1952 AHSME Problems/Problem 46

Problem

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:

$\text{(A) greater than the area of the given rectangle} \quad\\ \text{(B) equal to the area of the given rectangle}  \quad\\ \text{(C) equal to the area of a square with its side equal to the smaller side of the given rectangle}  \quad\\ \text{(D) equal to the area of a square with its side equal to the greater side of the given rectangle}  \quad\\ \text{(E) equal to the area of a rectangle whose dimensions are the diagonal and the shorter side of the given rectangle}$

Solution

Let the larger side of the rectangle be y and the smaller side of the rectangle be x. Then, by pythagorean theorem, the diagonal of the rectangle has length $\sqrt{x^{2}+y^{2}}$.

By the definition of the problem, the area of the new rectangle is $(\sqrt{x^{2}+y^{2}}+y)(\sqrt{x^{2}+y^{2}}-y)$

Expanding this gives the area of the new rectangle to be $x^{2}$, or $\fbox{C}$.

See also

1952 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 45
Followed by
Problem 47
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