1973 AHSME Problems/Problem 16

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.

1973 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
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1973 AHSME Problems/Problem 16

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