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anonymous
 4 years ago
Another problem, (geometry this time),
Find the area of a right angle triangle whose circumradius and inradius is given by \(10\) and \(4\) units receptively.
PS: This is very easy.
anonymous
 4 years ago
Another problem, (geometry this time), Find the area of a right angle triangle whose circumradius and inradius is given by \(10\) and \(4\) units receptively. PS: This is very easy.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is circumradius?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is so easy, took me 1 mint to solve and then 3 mints to confirm the solution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but note there are not so popular properties that I have used.

precal
 4 years ago
Best ResponseYou've already chosen the best response.0Is there a circle within your triangle?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes gogind, now tell me what have you used?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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: \[\ c=20=ar+br= a+b 8 \] so:\[\ a+b = 28\] the rest is easy!
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