Community Edition 2.0.5.0Free1 On Linux Mint 10 Julia.
I'm trying to draw a triangle  a right angled triangle where I know the length of the hypotenuse (h) and one other side(x), but not the length of the other side, nor the other 2 angles.
The equation, as I'm sure we all know, is h^2 = SQRT(x^2 + y^2)
Obviously I can resolve y manually (well with a calculator since in my case the numbers were horrible), but I'm thinking that's exactly the sort of calculation QCad should be doing for me. Is there a way?
Regards
struggling with a hypotenuse!
Moderator: andrew

 Registered Member
 Posts: 1
 Joined: Sat Feb 19, 2011 10:11 pm
Thales comes to mind:
http://en.wikipedia.org/wiki/Thales%27_theorem
1. draw hypotenuse (h)
2. draw circle with center at middle point of h
3. draw circle with center at end point of hypotenuse and radius (x)
4. intersection point between two circles is the third point of the triangle.
Such auxiliary constructions are usually drawn on a separate layer (e.g. "aux") which is kept for future reference but hidden for printing.
http://en.wikipedia.org/wiki/Thales%27_theorem
1. draw hypotenuse (h)
2. draw circle with center at middle point of h
3. draw circle with center at end point of hypotenuse and radius (x)
4. intersection point between two circles is the third point of the triangle.
Such auxiliary constructions are usually drawn on a separate layer (e.g. "aux") which is kept for future reference but hidden for printing.
Re: struggling with a hypotenuse!
My favorite site assistant is "Euclid" .
" The square on the hypotenuse = the sum of the squares on the other two sides"
So where you know the length of the hypotenuse, the length of the other two sides are predetermined and identifiable.
To configure a right angle, the sides measuring 3 and 4, include the right angle opposing the hypotenuse measuring 5.
It's 60 + years since I learned that trick, and I don't think it has changed since.
Just throwing my thoughts into the discussion.
Hope it helps someone.
tmno2
" The square on the hypotenuse = the sum of the squares on the other two sides"
So where you know the length of the hypotenuse, the length of the other two sides are predetermined and identifiable.
To configure a right angle, the sides measuring 3 and 4, include the right angle opposing the hypotenuse measuring 5.
It's 60 + years since I learned that trick, and I don't think it has changed since.
Just throwing my thoughts into the discussion.
Hope it helps someone.
tmno2

 Registered Member
 Posts: 2
 Joined: Fri Jan 04, 2013 4:19 am
Re: struggling with a hypotenuse!
Erm, didn't you get the Pythagorean theorem wrong in the OP? Shouldn't it be h=SQRT(x^2+y^2)? You squared the h.
Anyway, thanks for sharing Thales's theorem! Never heard of it. I tried this using an auxiliary layer as you suggested, and it worked well.
Fun note: I did this with another triangle of different dimensions just for practice, and the legs ended up matching the dimensions of my monitor! (mine is the first one listed here: [uh, this got blocked. oh well, don't want to bother with the permission.]). Loving my new QCAD, thanks for all your work guys.
Anyway, thanks for sharing Thales's theorem! Never heard of it. I tried this using an auxiliary layer as you suggested, and it worked well.
Fun note: I did this with another triangle of different dimensions just for practice, and the legs ended up matching the dimensions of my monitor! (mine is the first one listed here: [uh, this got blocked. oh well, don't want to bother with the permission.]). Loving my new QCAD, thanks for all your work guys.
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