Surface area of roof

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My new shop/garage will be 28'X42' (12' of long side will be a garage).
The roof slope will be 5/12.
How do I figure the area of the roof?
Thanks
Jim
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On Monday, April 11, 2016 at 11:30:03 AM UTC-4, swalker wrote:

http://www.calculator.net/roofing-calculator.html
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I bet you wish you paid closer attention in geometry class in high skool.
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On Mon, 11 Apr 2016 11:06:36 -0500, Gordon Shumway

Well I did pay attention in class and found geometry to be fun and made a good grade.
I tried to solve this by using a^2 = (B^2)+(C^2) to find the long leg. The answer didn't seem right so I am looking to check it.
From the calculator above I made a mistake. Some where. Thanks for the calculator.
One thing I noted about geometry was that guys good at the pool table weren't necessarily good at geometry.
Jim
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On 4/11/2016 12:26 PM, swalker wrote:

5/12 is a particularly convenient slope actually. It's one of the famous Pythagorean Triples. Sides of 5, 12 and 13 make a right triangle. So if your horizontal and vertical measurements were actually 12 and 5, the "hypotenuse" length would be 13. If it's something else, just multiply the horizontal measurement by 13/12.

Your equation should work. Use the numbers above as an example 13^2 (169) = 5^2 (25) + 12^2 (144)
I wonder if you added up the squares of the two legs but forgot to take the square root of the result.

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On 4/11/2016 2:00 PM, Greg Guarino wrote:

I knew that in high school when I had a 98 average in geometry. Fifty three years later, not so much. I wonder if I forgot anything else. Nah.
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On 4/11/2016 3:17 PM, Ed Pawlowski wrote:

The most famous Pythagorean Triple is 3-4-5, reportedly used by the Ancient Egyptians to lay out accurate right-angle corners.
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On 4/11/2016 3:17 PM, Ed Pawlowski wrote:

infromation like your social security number and the names of all of you doctors
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Happens to all of us. I used to be familiar with Calculus, but after college never calculated a derivative other than to show off. Now I'm filling the space with more important things... Like how to properly hit a softball. Let me tell you it's not about swinging your arms. Your entire body is involved.
Puckdropper
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On 11 Apr 2016 20:04:27 GMT, Puckdropper <puckdropper(at)yahoo(dot)com> wrote:

always said that we'd probably never use the skills again after graduation but that I should know how to use integration tables. Unfortunately, they never taught how to use them. I never really needed them, either but at times it would have been good to know.
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On Monday, April 11, 2016 at 4:33:53 PM UTC-4, krw wrote:

I think it's less about using the skills you learned in high school and more about learning, period.
The ability to learn what you need know to accomplish what you are trying to do Is more important than any given subject matter.
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On Mon, 11 Apr 2016 18:54:54 -0700 (PDT), DerbyDad03

Sorry, that should have been "four semesters".

I don't agree. Secondary school should be about learning things that you will need to become a productive citizen. That includes learning how to learn but it also includes at least "business" math, some history and civics. Unfortunately it's now more about indoctrination.
I didn't have any "calculus" in high school, though. We were taught all of the foundations for it but not the grind. ;-)

The foundations are pretty important. Without arithmetic (and I'll include at least beginning algebra) you're toast in any technical field (everything is now). Without learning some sense of history and civics, we're doomed (they don't and we are - hopefully I won't see it but I'm not so sure).
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krw wrote:

Since I posted I pictures last year, I didn't mention it this year, but I made my 2nd pilgrimage to Colonial Williamsburg a few weeks ago. The antique furniture in the museum is starting to feel like old friends and I spent enjoyable time in the cabinetmakers shop. Recommended, if you're on the eastern side of the country. This time, my wife brought plenty of apples for some of the horses (the ones that weren't currently working).
But back to history, after my visit above, I became more curious about the related history prior to the mid 18th century. Wikipedia contained the following reference which I found interesting (Note: I didn't finish it yet).
http://www.jstor.org/stable/659023
In a nutshell, it describes some of the dynamics between the new "settlers" and the indigenous (Indian) population. Wikipedia, of course, contains a lot of historical facts in bite-size pieces. I owe credit Williamsburg for fostering my interest (making me a slightly more-informed individual). I still have a lot of catching up to do, I suppose. ; )
Bill
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On Monday, April 11, 2016 at 3:17:09 PM UTC-4, Ed Pawlowski wrote:

I've forgotten more stuff than most people know. At least I think I have...I don't really remember.
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On Monday, April 11, 2016 at 11:27:20 AM UTC-5, swalker wrote:

Shooting pool ain't just about geometry. Left spin, right spin, back spin, over the top spin, force of the shot, hard kiss, soft kiss, it all adds up to making the ball go everywhere but in the pocket. Doubt old Pythagoras ever shot a game of pool in his life.
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On Monday, April 11, 2016 at 11:30:03 AM UTC-4, swalker wrote:

I forgot to mention that you need to add in the eaves and other overhangs. It's not just about the square footage of the building.
http://www.calculator.net/roofing-calculator.html
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On Mon, 11 Apr 2016 09:10:00 -0700 (PDT), DerbyDad03

Yeah, I got that figured in.
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On 4/11/2016 10:29 AM, swalker wrote:

That will depend on how the roof is built. One plane sloping from one side to other other will be less than a hip design. If you have gables on one or both ends that will change a little too.
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swalker wrote:

a2 + b2 = c2 ("2" = squared). Don't forget the overhang.
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Don't forget that Pythagorus requires a right triangle.
Take the triangle formed by the roof (22.5/22.5/135 degrees), bisect into two right-triangles, then use trig to calculate the hypotenuse length, since you know the length of the adjacent side (1/2 the length of the long side of the original triangle - the long side is opposite the 135 degree angle).
cos(90) = adjacent/hypotenuse
hypotenuse = cos(90)/adjacent
area = 2 x (hypotenuse x length of roof).
qed.
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