This is part c) of a problem using circles within squares - the first 2 parts included information necessary for part c), which included the length of the rope if the total area "eatable" was . However, I don't know how to approach this part of the question. Please help! Thanks
Well, removing the goats and company out of there, you've got
2 intersecting circles, one with center (0,24) and radius 19.15 [1],
the other with center (24,0) and radius 19.15 [2].
Their equations are:
(x - 0)^2 + (y - 24)^2 = 19.15^2 [1]
(x - 24)^2 + (y - 0)^2 = 19.15^2 [2]
Now you can easily (easily enough!) calculate the 2 intersecting points...
Get my drift?
Thanks all of your for your input, I've worked it out now.
Let the 4 corners be A, B, C, D and and the obtuse angles of the rhombus as E and F.
By pythagoras, the diagonal is 33.94m, and the 2 radii from either circle (length 19.15m) make a triangle with those dimensions. By cosine rule, the obtuse angle AED is 124.79 degrees, and as opposite angles are equal, so is the other side.
Therefore, each acute angle is 55.21 degrees, so the area of the sector EDF is . Similarly, the area of the triangle EDF is , therefore the extra circular area in the sector is equal to , which is half of the shared area, so the total shared area is .
Once again, thanks a lot for your help BG