A Math Forum Project

Attaching a Shed - January 22-26, 1996 Next Problem || Previous Problem || Contents || All Problems || Search POWs In addition to our kitchen remodeling project, we are also going to build a shed off the back of the house to store the garden tools, lawnmower, and motorcycle. It will be attached to the back of the mud room, and the roof will have the same slope as the roof of the mud room. The mud room roof has a 2-inch pitch. The "pitch" of a roof stands for how many inches it rises for each foot it runs. (A 2-inch pitch isn't very steep if you have a shingle roof, and is one reason why our roof leaks in a heavy rain.) I need to decide where to attach the shed to the mud room wall, so let's work backwards. I want the shed to be 12 feet long, so it will run 12 feet out into the yard. I want the end wall to be 6 feet tall. How far up the mud room wall will the shed roof need to be attached? (Explain your answer! There are a lot of ways to figure this one out, so I want to know how you did it.) -------- -------- | | | | | | mud room | -------- | -------- | -------- | -------- | -------- | | | | | ?? | | | | shed | 6' | | | | | | | | ___________________|_______________________________________________| 12' The roofline of the mud room is 12'6" off the ground. If I just continued the roof of the shed from the roof of the mud room, how far out into the yard would it go before it was 8 feet off the ground? - Annie Fetter Solutions Annie says: I like this problem because there are a number of ways to solve it. The easiest, and the one that most people used, is to just sort of count - "a 2-inch pitch means 2 inches for every foot, so 12 feet is 24 inches of drop, add that to 6, and you get 8." Two other ways are a little more rigorous and involve setting up equations to solve the problem. One way is to write an equation for the roofline and plug numbers in. This is exactly what Bilal Seyal did, and did well. Megan Booth set up a proportion and similar triangles. I like to see that sort of variety in the solutions. This problem also lent itself to some creative math, such as "I did 6 inches left to go divided by the 2-inch pitch and got 3 feet" and "[you take] 54 inches, then you divide by 2 for the pitch and you get 27 feet." What happened to the units here? A little more explanation might be in order, don't you think? Of course, if you attach units to things, that all works out, but it is still a funny way to write it. I like Cassie Gorish's approach. A number of people did the same sort of thing, breaking down the distances into parts, but it suddenly made a lot of sense to me when I saw her solution. Below are some highlights, and the names of all the people who submitted correct solutions and most of the solutions are available. This week we also welcome four new schools. Bilal Seyal Grade 9 Fairfield High School, Fairfield, Connecticut I solved this problem by graphing. The wall between the mud room and the shed would be the y-axis and the ground would be the x-axis. I scaled the units to one foot apart and went to the point (12, 6) where the roof ends on the shed. Since there is a two-inch pitch, the slope is 2/12 or 1/6, but the line has a negative slope so it is -1/6. When you graph that line, the y-intercept is (0,8). So the shed roof will have to be attached 8 feet up the mud room wall. Since the mud room is 12'6" off the ground, the top would be the point (0, 12.5). The slope is -1/6 so the equation of the line is y = -1/6 s + 12.5, which is the roof of the mud room. If this roof is extended so that it is also the roof of the shed, and if the end of the roof is 8 feet off the ground, then, by graphing, the shed will have to go out 27 feet into the yard. Megan Booth Grade 10 Fairfield High School, Fairfield, Connecticut 1. If you want the slope of the shed of your roof to have a 2-inch pitch, the roof will rise 2 feet, making the mud room wall 8 feet tall. Since 12' = 144", 12"/2" = 144"/x; x = 2'. Add this to 6'. If the roof were flat, the shed would be a rectangle, and the mud room wall would be 6 feet tall. When you include the pitch, the wall is 8 feet. 2. By using a proportion, since pitch is proportional, 2"/12" = 54" (the roof line - end wall)/x (the length). x = 324" or 27'. P.S. Why not change the pitch of the roof on your shed so it won't leak on your motorcycle when it rains? From: Cassie Gorish Grade: 8 School: Murray Junior High 1. It has to be 8 ft. up the wall. I calculated that if for every foot 2 in. goes down, then in 12 ft. 24 in. goes down, or 2 ft. 6 + 2 = 8 (of course!) 