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?

Correct 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 5.8424=11.685 cmt_{1}=8 \ \text{cm} \ \\ t_{2}=12 \ \text{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 \ \text{cm} \ \\ x=\sqrt{ t_{1}^2 - 4 \cdot \ y^2 }=\sqrt{ 8^2 - 4 \cdot \ 2.7325^2 } \doteq 5.8424 \ \text{cm} \ \\ \ \\ a=2 \cdot \ x=2 \cdot \ 5.8424=11.685 \ \text{cm}
b=2 y=2 2.7325=5.465 cmb=2 \cdot \ y=2 \cdot \ 2.7325=5.465 \ \text{cm}
a2+b2=c2  c=a2+b2=11.68472+5.4652=12.9 cma^2+b^2=c^2 \ \\ \ \\ c=\sqrt{ a^2 + b^2 }=\sqrt{ 11.6847^2 + 5.465^2 }=12.9 \ \text{cm}

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