Question 117871
If you create a diagonal in a rectangle, then you have created a right triangle with the diagonal as a hypotenuse.  Let's say the short leg of the triangle is x, and then the long leg (the long side of the rectangle) has to be x + 1.  And we are given that the hypotenuse is 4 cm.

Pythagoras tells us that the sides and hypotenuse of a right triangle are related thus:


{{{a^2+b^2=c^2}}}, or {{{sqrt(a^2+b^2)=c}}}.


Replace a, b, and c with the values we are given:


{{{sqrt(x^2+(x+1)^2)=4}}}


Expand the binomial and combine terms under the radical


{{{sqrt(x^2+x^2+x+x+1)=4}}}
{{{sqrt(2x^2+2x+1)=4}}}


Square both sides:


{{{2x^2+2x+4=16}}}


Add -16 to both sides:


{{{2x^2+2x-15=0}}}


Use the quadratic formula

{{{x = (-2 +- sqrt( 2^2-4*2*(-15)))/(2*2) }}} 
{{{x = (-2 +- sqrt( 124))/4 }}} 
{{{x = (-2 +- 2sqrt( 31))/4 }}}, so

 
{{{x[1] = (-1 + sqrt( 31))/2 }}}, or
{{{x[2] = (-1 - sqrt(31))/2}}}.


{{{x[2] < 0}}} so we can exclude this answer because we are looking for a positive number length.


Since {{{sqrt(31)>5}}}, {{{x[1]>0}}}, so {{{x[1] = (-1 + sqrt( 31))/2 }}} is the answer we are seeking and is the width of the rectangle.  The length of the rectangle is {{{x+1}}}, so the length must be {{{((-1 + sqrt( 31))/2)+1 =(1 + sqrt( 31))/2)}}} 


Hope this helps,
Happy Holidays,
John