2. 12'6" - 8' = 4'6" _2_ft._ + _2_ft._ + _6_in._ = 4'6" 12 ft 12 ft. 3 ft 12 + 12 + 3 = 27 ft. It would go 27 ft. into the yard. Christina Capacci Grade: 9 School: Juneau Douglas High School - Phoenix Program You should probably reconsider your roof's slope because it is VERY shallow. Problem one The roof must connect from a six-foot-high, twelve-foot-out wall. The roof rises vertically 2 inches for every horizontal foot. 12 x 2 = 24 inches the roof gains entirely feet inches per foot of horizontal gain The roof climbs 2 feet over its original height and thus intersects the mud room wall at eight feet. Problem two 12 1/2 - 8 = 4 1/2 existing feet, the feet of vertical feet of roof desired height difference above ground end point above ground 4 1/2 / 2 = 27 feet horizontally feet of inches, needed to be eight feet vertical the amount of above the ground difference vertical growth per one foot of horizontal gain It's good to think of this problem on an x and y axis graph. Brian MCCloskey Grade:10 School: Smoky Hill High School, Aurora,Colorado Before I started, I looked at all my givens. The roof is 6' tall and 12' long, the roof rises 2" for every 1 foot, and I have to find the difference in height between the house and the yard sides of the shed. Therefore I came up with the ratio of 12:2 for the rise of the roof. Since the length of the shed is 12', the rise of the roof would be 2'. I got this by thinking that if for every 12", the roof rises 2", then for every 12', the roof would rise 2', for every 120', the roof would rise 12', and so on. I basically just multiplied each number by 10. Getting back to the problem, if the roof rises 2', it will meet up with the house at 8'. To check my answer, I asked myself if the roof was 8' high next to the mud room and 6' high in the yard, what would be the length of that stretch of roof, with the ratio of 12:2? I subtracted 6 from 8 and came up with 2'. This is the rise of the roof. If for every foot the roof rises 2", then at how many feet would the roof raise 24"? I divided 24" by 12" to get my answer and came up with 12'. Since that was one of my givens, I knew that my answer was correct. Now for the second problem. The first thing I did here was list my givens: roof is 12'6" at the house, want to get where the roof is 8' with the same ratio of 12:2. I then subtracted 8' from 12'6" to find out the distance of descent I would be accounting for and came up with 4'6". I then converted 4'6" into inches, because the ratio of the slope of the roof is in inches, and I wanted to make everything have the same unit of measurement. I came up with 54". I then divided 54" by 2" and came up with 27'. This worked out the way it did because the roof had descended 54" and if the slope of the roof is 2" to every foot, then to get the number of feet, I have to divide the amount of descent by the number of inches per foot, since that was the given for the slope of the roof. My answer is that at 27' into the yard, the roof will be 8' off the ground. From: Ruth Carver Jill Sommer Grade 10 Mt. St. Joseph Academy, Flourtown, PA Since for every foot, the roof goes up 2 inches, the roof would have to rise 24 inches to compensate for the 12 feet length of the shed. I added that to the 6 feet and decided that the shed roof would have to begin 8 feet up the mud room wall. If the roof needed to be continuous from that of the mudroom and end up 8 feet off the ground, it would need to extend 27 feet from the end of the mudroom roof. To get this answer, I converted 12'6" and 8' to inches and got 150' and 96'. Then I subtracted 96 from 150 and got 54. Since for every 2 inches the roof went down, it went across 1 foot, I divided 54 by 2 and got 27. So, the roof would have to extend 27 feet from the mudroom in order for it to run continuously until it was 8' high off the ground. [Privacy Policy] [Terms of Use] Home || The Math Library || Quick Reference || Search || Help  © 1994-2008 Drexel University. 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