1950 AMC 12 Problem 46

Attempt Problem 46 of the 1950 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 1950 AMC 12 solutions, or check the answer key.

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

46.

In triangle ABC,ABC, AB=12,AB=12, AC=7,AC=7, and BC=10.BC=10. If sides ABAB and ACAC are doubled while BCBC remains the same, then:
The area is doubled
The altitude is doubled
The area is four times the original area
The median is unchanged
The area of the triangle is 00
Answer: E
Concepts:triangle inequalityarea
Small hint:
Write down the three new side lengths
Big hint:
Compare the largest new side with the sum of the other two
Solution:
The new side lengths are 24,24, 14,14, and 10.10. Because 14+10=24, 14+10=24, they form a degenerate triangle: all three vertices are collinear. Its area is therefore 0.0.
Thus, the correct answer is E.
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