1988 AHSME Problems/Problem 6

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Problem

A figure is an equiangular parallelogram if and only if it is a

$\textbf{(A)}\ \text{rectangle}\qquad \textbf{(B)}\ \text{regular polygon}\qquad \textbf{(C)}\ \text{rhombus}\qquad \textbf{(D)}\ \text{square}\qquad \textbf{(E)}\ \text{trapezoid}$

Solution

The definition of an equiangular parallelogram is that all the angles are equal, and that pairs of sides are parallel. It may be a rectangle, because all the angles are equal and it is a parallelogram. It is not necessarily a regular polygon, because if the polygon is a pentagon, it is not a parallelogram. It is not necessarily a rhombus, because all the angles are not necessarily equal. It may be a square, since it is a parallelogram and all the angles are equal. It is not necessarily a trapezoid, because the angles are not necessarily equal. We have that it could be a square or a rectangle. A square is a rectangle, but a rectangle is not necessarily a square. We want the all-encompassing answer so the answer is a rectange $\implies \boxed{\text{A}}$.

See also

1988 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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