document.write( "Question 82494This question is from textbook Intermediate Algebra
\n" ); document.write( ": The following slab of concrete is twice as long as it is wide. The area in which it is placed includes a 1 ft. wide border of 70 square ft. Find the dimensions of the slab. \n" ); document.write( "
Algebra.Com's Answer #59221 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
The following slab of concrete is twice as long as it is wide. The area in which it is placed includes a 1 ft. wide border of 70 square ft. Find the dimensions of the slab.
\n" ); document.write( ":
\n" ); document.write( "I assume this is rectangle with a 1 ft path around it which is 70 sq/ft
\n" ); document.write( ":
\n" ); document.write( "Let x = width of the concrete
\n" ); document.write( "Then 2x = length of the concrete
\n" ); document.write( ":
\n" ); document.write( "The overall dimensions(including the 1 ft path):
\n" ); document.write( "Length = (2x+2)
\n" ); document.write( "Width = (x+2)
\n" ); document.write( ":
\n" ); document.write( "(Over all area) - (concrete area) = path area (given as 70 sq/ft)
\n" ); document.write( ":
\n" ); document.write( "(2x+2)*(x+2) - 2x*x = 70
\n" ); document.write( ":
\n" ); document.write( "2x^2 + 6x + 4 - 2x^2 = 70; FOILed (2x+2)(x+2)
\n" ); document.write( ":
\n" ); document.write( "6x + 4 = 70; the 2x^2's cancel
\n" ); document.write( ":
\n" ); document.write( "6x = 70 - 4
\n" ); document.write( ":
\n" ); document.write( "6x = 66
\n" ); document.write( ":
\n" ); document.write( "x = 66/6
\n" ); document.write( ":
\n" ); document.write( "x = 11 ft is the width of the slab; 2*11 = 22 ft is the length
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check our solution, using the overall dimensions of 24 by 13
\n" ); document.write( "(24*13) - (22*11) =
\n" ); document.write( " 312 - 242 = 70
\n" ); document.write( ":
\n" ); document.write( "Make sense to you? Any questions? \n" ); document.write( "
\n" );