# RT and circles

Solve right triangle if the radius of inscribed circle is r=9 and radius of circumscribed circle is R=23.

Correct result:

a =  37.83
b =  26.17
c =  46

#### Solution:

$R = \dfrac{c}{2} \ \\ c = 2 R = 46 \ \\ \ \\ r = \dfrac{ a+b-c}{2} \ \\ a + b = 64 \ \\ a^2 + b^2 = 2116 \ \\ \ \\ 2a^2 -128a +1980 =0 \ \\ \ \\ p=2; q=-128; r=1980 \ \\ D = q^2 - 4pr = 128^2 - 4\cdot 2 \cdot 1980 = 544 \ \\ D>0 \ \\ \ \\ a_{1,2} = \dfrac{ -q \pm \sqrt{ D } }{ 2p } = \dfrac{ 128 \pm \sqrt{ 544 } }{ 4 } = \dfrac{ 128 \pm 4 \sqrt{ 34 } }{ 4 } \ \\ a_{1,2} = 32 \pm 5.8309518948453 \ \\ a_{1} = 37.830951894845 \ \\ a_{2} = 26.169048105155 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2 (a -37.830951894845) (a -26.169048105155) = 0 \ \\$
$b=(-(-128) - (23.323807579381))/ (2 \cdot \ (2))=26.17$
$c=2 \cdot \ 23=46$

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