<s>[INST] finding sides of a triangle when circumradius and inradius are given 
The radius of the circumscribed circle of a right triangle is $15 cm$ and the radius of its inscribed circle is $6 cm$. Find sides of triangle.

From another site I got,  $c=30$, $a+b=2(15+6)=42$. $a+b+c=72$. $ab=6\times 72=432$. So, sides are $18$, $24$.
I didn't get how we wrote $a+b$ and $ab$ equations. What's the relation of sum and product of sides with circumscribed and inscribed radii.
 [/INST] Because radius of circumcircle is $15$, it means yes that $c$ or hypotenuse is $30$. 
Also,
$r=(a+b-30)/2$
Putting values we get,
$a+b=42$
Now we have
$a^2+b^2=900$
$a+b=42$
From which, we have $b=42-a$.
We get,
$a^2+(42-a)^2=900$
$a^2+1764-84\cdot a+a^2=900$
$2\cdot a^2-84\cdot a+864=0$
Or,  $a^2-42\cdot a+432-0$
$D=1764-1728=36$
Could you continue please? Also reject negative values.
EDITED:
$a_1=(42+6)/2=24$
and  $a_2=(42-6)/2=18$
Therefore, 
$b_1=18$ and $b_2=24$
Now you know what is the relationship between small radius and sides, and also hypotenuse and big radius. Sure you can find a general formula for relationship between sides combination and  radius, but it would be a little tricky, you should express from Pythagorean theorem sides and put in radius calculation formulas, or at least use angles formula, which you can find easily on the internet.
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