Triangle hypotenuse circle

In a right-angled triangle ABC with a right angle at the vertex C, the magnitudes of the hypotenuses are given ta=5, tb=2√10.
Calculate the side sizes of triangle ABC and the circle's radius described by this triangle.

Final Answer:

a =  6
b =  4
c =  7.2111
R =  3.6056

Step-by-step explanation:

$$\begin{gathered} t_1=5 \\ {t}_2=2\cdot \sqrt{10}=2\sqrt{10}\doteq 6.3246 \\ {t}_1^2=(a/2{)}^2+b^2 \\ {t}_2^2=(b/2{)}^2+a^2 \\ {t}_1^2=(t_2^2-(b/2{)}^2)/4+b^2 \\ 4t_1^2=t_2^2-(b/2{)}^2+4b^2 \\ 4t_1^2-t_2^2=4b^2-b^2/4 \\ 16t_1^2-4t_2^2=16b^2-b^2 \\ 16t_1^2-4t_2^2=15b^2 \\ b=\sqrt{\frac{16\cdot t_1^2-4\cdot t_2^2}{15}}=\sqrt{\frac{16\cdot 5^2-4\cdot 6.3246^2}{15}}=4 \\ a=\sqrt{t_2^2-(b/2{)}^2}=\sqrt{6.3246^2-(4/2{)}^2}=6 \end{gathered}$$

$c=\sqrt{a^2+b^2}=\sqrt{6^2+4^2}=2\sqrt{13}=7.2111$

$R=c/2=7.2111/2=\sqrt{13}=3.6056$

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