We can try to see if there is a rectangle which meets this criteria.
The Area of the Rectangle would be: #l xx w = 26#
The Perimeter of the Rectangle would be: #2l + 2w = 26#
First, we could solve the Perimeter formula for #l#:
#2l + 2w - color(red)(2w) = 26 - color(red)(2w)#
#2l + 0 = 26 - 2w#
#2l = 26 - 2w#
#(2l)color(red)(2) = (26 - 2w)/color(red)(2)#
#l = 13 - w#
Next, we can substitute #(13 -w)# into the formula for Area and solve for #w#
#l xx w = 26# becomes
#(13 - w) xx w = 26#
#13w - w^2 = 26#
#-w^2 + 13w = 26#
#-w^2 + 13w - color(red)(26) = 26 - color(red)(26)#
#-w^2 + 13w - 26 = 0#
#-1(-w^2 + 13w - 26) = -1 xx 0#
#w^2 - 13w + 26 = 0#
We can now use the quadratic equation to solve this problem:
The quadratic formula states:
For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:
#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#
Substituting:
#color(red)(1)# for #color(red)(a)#
#color(blue)(-13)# for #color(blue)(b)#
#color(green)(26)# for #color(green)(c)# gives:
#w = (-color(blue)(-13) +- sqrt(color(blue)(-13)^2 - (4 * color(red)(1) * color(green)(26))))/(2 * color(red)(1))#
#w = (13 +- sqrt(169 - 104))/2#
#w = (13 +- sqrt(65))/2#
So we can have a rectangle where:
The Length Is: #(13 - sqrt(65))/2#
The Width Is: #(13 + sqrt(65))/2#
Then the Perimeter would be 26 and the Area would be 26
