Extra credit, missing area

>>229595
You have to find the area of the shaded region.

>>229597
ah thanks la' didnt see that bit.

would like a real response from someone if im being honest

>>229595
find area of circle, subtract area of sector, add back in area of triangle. done

Ten seconds of Google gave the formula for the area of a segment.

>>229606

>bumping after 6 minutes when you're still on the first page of a board where it takes 5 hours to get a reply

wew lad. You're just supposed to find the area of the blue stuff, which is the area of the triangle plus the area of the rest. You already have 4 out of 6 pieces of info on the triangle, so you solve it first and now you just have to find the area of the other stuff.

Make the center of the circle (0,0) and use the known radius to calculate a parabola that represents half of the circle, then use definite integration to know the area between that parabola and the x axis. Now you only have 2 unknown regions that are neither the triangle nor the half-circle. These are perfect slices of a circle of known radius, and since you know that there's already 130 degrees + 180 degrees covered, each slice is half of what's left and you can directly know their areas. Or you could do it like that from the beginning if you want to.

>>229595
294.49m^2

Area of Circle - Area of segment (white part)

>Area of Circle
Google area of circle formula
>Area of Segment
http://planetcalc.com/1421/

>>229595
Dunno if you figured it out yet, but here's an easy way to look at it:

1) If the angle was 180 degrees, you'd know it cut the circle in half. If it was 90 degrees, you'd know the piece of circle cut out by the angle's 2 short sides would be 1/4th the entire circle. So it makes sense that the triangle PLUS the white region is Area x (13/36).

2) The next part is figuring out the length of the triangle side opposite the angle. Well, sin(x) = opposite/hypotenuse, right? You have an angle of 130, but if you imagine a line bisecting the triangle in half, you now have one half-triangle (another triangle!) with an angle of 65. And sin(65) = A / 11.1, where "A" is one-half your chord length. Do the same with the cosine function to get the height of this mini-triangle "B", then multiply A and B. (A rectangle with dimensions AxB has the same area as your two mini-triangles)

3) Now that you know the area of the triangle, AND how to find a proportion of the circle's area, you're all set.