SOLUTION: The vertices of a square are the centers of four circles as shown below. The two big circles touch each other and also the two little circles. With which factor do you have to mult

Algebra ->  Circles -> SOLUTION: The vertices of a square are the centers of four circles as shown below. The two big circles touch each other and also the two little circles. With which factor do you have to mult      Log On


   

Question 1198960: The vertices of a square are the centers of four circles as shown below. The two big circles touch each other and also the two little circles. With which factor do you have to multiply the radius of the little circles to obtain the radius of the big circle?
Found 3 solutions by lotusjayden, greenestamps, math_tutor2020:
Answer by lotusjayden(18) About Me  (Show Source):
You can put this solution on YOUR website!
Here is a diagram to visualize it, although I believe it is not to scale:

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a diagram:



Let 2x be the side length of the square. Then...
the diagonal of the square is 2x%2Asqrt%282%29,
so the radius of the larger circles is x%2Asqrt%282%29,
so the radius of the smaller circles is 2x-x%2Asqrt%282%29=x%282-sqrt%282%29%29

The problem asks for the factor by which the radius of the smaller circles has to be multiplied to get the radius of the larger circles. That factor is



ANSWER: sqrt%282%29%2B1

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

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 %28x%2By%29%2Asqrt%282%29 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,
%28x%2By%29%2Asqrt%282%29+=+2x

%28x%2By%29%2Fx+=+2%2Fsqrt%282%29

x%2Fx+%2B+y%2Fx+=+2%2Fsqrt%282%29

1+%2B+y%2Fx+=+2%2Fsqrt%282%29

y%2Fx+=+2%2Fsqrt%282%29-1

y%2Fx+=+2%2Fsqrt%282%29-sqrt%282%29%2Fsqrt%282%29

y%2Fx+=+%282-sqrt%282%29%29%2Fsqrt%282%29

x%2Fy+=+sqrt%282%29%2F%282-sqrt%282%29%29

Multiplying top and bottom by (2+sqrt(2)) so the denominator is rationalized.

Expand the numerator. Use the difference of squares rule in the denominator.

x%2Fy+=+%282%2Asqrt%282%29%2B2%29%2F%284-2%29 The square root cancels out in the denominator.

x%2Fy+=+%282%28sqrt%282%29%2B1%29%29%2F%282%29

x%2Fy+=+sqrt%282%29%2B1

x%2Fy+=+1%2Bsqrt%282%29


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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%2Fsqrt%282%29+=+d%2Asqrt%282%29%2F2 which is the side length of the square.

The smaller radius is

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

Ratio+=+%28LargerRadius%29%2F%28SmallerRadius%29

Ratio+=+%28d%2F2%29%2F%28%28d%28sqrt%282%29-1%29%29%2F2%29

Ratio+=+%28d%2F2%29%2A%282%2F%28d%28sqrt%282%29-1%29%29%29

Ratio+=+%28cross%28d%29%2F2%29%2A%282%2F%28cross%28d%29%28sqrt%282%29-1%29%29%29 The 'd' terms cancel

Ratio+=+%281%2Fcross%282%29%29%2A%28cross%282%29%2F%28sqrt%282%29-1%29%29 These '2's cancel also

Ratio+=+1%2F%28sqrt%282%29-1%29

Ratio+=+%28sqrt%282%29%2B1%29%2F%28%28sqrt%282%29-1%29%28sqrt%282%29%2B1%29%29 Multiplying top and bottom by (sqrt(2)+1) so the denominator is rationalized.

Ratio+=+%28sqrt%282%29%2B1%29%2F%28%28sqrt%282%29%29%5E2-1%5E2%29

Ratio+=+%28sqrt%282%29%2B1%29%2F%282-1%29

Ratio+=+%28sqrt%282%29%2B1%29%2F%281%29

Ratio+=+sqrt%282%29%2B1

Ratio+=+1%2Bsqrt%282%29