Medians and sides

Triangle ABC in the plane Oxy; are the coordinates of the points:
A = 2.7
B = -4.3
C-6-1

Try calculate lengths of all medians and all sides.

Result

a =  10.77
b =  8.944
c =  7.211
t₁ =  6.083
t₂ =  8
t₃ =  9.219

Solution:

$a=|BC| \ \\ a=\sqrt{ (-4-6)^2+(3-(-1))^2 }=2 \ \sqrt{ 29 } \doteq 10.7703 \doteq 10.77$

$b=|AC| \ \\ b=\sqrt{ (2-6)^2+(7-(-1))^2 }=4 \ \sqrt{ 5 } \doteq 8.9443 \doteq 8.944$

$c=|AB| \ \\ c=\sqrt{ (2-(-4))^2+(7-3)^2 }=2 \ \sqrt{ 13 } \doteq 7.2111 \doteq 7.211$

Try calculation via our triangle calculator.

$t_{1}=\sqrt{ 2 \cdot \ b^{ 2 }+2 \cdot \ c^{ 2 } - a^{ 2 } }/2=\sqrt{ 2 \cdot \ 8.9443^{ 2 }+2 \cdot \ 7.2111^{ 2 } - 10.7703^{ 2 } }/2 \doteq 6.0826 \doteq 6.083$

$t_{2}=\sqrt{ 2 \cdot \ c^{ 2 }+2 \cdot \ a^{ 2 } - b^{ 2 } }/2=\sqrt{ 2 \cdot \ 7.2111^{ 2 }+2 \cdot \ 10.7703^{ 2 } - 8.9443^{ 2 } }/2 \doteq 7.9998 \doteq 8$

$t_{3}=\sqrt{ 2 \cdot \ b^{ 2 }+2 \cdot \ a^{ 2 } - c^{ 2 } }/2=\sqrt{ 2 \cdot \ 8.9443^{ 2 }+2 \cdot \ 10.7703^{ 2 } - 7.2111^{ 2 } }/2 \doteq 9.2192 \doteq 9.219$

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