Medians in right triangle

It is given a right triangle, angle C is 90 degrees. I know it medians t1 = 8 cm and median t2 = 12 cm. .. How to calculate the length of the sides?

Result

a =  11.685 cm
b =  5.465 cm
c =  12.9 cm

Solution:

t1=8 cm t2=12 cm  t12=x2+(2y)2 t22=y2+(2x)2  x2=t124y2  y=4 t12t2215=4 82122152.7325 cm x=t124 y2=824 2.732525.8424 cm  a=2 x=2 5.842411.6847=11.685  cm t_{ 1 } = 8 \ cm \ \\ t_{ 2 } = 12 \ cm \ \\ \ \\ t_{ 1 }^2 = x^2 + (2y)^2 \ \\ t_{ 2 }^2 = y^2 + (2x)^2 \ \\ \ \\ x^2 = t_{ 1 }^2-4y^2 \ \\ \ \\ y = \sqrt{ \dfrac{ 4 \cdot \ t_{ 1 }^2-t_{ 2 }^2 }{ 15 } } = \sqrt{ \dfrac{ 4 \cdot \ 8^2-12^2 }{ 15 } } \doteq 2.7325 \ cm \ \\ x = \sqrt{ t_{ 1 }^2 - 4 \cdot \ y^2 } = \sqrt{ 8^2 - 4 \cdot \ 2.7325^2 } \doteq 5.8424 \ cm \ \\ \ \\ a = 2 \cdot \ x = 2 \cdot \ 5.8424 \doteq 11.6847 = 11.685 \ \text { cm }
b=2 y=2 2.73255.465=5.465  cm b = 2 \cdot \ y = 2 \cdot \ 2.7325 \doteq 5.465 = 5.465 \ \text { cm }
a2+b2=c2  c=a2+b2=11.68472+5.465212.8998=12.9  cm a^2+b^2 = c^2 \ \\ \ \\ c = \sqrt{ a^2 + b^2 } = \sqrt{ 11.6847^2 + 5.465^2 } \doteq 12.8998 = 12.9 \ \text { cm }

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