I have a center for a quilt that is on point and I want to put setting triangles on the sides to square it up. Good idea, right? Well ........ :oops:

I don't remember enough geometry for how to go about determining the sides of a triangle when I know the hypoteneuse. In other words, I know the diagonal for a half-square triangle, but I don't know the sides. :shock:

Can anyone help me? I need to get this done this week, please! :(

Thanks so much,

Platinum Odessa
:roll: :oops: :roll:

Sorry, I'm horrid at math too. :cry: I'm sure someone on here can help but until then, you've got my sympathy.

I hope this helps:

http://www.quiltbug.com/Articles/on-point.htm#

Try this......you just put in the info you know about your triangle and it figures out the rest. I asked my husband the math whiz and he said it is accurate. Goodness knows I wouldn't know for myself.

http://easycalculation.com/trigonome...gle-angles.php

I had to figure this out for myself yesterday. The way I did it was to measure from the center of my quilt to the corner, and that measurement was the width and height of my triangle. I measured it out and cut a square of my fabric (plus a half inch for error) and then cut from corner to corner to make 2 triangles. The triangles fit perfectly.

does that make sense?

I just look it up on my 20 1/2 inch ruler. it has all that info on it.

Here's another link that might help: http://quiltville.com/onpointmath.shtml

Lisa, I had run across this website before, too, but the problem is that I only know one side (the hypoteneuse or diagonal of the triangle). What I need to know are the other two legs, and this formula wanted at least 2 sides. I know it should be very easy, especially when you consider the two legs on the right angle are supposed to be equal in length, but the formula kept asking to input ALL of the values I knew, which I did, but when it's only one value, then I'm stuck!

My 10th grade Geometry teacher would be SOOOOOOOOOOo disappointed! Oh, well!

Thanks one and all for the help, though, I think I will be able to manage it tonight.

Okay, Geometry teacher here, let's see what I can do:

On a 45-45-90 triangle (where one side is a right angle, and the legs are equal), the sides are x, x, and x*sqrt(2). So, if I know that one leg is 5, the other leg is 5, meaning the hyp is 5*sqrt(2) or about 7.07.

So if you know the LEG and want to find the HYPOTENUSE, multiply the leg by sqrt (2).

If you know the HYPOTENUSE, then DIVIDE by sqrt (2) to get the legs. Ex: If my hyp is 7, each leg is 7 / sqrt(2) or about 4.95 (and if you think about this, this makes sense since it's close to what we figured out in the above one).

What I do, is I find out about how long the legs are, and then add 1/2" for seam allowance. It's not exact, but I'd rather have too much then too little. I have a bit of trouble working the seam allowance into my calculations because you lose part of the point, so I just add a little extra.

Hope this helps. Let me know if you need more

Lucy, thank you! I tried to do it that way in the first place, but was not coming up with what I felt to be a credible answer. When I get home tonight I will make that attempt one more time and see how it all plays out.

You guys are truly awesome!


Odessa