Another problem, (geometry this time),

Find the area of a right angle triangle whose circumradius and in-radius is given by $10$ and $4$ units receptively.

PS: This is very easy.

Question

Another problem, (geometry this time),

Find the area of a right angle triangle whose circumradius and in-radius is given by $10$ and $4$ units receptively.

PS: This is very easy.

anonymous · Answer

What is circumradius?

anonymous · Answer

This is so easy, took me 1 mint to solve and then 3 mints to confirm the solution.

anonymous · Answer

but note there are not so popular properties that I have used.

precal · Answer

Is there a circle within your triangle?

anonymous · Answer

I got 96

anonymous · Answer

Yes gogind, now tell me what have you used?

anonymous · Answer

I sliced the triangle up in 3 parts with lines that divide the angles in 2 equal parts. So the total area of the triangle has to be  the sum of the areas of these 3 parts. The height of each of those triangles is the radius of the incircle
So: 
$$\ \frac{1}{2}ra +\frac{1}{2}rb+\frac{1}{2}rc = \frac{1}{2}ab=P$$

The Hypotenuse of the triangle is just 2R (2*10) = 20, and I deduced(I draw a picture) that it has to be equal to:
$$\ c=20=a-r+b-r= a+b -8 $$ so:$$\ a+b = 28$$

the rest is easy!