Question 945714
It sounds like the pavement is the
same width all the way around
Let {{{ x }}} = the width of the pavement
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{{{ 16*15 = 240 }}} ft2 is the outside area
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{{{ ( 16 - x )*( 15 - x ) = 132 }}}
{{{ 240 - 15x - 16x + x^2 = 132 }}}
{{{ x^2  - 31x + 108 = 0 }}}
Use the quadratic equation
{{{ x = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = -31 }}}
{{{ c = 108 }}}
{{{ x = ( -(-31) +- sqrt( (-31)^2 - 4*1*108 )) / (2*1) }}}
{{{ x = ( 31 +- sqrt( 961 - 432 )) / 2 }}}
{{{ x = ( 31 +- sqrt( 529 )) / 2 }}}
{{{ x = ( 31 - 23 ) / 2 }}}
{{{ x = 8/2 }}}
{{{ x = 4 }}} ( the positive square root makes no sense )
The width of the pavement is 4 ft
check:
{{{ ( 16 - x )*( 15 - x ) = 132 }}}
{{{ ( 16 - 4 )*( 15 - 4 ) = 132 }}}
{{{ 12*11 = 132 }}}
{{{ 132 = 132 }}}
OK