1956 AHSME Problems/Problem 29

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Problem 29

The points of intersection of $xy = 12$ and $x^2 + y^2 = 25$ are joined in succession. The resulting figure is:

$\textbf{(A)}\ \text{a straight line}\qquad \textbf{(B)}\ \text{an equilateral triangle}\qquad \textbf{(C)}\ \text{a parallelogram} \\ \textbf{(D)}\ \text{a rectangle} \qquad \textbf{(E)}\ \text{a square}$


Solution

Graphing both equations, we see that the graphs intersect at $(-4, -3),$ $(-3, -4),$ $(3, 4),$ and $(4, 3).$ These points form a rectangle, so the answer is $\boxed{\textbf{(D)}}.$

~coolmath34

See Also

1956 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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