Question 1127994
given:

{{{ABC}}} is right triangle
{{{AB}}} is longer than side {{{BC}}}, means {{{AB >BC}}}

Boxed numbers are: 
{{{7}}},{{{ 8}}}
{{{15}}},{{{ 17}}}
{{{18}}}, {{{20 }}}
{{{24}}}, {{{25}}} 

If  triangle {{{ABC}}} is right triangle, we will use Pythagorean theorem {{{a^2 + b^2 = c^2}}}.

Before we try out all possible combinations,  we can simply use Pythagorean triples and see which ones fit your boxed numbers.

here are first few Pythagorean triples to see where you can find given sides
{{{3}}},{{{4}}},{{{5}}}
{{{5}}},{{{12}}},{{{13}}}
{{{7}}},{{{24}}},{{{25}}}-> this one is your {{{solution}}}
{{{8}}},{{{15}}},{{{17}}}-> and this one is your {{{solution}}}
{{{9}}},{{{40}}},{{{41 }}}

so, you have 
{{{7}}},{{{24}}},{{{25}}} where


{{{AB=c=25}}} 
{{{BC=a=24}}}
{{{AC=b=7}}} 

check:
{{{24^2 + 7^2 = 25^2}}}
{{{576 +49 = 625}}}
{{{625 = 625}}}

and {{{8}}},{{{15}}},{{{17}}}->

{{{AB=c=17}}} 
{{{BC=a=15}}}
{{{AC=b=8}}} 

check:
{{{15^2 + 8^2 = 17^2}}}
{{{225 +64 = 289}}}
{{{289 = 289}}}