1973 AHSME Problems/Problem 16

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Problem

If the sum of all the angles except one of a convex polygon is $2190^{\circ}$, then the number of sides of the polygon must be

$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 19 \qquad \textbf{(E)}\ 21$

Solution

Let $n$ be the number of sides in the polygon. The number of interior angles in the polygon is $180(n-2)$. We know that the sum of all but one of them is $2190^{\circ}$, so the sum of all the angles is more than that. \[180(n-2) > 2190\] \[n-2 > 12 \tfrac{1}{6}\] \[n > 14 \tfrac{1}{6}\]

The sum of the angles in a 15-sided polygon is $2340^{\circ}$, making the remaining angle $150^{\circ}$. The angles of a convex polygon are all less than $180^{\circ}$, and since adding one more side means adding $180^{\circ}$ to the measure of the remaining angle, we can confirm that there are $\boxed{\textbf{(B)}\ 15}$ sides in the polygon.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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