Question 1198960
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The tutor @greenestamps has a great efficient method. 


I'll show two other methods. 
They aren't as efficient, but they're still handy to see how to find alternative pathways.


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Method 1


x = larger radius
y = smaller radius
The goal is to find the ratio x/y
Multiplying the smaller radius y by the scale factor x/y, will get you the larger radius x.


Based on the diagram the student @lotusjayden has posted, the square has side length x+y.


Use the pythagorean theorem to find the diagonal is {{{(x+y)*sqrt(2)}}} units long. 
Or you could note that there are two 45-45-90 right triangles that make up the square.


Draw a diagonal from the bottom left corner of the square to the top right corner. 
This diagonal is composed of the radii of the larger tangent circles, so each diagonal is also x+x = 2x units long.



So,
{{{(x+y)*sqrt(2) = 2x}}}


{{{(x+y)/x = 2/sqrt(2)}}}


{{{x/x + y/x = 2/sqrt(2)}}}


{{{1 + y/x = 2/sqrt(2)}}}


{{{y/x = 2/sqrt(2)-1}}}


{{{y/x = 2/sqrt(2)-sqrt(2)/sqrt(2)}}}


{{{y/x = (2-sqrt(2))/sqrt(2)}}}


{{{x/y = sqrt(2)/(2-sqrt(2))}}}


{{{x/y = (sqrt(2)*(2+sqrt(2)))/((2-sqrt(2))(2+sqrt(2)))}}} Multiplying top and bottom by (2+sqrt(2)) so the denominator is rationalized.


{{{x/y = (2*sqrt(2)+sqrt(2)*sqrt(2))/(2^2-(sqrt(2))^2)}}} Expand the numerator. Use the difference of squares rule in the denominator.


{{{x/y = (2*sqrt(2)+2)/(4-2)}}} The square root cancels out in the denominator.


{{{x/y = (2(sqrt(2)+1))/(2)}}}


{{{x/y = sqrt(2)+1}}}


{{{x/y = 1+sqrt(2)}}}



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Method 2


d = diagonal of the square
d/2 = radius of each larger circle


The square splits into two congruent 45-45-90 right triangles
The hypotenuse is d, so each leg is {{{d/sqrt(2) = d*sqrt(2)/2}}} which is the side length of the square.


The smaller radius is {{{(SquareSide)-(LargerRadius) = d*sqrt(2)/2 - d/2 = (d*sqrt(2)-d)/2 = (d(sqrt(2)-1))/2}}}


To find the ratio of the larger radius to smaller radius, we divide the two items


{{{Ratio = (LargerRadius)/(SmallerRadius)}}}


{{{Ratio = (d/2)/((d(sqrt(2)-1))/2)}}}


{{{Ratio = (d/2)*(2/(d(sqrt(2)-1)))}}}


{{{Ratio = (cross(d)/2)*(2/(cross(d)(sqrt(2)-1)))}}} The 'd' terms cancel


{{{Ratio = (1/cross(2))*(cross(2)/(sqrt(2)-1))}}} These '2's cancel also


{{{Ratio = 1/(sqrt(2)-1)}}}


{{{Ratio = (sqrt(2)+1)/((sqrt(2)-1)(sqrt(2)+1))}}} Multiplying top and bottom by (sqrt(2)+1) so the denominator is rationalized.


{{{Ratio = (sqrt(2)+1)/((sqrt(2))^2-1^2)}}} 


{{{Ratio = (sqrt(2)+1)/(2-1)}}} 


{{{Ratio = (sqrt(2)+1)/(1)}}}


{{{Ratio = sqrt(2)+1}}}


{{{Ratio = 1+sqrt(2)}}}
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