Question 47741
The length of a rectangle is 1 cm longer than its width.
LET WIDTH =W
LENGTH=W+1

 If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?
DIAGONAL^2=L^2+W^2
4^2=(W+1)^2+W^2
W^2+2W+1+W^2=16
2W^2+2W-15=0
{{{w = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{w = (-2 +- sqrt( 2^2-4*2*(-15) ))/(2*2) }}}
{{{w = (-2 +- sqrt( 4+120 ))/(4) }}}
{{{w = (-2 +2*sqrt( 31 ))/(4) }}}=2.28
W=2.28 AND L=3.28
Thank You!