Radius Small Circle: Measurement & More

[Albert1]

View attachment 3046

 

[chisigma]

[sp]Let suppose that the side of the square is 2. In this case, if x is the radius of the 'small circle', for the theorem of Pythagoras it must be...

$\displaystyle (1-x)^{2} + 1 = (1+x)^{2}$

... so that is $\displaystyle x = \frac{1}{4}$...[/sp]

Kind regards

$\chi$ $\sigma$

 

[RLBrown]

hint:

Spoiler

see Ford Circles

 

[Albert1]

[sp]Let suppose that the side of the square is 2. In this case, if x is the radius of the 'small circle', for the theorem of Pythagoras it must be...

$\displaystyle (1-x)^{2} + 1 = (1+x)^{2}$

... so that is $\displaystyle x = \frac{1}{4}$...[/sp]

Kind regards

$\chi$ $\sigma$

very good solution !