Question 1179290
you actualy need one of pythagorean triples

Pythagorean Triples if it satisfies the equation,

{{{a^2+b^2=c^2}}}

List of Primitive Pythagorean Triples:
({{{3}}},{{{ 4}}}, {{{5}}})
({{{5}}},{{{12}}}, {{{13}}})
({{{7}}}, {{{24}}}, {{{25}}})
({{{8}}}, {{{15}}}, {{{17}}})
({{{9}}}, {{{40}}},{{{ 41}}})
({{{11}}}, {{{60}}}, {{{61}}})=> first one that you can use, or one of the following triples
({{{12}}}, {{{35}}},{{{ 37}}})
({{{13}}},{{{ 84}}},{{{ 85}}})

so, let {{{a=12}}} and {{{b=35}}} ( each is greater than {{{10}}})

then  plug it in

{{{a^2+b^2=c^2}}}
{{{12^2+35^2=c^2}}}
{{{1369=c^2}}}
{{{c=sqrt(1369)}}}
{{{c=37}}} 

it confirms we have triples ({{{12}}}, {{{35}}},{{{ 37}}})