Question 125960
{{{A= a^2}}}

{{{A1= A-36 = a^2 -36}}}…..(1)

{{{A1= (a-2)^2}}}………..(2)

From (1) and (2), we have:

{{{ a^2 -36 = (a-2)^2 }}}

{{{ a^2 =  a^2-4a + 4 + 36}}}

{{{ cross(a^2) =  cross(a^2) -4a  + 40}}}

{{{ 0 =  -4a  + 40}}}

{{{ 4a  = 40}}}

{{{ highlight(a  = 10)}}}............the dimensions of the original square

Check:

{{{A= a^2}}}
{{{A= 10^2}}}
{{{A= 100}}}

{{{A1= (10-2)^2 }}}
{{{A1= (8)^2 }}}
{{{A1= 64 }}}
	

{{{A-36=A1}}}

{{{100-36=64}}}