Circle and square

An ABCD square with a side length of 100 mm is given. Calculate the radius of the circle that passes through the vertices B, C and the center of the side AD.

Correct result:

r =  62.5 mm

Solution:

a=100 mm r2=x2+(a/2)2 r+x=a  r2=(ra)2+(a/2)2  r2=r22 ra+a2+(a/2)2  2 r a=a2+(a/2)2  r=a2+(a/2)22 a  r=58 a=58 100=1252=62.5 mma=100 \ \text{mm} \ \\ r^2=x^2 + (a/2)^2 \ \\ r+x=a \ \\ \ \\ r^2=(r-a)^2 + (a/2)^2 \ \\ \ \\ r^2=r^2 - 2 \cdot \ ra + a^2 + (a/2)^2 \ \\ \ \\ 2 \cdot \ r \cdot \ a=a^2 + (a/2)^2 \ \\ \ \\ r=\dfrac{ a^2+(a/2)^2 }{ 2 \cdot \ a } \ \\ \ \\ r=\dfrac{ 5 }{ 8 } \cdot \ a=\dfrac{ 5 }{ 8 } \cdot \ 100=\dfrac{ 125 }{ 2 }=62.5 \ \text{mm}



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