<PRE><B>Question 273089</B>
The area of a rectangle is 12sq. units and the diagonal of the rectangle has a length of 2 then the square root and underneath that is a 10.
 Find the perimeter of the rectangle?
:
Area = 12 sq/units; Diagonal = {{{2sqrt(10)}}}
:
Find legs of the right triangle; a^2 + b^2 = c^2
{{{sqrt(L^2 + W^2)}}} = {{{2sqrt(10)}}}
{{{sqrt(L^2 + W^2)}}} = {{{sqrt(4*10)}}}
{{{sqrt(L^2 + W^2)}}} = {{{sqrt(40)}}}
 square both sides
L^2 + W^2 = 40
:
The area equation
L*W = 12
L = {{{12/W}}}; for substitution below
:
{{{(12/W)^2}}} + W^2 = 40
{{{(144/W^2)}}} + W^2 = 40
multiply equation by W^2
144 + W^4 = 40W^2
:
A quadratic equation
W^4 - 40W^2 + 144 = 0
:
this will factor
(W^2 - 36)(W^2 - 4) = 0
:
Two solutions
W^2 = 36
W = 6, call this one the length
and
W^2 = 4
W = 2,  the width
:
P = 2(6) + 2(2) 
P = 12 + 4
P = 16 units is the perimeter
:
A = 6*2 = 12 confirm using the area</PRE>