Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 10 - Area - 10-3 Areas of Regular Polygons - Practice and Problem-Solving Exercises - Page 632: 17


$A\approx841.8 ft^{2}$\triangle

Work Step by Step

$360^{\circ}/6=60^{\circ}$ All of the angles forming the center of the hexagon added together equal 360 degrees. To find one of the triangles, you'd need to divide the total degrees, 360, by the total number of triangles, 6. Therefore, the innermost angle in any of the triangles is 60 degrees. $60^{\circ}/2=30^{\circ}$ In order to make an equilateral triangle a 30-60-90 triangle, you must split it in half. So, the 60 degree angle is half of what it originally was, and is now 30 degrees. The line forming the split creates a 90 degree angle. Because the angles of a triangle must add up to 180 degrees, the last angle in the triangle is 60 degrees. $A=9\sqrt 3 ft$ Because the triangle is a special 30-60-90 triangle, the length of the longer leg is the square root of 3 times the length of the shorter leg. Plugging the information we already have, we can conclude that the apothem is 9 * the square root of 3 ft. $18 ft\times6=108 ft$ In order to find the perimeter of the regular hexagon, you must multiply the length of one side, 18, by the number of sides, 6. So, the perimeter is 108 ft. $A=\frac{1}{2}(p)(a)$ $A=\frac{1}{2}(108)(9\sqrt 3)$ $A=486\sqrt 3$ $A\approx841.8 ft^{2}$ To find the area, just plug the apothem and perimeter into the regular polygon area formula and solve.
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