document.write( "Question 884734: The width of a rectangle is 1 less than twice its length. If the area of the rectangle is 88 cm^2, what is the length of the diagonal? \n" ); document.write( "

Algebra.Com's Answer #534619 by lwsshak3(11628) $\"\"$ $\"About$

You can put this solution on YOUR website!
The width of a rectangle is 1 less than twice its length. If the area of the rectangle is 88 cm^2, what is the length of the diagonal?
\n" ); document.write( "***
\n" ); document.write( "let x=length of rectangle
\n" ); document.write( "2x-1=width of rectangle
\n" ); document.write( "..
\n" ); document.write( "length*width=area
\n" ); document.write( "x(2x-1)=88
\n" ); document.write( "2x^2-x=88
\n" ); document.write( "2x^2-x-88=0
\n" ); document.write( "solve for x by quadratic formula:
\n" ); document.write( " $\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"$
\n" ); document.write( "a=2, b=-1, c=-88
\n" ); document.write( "ans:
\n" ); document.write( "x=length≈6.89
\n" ); document.write( "width≈2*6.89-1≈12.78
\n" ); document.write( "diagonal≈√[(6.89)^2+(12.78)^2]≈√(47.47+163.33)≈√210.80≈14.52
\n" ); document.write( "length of the diagonal≈14.52 cm \n" ); document.write( "

\n" );