document.write( "Question 1146409: A circle of radius 6 cm is inscribed in a square. A smaller circle is drawn tangent to the two sides of the square and the bigger circle. What is the radius of the smaller circle? \n" ); document.write( "

Algebra.Com's Answer #767774 by greenestamps(13200) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "The solution from tutor @ankor is faulty.

\n" ); document.write( "He states that the diameter of the smaller circle is equal to half the diagonal of the square minus the radius of the large circle. That is not true; there is a piece of the diagonal of the square between the small circle and the corner of the square.

\n" ); document.write( "Here is my solution to the problem, copied from a response I made earlier to the same question.

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\n" ); document.write( "Draw the figure as described.

\n" ); document.write( "Draw 3 radii of the smaller circle -- perpendicular to the two sides of the square, and to the point of tangency of the two circles.

\n" ); document.write( "Calling r the radius of the small circle, the distance from the center of the large circle to the corner of the square is

\n" ); document.write( " $\"6%2Br%2Br%2Asqrt%282%29\"$

\n" ); document.write( "But that distance is $\"6%2Asqrt%282%29\"$

\n" ); document.write( "So

\n" ); document.write( " $\"6%2Br%2Br%2Asqrt%282%29+=+6%2Asqrt%282%29\"$
\n" ); document.write( " $\"6%2Br%281%2Bsqrt%282%29%29+=+6%2Asqrt%282%29\"$
\n" ); document.write( " $\"r%281%2Bsqrt%282%29%29+=+6%2Asqrt%282%29-6+=+6%28sqrt%282%29-1%29\"$
\n" ); document.write( " $\"r+=+%286%28sqrt%282%29-1%29%29%2F%28sqrt%282%29%2B1%29\"$
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\n" ); document.write( " $\"r+=+6%283-2%2Asqrt%282%29%29%2F%282-1%29\"$
\n" ); document.write( " $\"r+=+6%283-2%2Asqrt%282%29%29\"$
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