Balmoral Software
Solutions: 10
which reduces to![]()
Equivalently,![]()
The base area must be an integer multiple of π, so let r2 = k for some positive integer k. We then have![]()
so h is rational. The volume V must also be an integer multiple of π, and![]()
Therefore, k - 9 must divide 162, and so r2 ∈ {10,11,12,15,18,27,36,63,90,171}. The following are all 10 equable right cones, ordered by volume:![]()
The only equable solution where r is also rational is the all-integer solution h = 8 and r = 6, for which V = S = 96 π. That cone is a solid of revolution of an equable right triangle with base 6 and height 8, having perimeter and area equal to 24.
h r2 r Lateral surface area/π V/π=S/π 12 18 4.243 54 72 15 15 3.873 60 75 9 27 5.196 54 81 8 36 6 6.000 60 96 24 12 3.464 84 96 33 11 3.317 110 121 7 63 7.937 84 147 20/3 90 9.487 110 200 60 10 3.162 190 200 19/3 171 13.077 190 361
Copyright © 2019 Balmoral Software (http://www.balmoralsoftware.com). All rights reserved.