Calculate the Area of a Triangle

Hi All

I need a formula to calculate the area of a trinagle, but theres a twist. The triangle as a curved face. Heres a picture of what I am talking about

I need to calculate the area using only the dimmensions specified on the drawing, A, B & C.

Many Thanks
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Didn't I ansqwer this before?
Iknow you could do this with calculus, but thinking simpler, this is a sector of a circle.

Area of a circle = Pi*r^2.

Area of a secotor of a circle = 0.5*r^2*x

angle of sector = x = 2z

r = a = b

sin z = a/(0.5*c)

z = sin^-1(a/(0.5*c))

x = 2*(sin^-1(a/(0.5*c))) = angle of sector of circle


Area of a secotor of a circle = 0.5*r^2*x

= 0.5*r^2*(2*(sin^-1(a/(0.5*c))))

oops, r = a = b....

Area = 0.5*a^2*(2*(sin^-1(a/(0.5*c))))

Also, I assumed the angles created by the intersection of lines b & c to be 90 degrees.

the drawing is a sector of a circle.  is is also 2 identical sectors of the same circle.  I used one of the triangles created by these 2 identical sectors of the same circle to find the corresponding angle.  This angle was doubled to find the angle of the original sector (hence, x = 2z).

It has been a while, but I think this is correct.
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jclothierAuthor Commented:
Thanks to both of you.

Oceanbeach, The angle is not always going to be 90 degrees and it also is not always going to be a sector of a circle. The formula that you have given is correct based on that so I have awarded you some points.

ozo, Yes you have answered this for me before but after I accepted I couldn't get the formula to work.

A = (R²sin¯¹(W/(2R))+W(L-sqrt(R²-(W/2)²))/2)

I think that the problem was with this part sin¯¹

I note that this time you have changed it to arcsin

This may just be not knowing what I am talking about but this made it work.

Thanks again!

You know what they say about assumptions....

Aside from bad assumptions, I am glad my thought process was correct.  I sure hope I at least helped since you gave me an assist.

Happy number crunching!
sin¯¹(x) is a notation for arcsin(x)
which can be confusing, since it looks a lot like sin²(x)
jclothierAuthor Commented:
Thanks for clearing that up ozo.
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