Question 279050
gives me the following information:
:
Start by using L for Length, W for width
:
Perimeter of the Rectangle: 68cm
2L + 2W = 68
Simplify, divide by 2
L + W = 34
or
W = (34-L)
:
Area of the Rectangle: 144sq. cm
L * W = 144
Substitute (34-L) for W
L * (34-L) = 144
34L - L^2 = 144
Arrange as a quadratic equation
-L^2 + 34L - 144 = 0
We have to solve this using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
In this equation: x=L; a=-1; b=34; c=-144
{{{L = (-34 +- sqrt(34^2-4*-1*-144 ))/(2*-1) }}} 
:
{{{L = (-34 +- sqrt(1156-576 ))/(-2) }}}
:
{{{L = (-34 +- sqrt(580 ))/(-2) }}}
Two solutions
{{{L = (-34 + 24.083)/(-2) }}}
L = {{{(-9.917)/(-2)}}}
L = +4.958
and 
{{{L = (-34 - 24.083)/(-2) }}}
L = {{{(-58.083)/(-2)}}}
L = +29.0415
:
the larger 29.0415 cm is the length
the smaller 4.958 cm is the width
:
Check our solution by finding the perimeter with a calc
2(29.0415) + 2(4.958) = 67.999 ~ 68
:
and its asking for the length of the diagonal, now we have the dimensions:
Use pythag for this: c = {{{sqrt(a^2 + b^2)}}},
: 
c = {{{sqrt(29.0415^2 + 4.958^2)}}}
c = 29.462 is the diagonal