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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.

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  1. anonymous
    • 4 years ago
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    What is circumradius?

  2. anonymous
    • 4 years ago
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    This is so easy, took me 1 mint to solve and then 3 mints to confirm the solution.

  3. anonymous
    • 4 years ago
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    but note there are not so popular properties that I have used.

  4. precal
    • 4 years ago
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    Is there a circle within your triangle?

  5. anonymous
    • 4 years ago
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    I got 96

  6. anonymous
    • 4 years ago
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    Yes gogind, now tell me what have you used?

  7. anonymous
    • 4 years ago
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    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!

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