Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 7 - Section 7.3 - More about Regular Polygons - Exercises - Page 344: 35

Answer

There is no regular polygon with 8 diagonals.

Work Step by Step

We can find the number of vertices if a regular polygon has 8 diagonals: $\frac{n(n-3)}{2} = 8$ $n(n-3) = 16$ $n^2-3n- 16 = 0$ We can use the quadratic formula to find the solutions of the equation: $n = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $n = \frac{-(-3) \pm \sqrt{(-3)^2-(4)(1)(-16)}}{(2)(1)}$ $n = \frac{3 \pm \sqrt{9+64}}{2}$ $n = \frac{3 \pm \sqrt{73}}{2}$ $n = \frac{3 - \sqrt{73}}{2}~~$ or $~~n = \frac{3 + \sqrt{73}}{2}$ $n = -2.77~~$ or $~~n = 5.77$ Since $n$ is not a whole number, there is no regular polygon with 8 diagonals.
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