28002, so I started fiddling around and trying to do math assuming that this value was somehow based on angles. The value for the close-but-not-quite hexagon was. I used some VBA to check what the position of the handle was (with the hexagon selected): ?(1).Adjustments(1) To get a regular hexagon in PowerPoint, create a hexagon using shift, then run the following VBA command with that hexagon selected: (1).Adjustments(1) = 0.28706Īfter I created the approximate shape using the overlay on the close-but-not-quite equilateral triangle, I decided to get programmatic.
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