Got a question for everyone that's currently stumping me at work. Any help much appreciated
The largest possible circle is drawn to fit inside a square of side 20cm. 4 identical small circles just fit into the gaps in the four corners. Find the diameters of the small circles
Maths Club
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Re: Maths Club
3.45cm approximately - exactly 20(3-2sqrt2)
- Adam Dexter
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Re: Maths Club
Peter Mabey wrote:3.45cm approximately - exactly 20(3-2sqrt2)
Spoiler much? Got any working for that?
I've been trying to draw this nicely so that I can visualize it... this is the best I could do:
image hosting over 5mb
First of all I worked out the diagonal of the square to be SQRT (800) = 28.284cm (3dp)
Then took the diameter of the circle A from this to give the diameter of Circle B + a little bit for the edge to the corner (once divided by 2) = 4.142cm
Then I got rather stuck.
Incidentally, I listened to this whilst working... this is the first gig I ever went to.
https://myspace.com/mistysbigadventure/ ... 6-46013126
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Re: Maths Club
Full working here
Diagonal of square = sqrt(800), by Pythagoras.
Subtract diameter of large circle = sqrt(800)-20 = 20 * (sqrt(2)-1)
This is 2r, where r is the radius of the small circle, plus 2 * the length from the centre of the small circle to the nearest corner of the square.
This second length is sqrt(2*r^2), again by Pythagoras.
So 2r + 2 * sqrt(2*r^2) = 20 * (sqrt(2)-1), which gives r = 10*(sqrt(2)-1)/(sqrt(2) + 1), and diameter twice that, giving roughly 3.43cm
Diagonal of square = sqrt(800), by Pythagoras.
Subtract diameter of large circle = sqrt(800)-20 = 20 * (sqrt(2)-1)
This is 2r, where r is the radius of the small circle, plus 2 * the length from the centre of the small circle to the nearest corner of the square.
This second length is sqrt(2*r^2), again by Pythagoras.
So 2r + 2 * sqrt(2*r^2) = 20 * (sqrt(2)-1), which gives r = 10*(sqrt(2)-1)/(sqrt(2) + 1), and diameter twice that, giving roughly 3.43cm