Question 829223
That is lenth and width are {{{(1/3)}}} larger, not smaller, a factor of {{{4/3}}}.


Size of the enlarged copy:
{{{(1&5/8)*(4/3)*(1&3/4)*(4/3)}}} {{{f^2}}}, assuming you want size as AREA.
'
{{{(13/8)(7/4)(4/3)^2}}}
{{{(13*7)/(2*9)}}}
{{{91/18}}}
{{{highlight(highlight(5&1/18* ft^2))}}}