Question 68720
I DID NOT COME UP WITH THE SAME ANSWER AS YOU BUT WE ARE FAIRLY CLOSE.  MAYBE YOU CAN CHECK BOTH OUR SOLUTIONS----PTAYLOR


In a right triangle, the Pythagorean Theorem states{{{c^2=a^2+b^2)}}} where c is the hypotenuse and a and b are the respective sides.

Let x= width of rectangle

Then x+1=length of rectangle
We are told that the diagonal or{or hypotenuse of the triangle) is 4 cm.  So or equation to solve is:

{{{x^2+(x+1)^2=(4^2)}}}  clear parens
{{{x^2+x^2+2x+1=16}}} subtract 16 from both sides
{{{2x^2+2x-15=0}}} quadratic in standard form

Using the quadratic formula: {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-2 +- sqrt( 4+120 ))/(4) }}} 

{{{x = (-2 +- 11.135)/(4) }}} 


{{{x = (+9.135)/(4) }}} 

{{{x =2.2837 }}}cm -----------------------------width
{{{x+1=2.2837+1=3.2837}}}------------------------length

I think that we can discount the negative value for x


Hope this helps----ptaylor