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Sometimes a geometry problem can be superimposed on the x-y plane to enable an efficient solution. This is especially the case if there is a sequence of points that are connected by a path in which each segment has a known slope, and the endpoint is offset by a known amount in the x and y dimensions. Example 1The given house has a curb roof. The upper rafters have a pitch of 1/4 and the lower rafters have a pitch of 1. Find the lengths AB and CD. We are told the house is symmetrical about the vertical line, CH, and we are given the total width (469 in) and height (244 in) of the building.
A key to understanding this problem is knowing the architectural definition of "pitch", which, in a symmetrical house like this one, is half the slope. Now we know both the horizontal and vertical offset of point C from point A, which is (469/2,244). And we can find a connected path from A to C in which each segment has a known slope -- AB and BC. At this point, we arbitrarily call segment AG "x". That makes GB 2x, because the slope of AB is 2. Then, having traveled x inches already, the distance from B to F is AH - AG, which is 469/2-x. Now, since the slope of BC is 1/2, it follows that CF is half of BF.
Now, we have two expressions for the vertical distance CH: the sum of CF and BG on the one hand, and CH on the other.
Now knowing that x=169/2, we can fill in the remaining values:
Finally, the Pythagorean Theorem can be used to give us
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