The Greatest Piñata Ever Made, Part 5

Yesterday, due to prior commitments, was just a materials-acquisition day. Nothing worth writing home about.

Today I set out to build the palate (the roof of the mouth), which in the finished piñata will be the "shelf" on which most of the candy will ultimately rest. Unfortunately, I had no pattern for the palate, because the skull pattern I cribbed from The Paper Museum was without palate or tongue. So I had to create a pattern all my own.

And I never could have done it (easily) without my good and long-dead friend Pythagoras!

Using a sharpie and a measuring tape, I figured all the edge-lengths and widths of my palate-to-be. When I was done, I had 7 widths (across the jaw from left-to-right), and six side-lengths (down the jaw, front to back). I checked most of the side-lengths for symmetry, but since I traced the same pattern for the left and right sides, I was pretty confident everything was going to line up, and I would have myself a nice, tight-fitting piece of cardboard to support hundreds of pounds of candy.

When I was done, I had broken the roof of the mouth into six trapezoids, like the front three illustrated below:And I had the lengths of each trapezoid's sides:Now edges and side-lengths are great and all, but it's hard to draw a really symmetrical trapezoid with just those measurements . . . I'd need to break out a straight-edge, protractor and compass, and constantly check that my base lines were parallel . . . what a pain!

If only I could break it down into rectangles and right-triangles . . . that would be so much simpler!
Oh wait, I can.

Now the nose trapezoid, with one base of 8.875", and one base of 14.125", and sides of 6.25", is not a rectangle of width 8.875", and two right triangles, each with a hypotenuse of 6.25".

But how tall is the rectangle? And how long are the other two sides of the right triangles?
To find the short side-length of the yellow triangle, I just took half of the difference between the bases of the trapezoid. Then I had two sides of a right triangle (the base and hypoteneuse), and since that wonderful man Pythagoras told us that:

a² + b² = c²

I could easily calculate the height of both the rectangle and the yellow right triangle.

Now, instead of having 6 trapezoids to deal with, I have six rectangles and the right triangles that they border. All I really needed to draw was the rectangles (something very easy to draw in Photoshop) and "connect the dots", as it were. When I was all done, I had a pretty little pattern for the roof of the Tyrannosaurus mouth, (full size printed at 4.926 px/inch will be to scale with other patterns). I printed and cut the pattern, traced it onto a big sheet of cardboard, freehanded some tabs off the sides (so I have something substantial to connect it to the inside of the upper jaw), and cut the whole thing out.

It's beautiful . . . it fits very snugly, though I'm not going to glue it in just yet. I need to rig up the steel cords that are going to support the finished piñata's weight . . . but more on that tomorrow (when I do it)!