



 
Hi Phil. Is the sketch below what you are making? Let me cut away the solid sides and show you some of the useful internal triangles that will solve this problem: We can solve your question by using the Pythagoras' Theorem (a^{2} + b^{2} = h^{2}). It relates the lengths of the sides of a right triangle to the length of its hypotenuse. First consider triangle ABC. We know the length of AC is the height (1100) and the length of BC is 225 Now look at triangle DEF. We know ED is 1100 and DF is 1325 ( (3000  350) / 2 ). That makes the slant height of the narrow pieces (EF) equal to sqrt(1325^{2} + 1100^{2}) which is about 1722 mm. Finally, look at the triangle EFG. We know FG = 225 and (now) we know EF is 1722. So EG is sqrt(225^{2} + 1722^{2}) which is about 1737 mm. Now we have all the measurements we need. Here are the two patterns we use for the four pieces: Hope this helps,  


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