Question 322749
{{{A^2+B^2=13^2}}}
{{{A+B=17}}}
From eq. 2,
{{{B=17-A}}}
Substitute into eq. 1,
{{{A^2+(17-A)^2=169}}}
{{{A^2+289-34A+A^2=169}}}
{{{2A^2-34A+120=0}}}
{{{A^2-17A+60=0}}}
{{{(A-12)(A-5)=0}}}
Two solutions:
{{{A=12}}}
Then from eq. 2,
{{{12+B=17}}}
{{{B=5}}}
.
.
.
{{{A-5=0}}}
{{{A=5}}}
Then, 
{{{5+B=17}}}
{{{B=12}}}
.
.
.
The other two sides are 5 and 12 cm long.