1958 AMC 12 Problem 36

Attempt Problem 36 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.

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36.

The sides of a triangle are 30,30, 70,70, and 8080 units. If an altitude is dropped upon the side of length 80,80, the larger segment cut off on this side is:

6262

6363

6464

6565

6666

Answer: D
Concepts:altitudePythagorean Theoremright triangle
Difficulty rating: 1550
Small Hint:

Let the altitude divide the side of length 8080 into xx and 80x80-x

Big Hint:

Equate the two expressions for the square of the altitude

Solution:

Let xx be the segment adjacent to the side of length 30.30. If the altitude has length h,h, then 302x2=h2,h2=702(80x)2. \begin{aligned} 30^2-x^2&=h^2,\\ h^2&=70^2-(80-x)^2. \end{aligned} Cancelling x2x^2 and solving gives x=15.x=15. The other segment is 8015=65,80-15=65, which is the larger one.

Thus, the correct answer is D.

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Problem 36 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1959 AMC 12 · 1960 AMC 12 · 1961 AMC 12 · 1962 AMC 12 · 1963 AMC 12 · 1964 AMC 12 · 1965 AMC 12 · 1966 AMC 12 · 1967 AMC 12