Discover Geometry Theorems for Simple Roof Truss Design | Figure 2.10

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The discussion focuses on solving a geometry problem related to a simple roof truss design, specifically determining the lengths of XT and XZ using the Pythagorean theorem. The truss has a base VWXY of 20m divided into three equal segments, with a height of 2.5m. Participants suggest using right triangles formed by drawing perpendiculars from point T to the base to set up equations for XT and XZ. However, confusion arises regarding the calculation of lengths TZ and ZY, as two unknowns remain unsolved. The conversation emphasizes the need for additional information to resolve the equations effectively.
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A simple roof truss design is shown in figure 2.10( i have attached) The lower section, VWXY is made from three equal length segments. UW and XZ are perpendicular to VT and TY, respectively. If VWXY is 20m and the height of the truss is 2.5m, determine the lengths of XT and XZ.

Ok It has been years since I have taken geometry and am looking for a little guidance. Any theorems or help to get this problem started in the right direction would be great. I do not want full solutions that would be defeating the purpose :)
 

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VWXY is 20m long and divided into 3 equal segments so each segment has length 20/3 m. If you draw a perpendicular from T to VWXY, it will bisect WX at, say S, and so forms a right triangle with legs of length (20/3)/2= 10/3 and 2.5 m. You can use the Pythagorean theorem to find the length of the hypotenuse XT: XT2= (10/3)2+ 2.52.

Similarly, that perpendicular forms right triangle STY having legs of length 2.5m and 10m (half of the 20m length of VWXY). TY2= 2.52+ 10[/sup]. Call that x.

Finding XZ is a little harder. Let u= length of TZ. Then length of ZY is x- u. You now have two right triangle with leg XZ. The first right triangle is XTZ which has hypotenuse
XT and legs TZ and XZ: TX2= XZ2+ u2. The other right triangle is XZY which has hypotenuse 10/3 m and legs XZ and TY- u: (10/3)2= XZ2+ (TY- u)2. Since you have calculated TX and TY above that gives two equations to solve for XZ and u.
 


Edit: Deleting my solutions, per OP request. *sorry*
 
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HallsofIvy said:
Finding XZ is a little harder. Let u= length of TZ. Then length of ZY is x- u. You now have two right triangle with leg XZ. The first right triangle is XTZ which has hypotenuse
XT and legs TZ and XZ: TX2= XZ2+ u2. The other right triangle is XZY which has hypotenuse 10/3 m and legs XZ and TY- u: (10/3)2= XZ2+ (TY- u)2. Since you have calculated TX and TY above that gives two equations to solve for XZ and u.

Ok maybe I am missing something. I can solve to get TY to be approximately 10.31. But after that I cannot see how to solve for TZ and ZY. There is always 2 things left unsolved and we all know you can't solve for 2 different things without at least one of them. I am a little confused. I understand the problem setup and where you are trying to go though just seems to be missing some important numbers to help solve.
 


You can, as I showed, use triangles XZT and XZY to set up two equations for those two unknown numbers.
 

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