document.write( "Question 777839: the area of a rectangular box is 33cm2.If the width is 8 cm less than the length. what are the dimensions \n" ); document.write( "
Algebra.Com's Answer #474289 by paul2007(1)\"\" \"About 
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Area of the rectangle is 33 cm^2
\n" ); document.write( "the width is 8 cm less than the length
\n" ); document.write( "Let the length be, x cm
\n" ); document.write( "therefore, the width is (x-8)cm
\n" ); document.write( "we know, area of a rectangle is= length*width
\n" ); document.write( "according to the problem,
\n" ); document.write( " x*(x-8)=33
\n" ); document.write( "or, x^2-8x=33
\n" ); document.write( "or, x^2-8x-33=0
\n" ); document.write( "or, x^2-(11-3)x-33=0 [Middle term factor]
\n" ); document.write( "or, x^2-11x+3x-33=0
\n" ); document.write( "or, x(x-11)+3(x-11)=0
\n" ); document.write( "or, (x-11)(x+3)=0\r
\n" ); document.write( "\n" ); document.write( "now, between these two factor either one should be zero
\n" ); document.write( "so, 1st factor, x-11=0
\n" ); document.write( " or, x=11\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " 2nd factor, x+3=0
\n" ); document.write( " or, x=(-3) [this value is not possible, because length cannot have negative value]\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "so, x=11
\n" ); document.write( "length of the rectangle is 11 cm
\n" ); document.write( "therefore, width=(length -8)cm
\n" ); document.write( " width=(11-8)cm=3cm
\n" ); document.write( "Answer: Dimension of the rectangle is length=11cm, width=3cm
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