# Circles within a square

• Apr 27th 2010, 06:31 AM
BG5965
Circles within a square
http://i301.photobucket.com/albums/n...5965/MASBc.png

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 $288m^2$. However, I don't know how to approach this part of the question. Please help! Thanks :)
• Apr 27th 2010, 07:44 AM
bjhopper
circles in square
Quote:

Originally Posted by BG5965
http://i301.photobucket.com/albums/n...5965/MASBc.png

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 $288m^2$. However, I don't know how to approach this part of the question. Please help! Thanks :)

Hello BG,
I think this problem belongs in trig. If you can handle geometry I can help by telling you i calculated the sector angle common to the two goats 55.2 deg. Rest is geometry.

bjh
• Apr 27th 2010, 10:00 AM
Wilmer
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?
• Apr 28th 2010, 05:32 AM
BG5965
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 $176.69m^2$. Similarly, the area of the triangle EDF is $150.59m^2$, therefore the extra circular area in the sector is equal to $26.1m^2$, which is half of the shared area, so the total shared area is $52.2m^2$.

Once again, thanks a lot for your help :) BG
• Apr 29th 2010, 08:43 AM
slovakiamaths
diff. ans
Dear BG5965, i worked out a diff. ans of your problem. Please check the attachment and give remarks
• Apr 29th 2010, 12:58 PM
bjhopper
Quote:

Originally Posted by slovakiamaths
Dear BG5965, i worked out a diff. ans of your problem. Please check the attachment and give remarks

Hi slovakiamaths,

The answers given by BG5965 are correct. Area in question consists of two back-back circular segments.No relation to ellipse.

bjh