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.
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.
In radius=(a+b-c)/2
a+b=42...........1
a²+b²=30²(in right angle triangle).............2
Square on both side in equation 1so
ab=432
a=42-b
Put the value of a in equation 1
(42-b)b=432
b²-42b-432=0
(B-18)(b-24)=0
Then 
Sides are 18 and 24