Calculate the Area of a Triangle

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

therefore:

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

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

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.

 L=sqrt(D²-(W/2)²)
R=(4(Z-L)²+W²)/(8*(Z-L))
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!

jclothier,

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)

Thanks for clearing that up ozo.