document.write( "Question 241467: I just cannot figure this out. Please help me: a sign is in the shape of a square with a semicircle of a radius x adjoining one side and a semicircle of diameter x removed from the opposite side. If the side of the square are length 2x, then write the area of the sign as a function of x. \n" ); document.write( "
Algebra.Com's Answer #176840 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
I think it looks like this:\r
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\n" ); document.write( "The area of the semi-circle added to one side
\n" ); document.write( "is \"%281%2F2%29%2Api%2Ax%5E2\"
\n" ); document.write( "The area of the square is \"2x%2A2x+=+4x%5E2\"
\n" ); document.write( "So far the area of the sign is
\n" ); document.write( "\"4x%5E2+%2B+%281%2F2%29%2Api%2Ax%5E2\"
\n" ); document.write( "The semi-circle that is removed
\n" ); document.write( "has radius = \"x%2F2\"
\n" ); document.write( "It's area is \"%281%2F2%29%2Api%2A%28x%2F2%29%5E2\"
\n" ); document.write( "So, the area of the sign is
\n" ); document.write( "\"4x%5E2+%2B+%281%2F2%29%2Api%2Ax%5E2+-+%281%2F2%29%2Api%2A%28x%2F2%29%5E2\"
\n" ); document.write( "\"4x%5E2+%2B+%281%2F2%29%2Api%2Ax%5E2+-+%281%2F2%29%2Api%2A%28x%5E2%2F4%29\"
\n" ); document.write( "Factor out \"x%5E2\"
\n" ); document.write( "\"x%5E2%2A%284+%2B+%281%2F2%29%2Api+-+%281%2F8%29%2Api%29\"
\n" ); document.write( "Factor out \"pi\"
\n" ); document.write( "\"x%5E2%2A%284+%2B+%281%2F2+-+1%2F8%29%2Api%29\"
\n" ); document.write( "\"x%5E2%2A%284+%2B+%283%2F8%29%2Api%29\"
\n" ); document.write( "Hope I got it\r
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