document.write( "Question 238288: The length of a rectangle is 4 centimeters more than the width. the measure of the diagonal is 10 centimeters. Find the dimensions of the the rectangle\r
\n" ); document.write( "\n" ); document.write( "I would appreciate help =] \n" ); document.write( "
Algebra.Com's Answer #175130 by checkley77(12844)\"\" \"About 
You can put this solution on YOUR website!
L=W+4
\n" ); document.write( "L^2+W^2=10^2
\n" ); document.write( "(W+4)^2+W^2=100
\n" ); document.write( "W^2+8W+16+W^2=100
\n" ); document.write( "2W^2+8W+16-100=0
\n" ); document.write( "2W^2+8W-84=0
\n" ); document.write( "2(W^2+4W-42)=0
\n" ); document.write( "\"W+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"
\n" ); document.write( "W=(-4+-SQRT[4^2-4*1*-42])/2*1
\n" ); document.write( "W=(-4+-SQRT[16+168])/2
\n" ); document.write( "W=(-4+-SQRT184)/2
\n" ); document.write( "W=(-4+-13.565)/2
\n" ); document.write( "W=(-4+13.565)/2
\n" ); document.write( "W=9.565/2
\n" ); document.write( "W=4.78 ANS. FOR THE WIDTH
\n" ); document.write( "L=4.78+4=8.78 ANS. FOR THE LENGTH.
\n" ); document.write( "PROOF:
\n" ); document.write( "4.78^2+8.78^2=100
\n" ); document.write( "22.85+77.09=100
\n" ); document.write( "100~100\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );