## anonymous 4 years ago 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.

1. anonymous

2. anonymous

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

3. anonymous

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

4. precal

Is there a circle within your triangle?

5. anonymous

I got 96

6. anonymous

Yes gogind, now tell me what have you used?

7. anonymous

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!