Surface area of a triangular section of a cylinder?

[aeonwalker]

Hi there! My wife and I are building a roughly hemispherical stone house with a slant roof. We need the surface area of the wall portion under the roof (the area colored in green) to double-check the quantity surveyor's calculations for bricks, as they seem to have estimated wildly.



The house has one flat wall and the rest of the floor plan is roughly half the circumference of a large circle (or vertical cylinder, a section measuring at 25m, marked in black). The roof forms a 22.5º isosceles triangle with a height of 3m (pink). This wrapped ~triangular wedge of the cylinder has me stumped, as I know the a-sides slope inward due to the curved floor plan. End to end, the chord (blue), which is close to the diameter but not exactly, is 15m. I hope that makes sense!

Real world application that may save us $1000s! Thanks for any help you can give us!
Max

 

[blamocur]

Here is what I get: for a perfectly circular model the area [imath]S[/imath] of that part is equal to [imath]Rh(\pi-2)[/imath], where [imath]R[/imath] is the radius (7.5m in your case), and [imath]h[/imath] is the height in the center (3m in your case). The computed value I get is [imath]\approx 25.7 m^2[/imath], but there are a couple of disclaimers:
1)Because your house does not have a perfect shape some numbers do not match. For example, the length of the half circle for that radius would be 23.56m, not 25m. And the height of the roof for that radius and [imath]22.5\degree[/imath] slant would be about [imath]3.11m[/imath], not 3m. Since I don't know the details I cannot tell which way this number would have to be adjusted, and I would not expected more than 10% accuracy there.
2) There is a remote chance of a mistake in my computations, so take this with a grain of salt.

 

[aeonwalker]

Here is what I get: for a perfectly circular model the area [imath]S[/imath] of that part is equal to [imath]Rh(\pi-2)[/imath], where [imath]R[/imath] is the radius (7.5m in your case), and [imath]h[/imath] is the height in the center (3m in your case). The computed value I get is [imath]\approx 25.7 m^2[/imath], but there are a couple of disclaimers:
1)Because your house does not have a perfect shape some numbers do not match. For example, the length of the half circle for that radius would be 23.56m, not 25m. And the height of the roof for that radius and [imath]22.5\degree[/imath] slant would be about [imath]3.11m[/imath], not 3m. Since I don't know the details I cannot tell which way this number would have to be adjusted, and I would not expected more than 10% accuracy there.
2) There is a remote chance of a mistake in my computations, so take this with a grain of salt.

Thank you so much! That's very helpful. And I see what you're saying about the numbers being a bit off - I think the 15m might not quite be the diameter, which makes everything else a bit wonky. But one way or another, once we get the exact measurements we can plug them into your equations and get the precise number. Very much appreciated!

 

[blamocur]

But one way or another, once we get the exact measurements we can plug them into your equations and get the precise number.

Just don't forget about the assumptions, e.g., that the walls form a cylinder, and that there is exactly half-circle at the base of the area. I'm somewhat skeptical about precise numbers, but you might get a better approximation.