Question 1187047
<br>
Let x be the length of the longer side
Let y be the length of the shorter side<br>
The length of the diagonal is {{{sqrt(x^2+y^2)}}}<br>
The length is also equal to the semiperimeter minus 3/4 the length of the shorter side: {{{(x+y)-(3/4)y=x+y/4}}}<br>
So<br>
{{{sqrt(x^2+y^2)=x+y/4}}}
{{{4sqrt(x^2+y^2)=4x+y}}}
{{{16(x^2+y^2)=16x^2+8xy+y^2}}}
{{{16x^2+16y^2=16x^2+8xy+y^2}}}
{{{16y^2=8xy+y^2}}}
{{{15y^2-8xy=0}}}
{{{y(15y-8x)=0}}}<br>
{{{y=0}}}  or  {{{15y-8x=0}}}<br>
y=0 makes no sense in the problem, so 15y-8x=0.<br>
With the requirement that x and y be integers, the smallest solution is with x=15 and y=8.<br>
ANSWER: The minimum length of the longer side is x=15.<br>
CHECK:
{{{sqrt(x^2+y^2)=sqrt(15^2+8^2)=sqrt(225+64)=sqrt(289)=17}}}<br>
{{{(x+y)-(3/4)y = 23-6 = 17}}}<br>