document.write( "Question 29496: the perimeter of a rectangle is 32m and the area of the rectangle is 55m squared. Find the dimensions of the rectangle \n" ); document.write( "
Algebra.Com's Answer #16629 by sdmmadam@yahoo.com(530)\"\" \"About 
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the perimeter of a rectangle is 32m and the area of the rectangle is 55m squared. Find the dimensions of the rectangle\r
\n" ); document.write( "\n" ); document.write( "Let the length of therectangle be L meters
\n" ); document.write( "and let the width be B meter.
\n" ); document.write( "Given that the perimeter of a rectangle is 32m
\n" ); document.write( "That is 2(L+B)= 32 ----(1)
\n" ); document.write( "which on dividing by 2 gives (L+B)=16 ----(2)
\n" ); document.write( "Given that the area of the rectangle is 55m
\n" ); document.write( "That is LXB = 55----(3)
\n" ); document.write( "By formula (L-B)^2 = (L+B)^2-4LB
\n" ); document.write( "=(16)^2-4X(55) [using (2) and (3)]
\n" ); document.write( "=256-220
\n" ); document.write( "=36
\n" ); document.write( "(L-B)^2 = 36
\n" ); document.write( "taking the positive sqroot
\n" ); document.write( "(since length >width by convention, therefore L>B implying (L-B) >0 }
\n" ); document.write( "Therefore L-B = 6 ----(4)
\n" ); document.write( "We have L+B= 16----(2)
\n" ); document.write( "Adding (4) and (2)
\n" ); document.write( "(L+L) = (6+16)
\n" ); document.write( "2L= 22
\n" ); document.write( "L=22/2 = 11
\n" ); document.write( "L=11 in (2) gives B = 5
\n" ); document.write( "Therefore Length of the rectangle = 11m
\n" ); document.write( "and width of the rectangle = 5 ms
\n" ); document.write( "Verification: Area = length X width = 11X5=55 sq meters which is correct
\n" ); document.write( "Therefore our values are correct.
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