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.
 A: 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.
A: 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
A: Cracked by Adarsh (8th grade)
Step by step process: 
1) For a Right Angled Triangle, 
if Circumradius (R) = 15 then 
Hypotenuse (c) = 2*R = 2*15=30 CM
2) For a Right Angled Triangle, 
Inradius (r) = (a+b-c)/2 
==> 6 = (a+b-30)/2 
==> a+b=42   
3) Area = s*r = (a+b+c)*r/2= (a+b+30)*6/2 = (a+b+30)*3 = (42+30)*3 = 216 sq.Cm
4) Area = ab/2 
==> ab = 2* Area = 2 *216 = 432 Sq.Cm
5) a^2 + b^2 = c^2 
==> a^2 + b^2 = 900 
6) (a+b)^2 = a^2 + b^2 + 2a*b = 900 + (2*432) = 1764 = 42^2 
==> a+b=42
7) (a-b)^2 = a^2 + b^2 - 2a*b = 900 - (2*432) = 36 = 6^2 
==> a-b=6
8) solving 6 and 7, 
a = 24 and b = 18 
c = 30 (ref step 1)
