Question 102565
{{{2*2.5 = 5}}}
If the exposed area is {{{4}}}, then the area of the
frame is {{{1}}}
Call the width of the frame {{{x}}}
Then {{{2.5x + 2.5x + x(2 - 2x) + x(2 - 2x) = 1}}}
{{{5x + 2x(2 - 2x) = 1}}}
{{{5x + 4x - 4x^2 = 1}}}
{{{4x^2 - 9x + 1 = 0}}}
This is in the form
{{{ax^2 + bx + c = 0}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-(-9) +- sqrt( (-9)^2-4*4*1 ))/(2*4) }}}
{{{x = (9 +- sqrt(81 - 16)) / 8}}}
{{{x = (9 +- sqrt(65)) / 8}}}
{{{x = (9 +- 8.062) / 8}}}
{{{x = (9 - 8.062) / 8}}} gives the only reasonable answer
{{{x = .117}}} m, or
{{{x = 11.7}}} cm answer