Circle question (1 Viewer)

[withoutaface]

I need some help with the attached question...

 

[tempco]

oh i did that as a 4u question.. just break up the shaded area into 2 equilateral triangles of 1 m radius and the curved part as the area of an arc with an angle of pi/3 and 1 m radius

Area of triangles:

0.5 x 0.5 x (root 3)/2

and there are two triangles so thats (root 3)/4

Area of arcs:

0.5 x 1 x (pi/3) - (root3)/8

So in total its

2pi/3 - root3/4

and there are 4 arcs, so thats

2pi/3 - root3/2


Yeh theres bound to be a mistake in there...

 

[CrashOveride]

Argh i was doing it as if the intersections wernt at the centres :/

Hey nekkid do u consider urself a circle geo. guru? I put up another question somewhere here and i got some others :/

 

[tempco]

Nah I'm not a guru... by far >_<.. but post em up and if I know how to do them I'll reply.

 

[Rorix]

(or area of minor segments = 1/2 r^2 (theta - sine theta)

 

[laohy]

is it changed from sum qs from 4u..
c 'untitled.jpg'