Topic: Trapezoid question

[paulmcz]

Calculate "len" side

[Kerry]

How about with just dim W ??

[Kerry]

Is this a [Question:] type question that you know the answer to, or a real <normal> question ?

[paulmcz]

I came across this problem recently and I had to find the answer. I found it quite interesting so I thought I would post it here as a bit of a challenge to some,

[hendie]

I'm a bit rusty on this stuff ~ are you looking for an actual figure or just the formula ?

If the 3 Len's are equal, then the trapezoid must be formed from 3 equilateral triangles, therefore Len must be equal to 0.5*W, the acute angles must be 60° and the obtuse angles = 120°
therefore, h is easy to work out using Pythagoras' theory.... h = SQRT (Len² - ½w²)

of course I could be completely miles away here

[Kerry]

Consider it to be HALF of an Inclosed Polygon.

As Hendie mentioned the angles are 60 and 120 degrees.

The length len is equal to the circle radius, half of length W.

[paulmcz]

formula, code, steps to draw it........ anything

[Kerry]

Whoa .. That will teach me to be presumptuous, shouldn't it

[hendie]

ok, I'll take the easy route.
Draw a line X any length from A to B
Draw a circle at A of diameter 0.5*X
Draw a circle at B of diameter 0.5*X
Draw a circle using TAN, TAN, RADIUS from Circle 1 to Circle 2 and enter same radius as previous
Now draw line from A to first & second circles intersection, the to second & third circles intersection, then to point B
Done

if I had time I would code it but ...

[Kerry]

No Hendie, That wont do it.

I'm guessing it would need a simultanius equation of some form.

Imagine if the base is less than length LEN. The formulae would have to work in that case too.

It's past my bedtime. I'll probably dream about this paulmcz, not good !!

[paulmcz]

It's past my bedtime. I'll probably dream about this paulmcz, not good !!

You made my day man.     It happens to me very often.

[paulmcz]

ok, I'll take the easy route.
Draw a line X any length from A to B
Draw a circle at A of diameter 0.5*X
Draw a circle at B of diameter 0.5*X
Draw a circle using TAN, TAN, RADIUS from Circle 1 to Circle 2 and enter same radius as previous
Now draw line from A to first & second circles intersection, the to second & third circles intersection, then to point B
Done

if I had time I would code it but ...

Kerry is right Hendie, it won't do it.

[CarlB]

Steps to draw:

Draw base of length w.  Draw circles with R=len, centered on baseline endpoints.  Draw vertical line up from midpoint of baseline, arbitrary legth.  offet the line 1/2 len to each side.  Where the offset lines intersect the circles are the endpoints of the top of the trapezoid.

[Tom]

Draw a line any length - base line
from one end say left draw a circle rad @ half the line length
from same end draw a line "A" at 60 deg to base line passing over circle
from intersecting point draw a line "B" parralell to base line
mirror "A" about mid point of base line
trim overlaps and thats it

[paulmcz]

Steps to draw:

Draw base of length w.  Draw circles with R=len, centered on baseline endpoints.  Draw vertical line up from midpoint of baseline, arbitrary legth.  offet the line 1/2 len to each side.  Where the offset lines intersect the circles are the endpoints of the top of the trapezoid.

Can't draw circle with R=len untill we know "len" dimension. The only known is height (h) and width (w) of the baseline. Your method is fine only if "len" is known.

[paulmcz]

Draw a line any length - base line
from one end say left draw a circle rad @ half the line length
from same end draw a line "A" at 60 deg to base line passing over circle
from intersecting point draw a line "B" parralell to base line
mirror "A" about mid point of base line
trim overlaps and thats it

Baseline length is given (w) as well as height (h). Angle is not necessarily 60 deg (see second picture above). You are describing the case where nothing is given, except angle 60 deg.

[CarlB]

Reading the question more closely this time..ready to take anothe stab at it;

Look at one of the triangles. Right triangle.  Call the hypotenuse L, height H, then the base is (W-L)/2.
Use the pythagorean theorum - the sum of the squares of the legs equals the square of the hypotenuse.  A little rearrangement gives you:

3L^2 +2LW - H^2 - W^2 = 0

Hey that's one equation, one unknown (L), should be solveable! And it's almost in the form of the quadratic equation,
ax^2 + bx + c = 0, which has the solution (going by memory here)...
x=[-b +/- sqrt(b^2 - 4ac]/2a

Now who's up to doing the math?

[paulmcz]

Reading the question more closely this time..ready to take anothe stab at it;

Look at one of the triangles. Right triangle.  Call the hypotenuse L, height H, then the base is (W-L)/2.
Use the pythagorean theorum - the sum of the squares of the legs equals the square of the hypotenuse.  A little rearrangement gives you:

3L^2 +2LW - H^2 - W^2 = 0

Hey that's one equation, one unknown (L), should be solveable! And it's almost in the form of the quadratic equation,
ax^2 + bx + c = 0, which has the solution (going by memory here)...
x=[-b +/- sqrt(b^2 - 4ac]/2a

Now who's up to doing the math?

CarlB, I don’t think that this will get you there.
You'll be surprised how simple this problem is. There is no need for quadratic equation here. Just simple trigonometry is needed, if you want to solve it through the math.
Graphical solution is much easier, though. It takes less than a minute to draw it on the AutoCAD with absolute precision. You don’t even need the CAD to draw this very accurately.

[CarlB]

Bob, it's got to get me there, it's pure math!

(but I messed up in my first expression after substituting in the pythagorean eqn.)

So I followed through on the math, got:

L=[-W + 2 x sqrt (W^2 + 3H^2)]/3

Seemed to check out for an example drawin in CAD.

I'd love to see how to solve it graphically 

[ronjonp]

Here is my not so elegant solution :

Code: [Select]

(defun C:trap (/ pt ang len line1 line2 line3)
  (setq pt (getpoint "\n Pick start point: "))
  (setq len (getstring "\n Enter length: "))
  (setq ang (getstring "\n Enter angle: "))
  (setq line1 (strcat "@" len "<" ang))
  (setq line2 (strcat "@" len "<0"))
  (setq line3 (strcat "@" len "<-" ang))
  (command "_pline" pt line1 line2 line3 "close")
)

[paulmcz]

Here is my not so elegant solution :

Code: [Select]

(defun C:trap (/ pt ang len line1 line2 line3)
  (setq pt (getpoint "\n Pick start point: "))
  (setq len (getstring "\n Enter length: "))
  (setq ang (getstring "\n Enter angle: "))
  (setq line1 (strcat "@" len "<" ang))
  (setq line2 (strcat "@" len "<0"))
  (setq line3 (strcat "@" len "<-" ang))
  (command "_pline" pt line1 line2 line3 "close")
)

Ronjonp, we don't know the angle. Just "w" and "h" (see the picture in the first post)

[paulmcz]

Bob, it's got to get me there, it's pure math!

(but I messed up in my first expression after substituting in the pythagorean eqn.)

So I followed through on the math, got:

L=[-W + 2 x sqrt (W^2 + 3H^2)]/3

Seemed to check out for an example drawin in CAD.

I'd love to see how to solve it graphically 

You got it CarlB!
Here are the solutions I found.
Thanks for the ride.
Paul.

[Kerry]

Nice Solution Paul .. Well done Carl