document.write( "Question 580602: If a wire is bent into the shape of a square, then the area enclosed by the square is 81 cm square. when the same wire is bent into semi-circle. find the area enclosed by the semi-circle \n" ); document.write( "
Algebra.Com's Answer #371295 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
\"+A+=+s%5E2+\"
\n" ); document.write( "\"+81+=+s%5E2+\"
\n" ); document.write( "\"+s+=+9+\"
\n" ); document.write( "The perimeter is \"+4s+\"
\n" ); document.write( "\"+P+=+4%2A9+\"
\n" ); document.write( "\"+P+=+36+\" cm
\n" ); document.write( "This is the length of the wire
\n" ); document.write( "The circumference of a circle is
\n" ); document.write( "\"+C+=+2%2Api%2Ar+\", or
\n" ); document.write( "\"+C+=+pi%2Ad+\" where \"d\" is the diameter
\n" ); document.write( "A semi-circle enclosed by a wire looks like
\n" ); document.write( "\"+%281%2F2%29%2Api%2Ad+%2B+d+\"
\n" ); document.write( "\"+%281%2F2%29%2Api%2Ad+%2B+d+=+36+\"
\n" ); document.write( "\"+d%2A%28+pi%2F2+%2B+1+%29+=+36+\"
\n" ); document.write( "\"+d+=+36+%2F+%28+pi%2F2+%2B+1+%29+\"
\n" ); document.write( "The radius is
\n" ); document.write( "\"+r+=+18+%2F+%28+pi%2F2+%2B+1+%29+\"
\n" ); document.write( "The area of the semi-circle is
\n" ); document.write( "\"+A%2F2+=+%281%2F2%29%2Api%2Ar%5E2+\"
\n" ); document.write( "\"+A%2F2+=+%281%2F2%29%2Api%2A%28+18+%2F+%28+pi%2F2+%2B+1+%29%29%5E2+\"
\n" ); document.write( "\"+A%2F2+=+%281%2F2%29%2Api%2A%28+18+%2F+%28+1.5708+%2B+1+%29%29%5E2+\"
\n" ); document.write( "\"+A%2F2+=+%281%2F2%29%2Api%2A%28+18+%2F+2.5708+%29%5E2+\"
\n" ); document.write( "\"+A%2F2+=+%281%2F2%29%2Api%2A%28+7.0017+%29%5E2+\"
\n" ); document.write( "\"+A%2F2+=+%281%2F2%29%2Api%2A49.024+\"
\n" ); document.write( "\"+A%2F2+=+%281%2F2%29%2A154.013+\"
\n" ); document.write( "\"+A%2F2+=+77.0067+\" cm
\n" ); document.write( "Hope I got it\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );