



 
Hi Joe, I assume this is the shape you want. If so then the answer is yes but you don't have a lot of choice. If the roof is as short as possible then $CDE$ is a straight line and the distance from $X$ to $Y$ is 4 feet. If the roof is as tall as possible then $BCD$ is a straight line and $BC=CD,$ although hey don't look equal in my diagram. Thus triangles $YBC$ and $ZCD$ are congruent and hence $XY = ZC= YB$ and thus the distance from $X$ to $Y$ is 3 feet. Hence choose a point Y on the base between 3 and 4 feet from the center point $X.$ Since the triangle $YBC$ is a right triangle the height $YC$ can then be found using Pythagoras theorem. \[BC^2 = YB^2 + YC^2\] so \[YC = \sqrt{4^2  YB^2}.\] In a similar fashion \[PD = \sqrt{4^2  PC^2}.\] I hope this helps,  


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