I want to make some Christmas ornaments out of old Christmas cards. The basis for the ornaments is circles that are folded and glued together, and I want to make a template to help with the folding.
The template would be an equilateral triangle, the largest that can fit in the circle. I need to know how large to make the triangle template. I have a vague recollection that this was something covered in basic geometry, but that was almost 50 years ago for me.
I don’t have a protractor or compass (again, geometry was almost a half-century ago). I of course have rulers and I’m sure I can find something to use to draw a right angle.
I hope I’ve stated my problem well enough that you can understand what formulae I need – I no longer remember some of the terminology used.
But the steps are fairly straightforward. Suppose that you have a circle of radius 1. Picture a hexagon, made up of six equilateral triangles, each of length 1. That hexagon just fits inside the circle, right? Now, the distance you want is the distance between two non-adjacent corners of that hexagon, or twice the height of one of the small triangles. Draw that line. You now have four right triangles nested together. The hypotenuse of one of those right triangles is 1, one side is 1/2, and so by the Pythagorean Theorem, the other side (which is the height of the equilateral triangles) is sqrt(3)/2. So the total side length you want, for your big triangle, is sqrt(3), or about 1.732 .
In other words, if the radius of your circle is 1, then the triangle template should have a side of 1.732 . So you measure your circles, in whatever unit you prefer, and then multiply that measurement by 1.732 . For instance, if your circles are 2 inches in radius, then your triangle should be 3.468 inches. If your circles are 5 cm, then your triangle should be 8.66 cm.
There are a few ways you can go about this. If you want I can prove it for you but the side of the triangle will be sqrt(3) x r approx. 1.732 x radius of the circle just like Chronos says.
Remember with a compass you could take the radius, and draw arcs starting from any arbitrary point, and it would take 6 of these radius to complete the circumference (which makes sense, because a circle is filled by 6 equilateral triangles centered on the circle’s center.
So without a compass…
Take a paper circle, and fold it in quarters (carefully) The straight edge is a radius long. Use this to mark off one arc AB along the circumference, then another BC- you have now marked off 120 degrees of the circle with AC (one radius along the circumference marks off 60 degrees.) Repeat two more times to get the equilateral triangle, wit points on the circumference spaced 120 degrees apart.
Assuming all circles are the same size…
If you need to do this for random circle sizes, create a 120-degree template with this trick. (but then the question is - do you know the center of each circle?)
If you don’t know where the center of a circle is, take anything with a right angle (like the corner of a piece of paper), and put the point of the angle somewhere on the circle. Mark the two places where the edges of the paper cross the circle. Those points are directly opposite each other; draw a line between them. Repeat this process to make another line. Where they intersect is the center.
I think most respondents so far has assumed that the person who could post the question , must be able to work that out , suitably accurately, such as for the purpose of measuring the diameter or radius. There’s so many ways to do it, even with just a measuring tape, and its all intuitive.
I was almost going to suggest Chronos work with the diamater, but heck … just use that multiply from Chronos,etc, and divide by two, and then use that to multiply the diameter… It was just intuitive that radius * M EQUALS diameter * M / 2. But thats just intuitive … obvious… prosaic… Just as its just obvious to all that a two decimal place approximation of M is good enough for the purpose at hand… unless its a bloody big circle.
A little less well known is that the centre is easy to find , if one can find diameters with a rule…
The centre must occur where two different diameters cross …
However it was redundant to state that, as the measuring of the diameter or radius gives the centre away just "well there it is " style,do we need to give the method and formula ??? IF the OP can construct the question straight forward and concisely, I think they can measure diameter and find centres… They just didn’t know about (or forgot about ) the sine or cosine of 30 degrees ! cosine is Adjacent over hypotenuse… sine opposite over hypotenuse… How does that help ? construct the right angle triangle … easy… just bisect the equilateral triangle… One side is the diameter of the circle, the wanted side is the hypotenuse…