S’(12,-20) T’(0,-8) U’(-12,-20)
The pitch of the roof is given by the ratio of the rise to the run of the roof, which is the slope of the rafter
![The \ pitch \ of \ the \ roof \ is \ \mathbf {\dfrac{6}{13}}](https://tex.z-dn.net/?f=The%20%5C%20pitch%20%20%5C%20of%20%20%5C%20the%20%5C%20%20roof%20%20%5C%20is%20%5C%20%5Cmathbf%20%7B%5Cdfrac%7B6%7D%7B13%7D%7D)
Please find attached the simple <u>diagram of the house</u> with all measurements labeled
The reason the above value for the pitch and the diagram are correct are as follows:
The given parameters are;
Width of the section of the house on which the gable of the roof is to be placed = 50 feet
Length of the over hang on the sides of the roof = 1 foot each
Total span of the roof = 52 feet
Horizontal length of each rafter, the run = 26 feet
The height of the ridge above the top of the frame, the rise = 12 feet
Required:
- To find the pitch of the roof
- To draw a diagram of the house showing measurements
Solution:
![Pitch = \dfrac{Rise}{Run}](https://tex.z-dn.net/?f=Pitch%20%3D%20%5Cdfrac%7BRise%7D%7BRun%7D)
Please find attached the <u>diagram</u> of the house showing the measurements for the roof parameters and the section of the house where the gable roof is to be placed
Learn more about finding the slope of a line here:
brainly.com/question/18990889
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Answer:
The writing isn't clear though hope it helps
Answer:
B (-2,-4)
Step-by-step explanation:
Let coordinates of B be (x,y)
![( \frac{ - 6+ x}{2} . \frac{4 + y}{2} ) = ( - 4.0) \\ x = 2( - 4) + 6 \\ x = - 2 \\ \\ y = 2(0) - 4 \\ y = - 4](https://tex.z-dn.net/?f=%28%20%5Cfrac%7B%20-%206%2B%20x%7D%7B2%7D%20.%20%5Cfrac%7B4%20%2B%20y%7D%7B2%7D%20%29%20%3D%20%28%20%20-%204.0%29%20%5C%5C%20x%20%3D%202%28%20-%204%29%20%2B%206%20%5C%5C%20x%20%3D%20%20-%202%20%5C%5C%20%20%5C%5C%20y%20%3D%202%280%29%20-%204%20%5C%5C%20y%20%3D%20%20-%204)
Circumference of circular track, C = 0.5 miles = 0.5×5280 feet = 2640 feet.
We know, circumference in circle is given by :
![C = 2\pi r](https://tex.z-dn.net/?f=C%20%3D%202%5Cpi%20r)
( Here, r is radius i.e distance of centre from track )
Putting values in above equation :
2 × 3.14 × r = 2640
6.28 × r = 2640
r = 420.38 feet
Therefore, distance between centre and track is 420.38 feet .
Hence, this is the required solution.