1952 AMC 12 Problem 46

Attempt Problem 46 of the 1952 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 1952 AMC 12 solutions, or check the answer key.

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46.

The base of a new rectangle equals the sum of the diagonal and the greater side of a given rectangle, while the altitude of the new rectangle equals the difference of the diagonal and the greater side of the given rectangle. The area of the new rectangle is:

Greater than the area of the given rectangle

Equal to the area of the given rectangle

Equal to the area of a square with its side equal to the smaller side of the given rectangle

Equal to the area of a square with its side equal to the greater side of the given rectangle

Equal to the area of a rectangle whose dimensions are the diagonal and shorter side of the given rectangle

Answer: C
Concepts:rectanglePythagorean Theoremdifference of squares
Difficulty rating: 1500
Small Hint:

Let dd be the diagonal and L,WL,W the greater and smaller sides

Big Hint:

The new area is (d+L)(dL)=d2L2(d+L)(d-L)=d^2-L^2

Solution:

Let the greater and smaller side lengths be LL and W,W, and let the diagonal be d.d. The new rectangle’s area is (d+L)(dL)=d2L2. (d+L)(d-L)=d^2-L^2. The Pythagorean theorem gives d2=L2+W2,d^2=L^2+W^2, so the new area is W2.W^2. This is the area of a square whose side equals the smaller side of the original rectangle.

Thus, the correct answer is C.

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Problem 46 in Other Years

1950 AMC 12 · 1951 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12