. The perimeter gives a+b=16.a+b=16. The altitude bisects the base, so a2−b2=82. a^2-b^2=8^2. Thus (a−b)(a+b)=64,(a-b)(a+b)=64, giving a−b=4.a-b=4. Hence b=6,b=6, so the base is 12.12. The area is 12(12)(8)=48. \frac12(12)(8)=48.
Therefore, the correct answer is B.
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