Hi Stephen,
We had a similar question some time ago from Kent but for a five pointed star. I think you might find it enjoyable to see what he asked and the final product he produced.
I made a paper model (not to scale) of three of the points of your star and labeled some points. You will see that I made "kite shaped" pieces rather than triangles. That is what Kent used and I find them easier to handle.
P is on the star at the highest point above the floor and Q, S, T and R are on the floor. Q is directly below P. From your description I have the distance from S to T to be 5 inches.
Consider the triangle PQR.
U is the point where the line segments ST and QR meet and angle VQR is a right angle. According to your description the length of VU is 2 inches, QR = 12 inches and QV = 2.5 inches. Triangles PQR and VUR are similar and hence
PQ/QR = VU/UR
thus
h = PQ = 12 2/9.5 = 2.526 inches.
By Pythagoras' Theorem PR^{2} = h^{2} + 12^{2} and thus
PR = 12.2630 inches.
Now consider triangle PQS.
h = 2.526 inches and k = 2.5 √2 = 3.5355 inches so, again by Pythagoras' Theorem
PS = 4.3452 inches
Finally consider triangle SUR.
Again this is a right triangle so
SR = sqrt(2.5^{2} + 9.5^{2}) = 9.8234 inches.
Thus the triangle you want to use, SPR has sides of length
PS = 4.3452 inches
PR = 12.2630 inches and
SR = 9.8234 inches
I would not use angles to construct this triangle. I would draw the line segment PR of length 12.26 inches and then with a compass with point at P and distance between the point and the pencil of 4.33 inches draw an arc at approximately where you think S is located. Then set the distance between the compass point and pencil at 9.82 inches set the point at R and draw an arc to intersect the arc previously drawn. The intersection of the two arcs is the point S.
For the remaining point of the star repeat the above calculation with QR = 24 inches.
I don't know what material you arre planning to use but I would make a paper model first.
Penny
