document.write( "Question 858299: A semicircle is situated on top of a rectangle so that the diameter (straight edge) is touching with the shorter side of the rectangle. The radius of the semicircle is 'r' and the perimeter of the whole shape is 6m. \r
\n" ); document.write( "\n" ); document.write( "The question asks to find the formula for 'A' (Area of whole shape) in terms of 'r'.\r
\n" ); document.write( "\n" ); document.write( "This was all the info give, hopefully that is enough. There was a diagram in the original question.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #517156 by rothauserc(4718)\"\" \"About 
You can put this solution on YOUR website!
let l be the length, r is radius of semi-circle, P is perimeter, A is area
\n" ); document.write( "A = (l * 2r) + 1/2 * pi * r^2
\n" ); document.write( "P = 2l + 2r + (1/2)pi2r
\n" ); document.write( "P = 2l + 2r + pi*r
\n" ); document.write( "6 = 2l + 2r + pi*r
\n" ); document.write( "2l = 6 - 2r - pi*r
\n" ); document.write( "l = 3 - r - (1/2)*pi*r
\n" ); document.write( "substitute for l in Area equation
\n" ); document.write( "A = [ (3 - r - (1/2)*pi*r)*2r] + 1/2 * pi * r^2
\n" ); document.write( "A = ( 6r - 2r^2 - pi*r^2) + (1/2 * pi * r^2)
\n" ); document.write( "A = 6r - 2r^2 - (1/2)*pi*r^2
\n" ); document.write( "A = r*(6 - 2r - (1/2)*pi*r)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );