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.

Correct result:

a =  10.7703
b =  8.9443
c =  7.2111
t₁ =  6.0828
t₂ =  8
t₃ =  9.2195

Solution:

$$\begin{gathered} a=\mathrm{\mid }BC\mathrm{\mid } \\ a=\sqrt{(-4-6{)}^2+(3-(-1){)}^2}=2\sqrt{29}=10.7703 \end{gathered}$$

$$\begin{gathered} b=\mathrm{\mid }AC\mathrm{\mid } \\ b=\sqrt{(2-6{)}^2+(7-(-1){)}^2}=4\sqrt{5}=8.9443 \end{gathered}$$

$$\begin{gathered} c=\mathrm{\mid }AB\mathrm{\mid } \\ c=\sqrt{(2-(-4){)}^2+(7-3{)}^2}=2\sqrt{13}=7.2111 \end{gathered}$$

Try calculation via our triangle calculator.

$t_1=\sqrt{2\cdot b^2+2\cdot c^2-a^2}\mathrm{/}2=\sqrt{2\cdot 8.9443^2+2\cdot 7.2111^2-10.7703^2}\mathrm{/}2=\sqrt{37}=6.0828$

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

$t_3=\sqrt{2\cdot b^2+2\cdot a^2-c^2}\mathrm{/}2=\sqrt{2\cdot 8.9443^2+2\cdot 10.7703^2-7.2111^2}\mathrm{/}2=\sqrt{85}=9.2195$

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