# Three altitudes

A triangle with altitudes 4; 5 and 6 cm is given. Calculate the lengths of all medians and all sides in a triangle.

Result

a =  7.524 cm
b =  6.019 cm
c =  5.016 cm
t1 =  4.067 cm
t2 =  5.642 cm
t3 =  6.335 cm

#### Solution:

$v_{1}=4 \ \text{cm} \ \\ v_{2}=5 \ \text{cm} \ \\ v_{3}=6 \ \text{cm} \ \\ \ \\ h=\dfrac{ 1/v_{1}+1/v_{2}+1/v_{3} }{ 2 }=\dfrac{ 1/4+1/5+1/6 }{ 2 } \doteq \dfrac{ 37 }{ 120 } \doteq 0.3083 \ \\ t=h \cdot \ (h-1/v_{1}) \cdot \ (h-1/v_{2}) \cdot \ (h-1/v_{3})=0.3083 \cdot \ (0.3083-1/4) \cdot \ (0.3083-1/5) \cdot \ (0.3083-1/6) \doteq 0.0003 \ \\ \ \\ s=4 \cdot \ \sqrt{ t }=4 \cdot \ \sqrt{ 0.0003 } \doteq 0.0665 \ \\ S=\dfrac{ 1 }{ s }=\dfrac{ 1 }{ 0.0665 } \doteq 15.0472 \ \text{cm}^2 \ \\ \ \\ S=\dfrac{ a \cdot \ v_{1} }{ 2 } \ \\ \ \\ a=2 \cdot \ S/v_{1}=2 \cdot \ 15.0472/4 \doteq 7.5236 \doteq 7.524 \ \text{cm}$

Try calculation via our triangle calculator.

$b=\dfrac{ 2 \cdot \ S }{ v_{2} }=\dfrac{ 2 \cdot \ 15.0472 }{ 5 } \doteq 6.0189 \doteq 6.019 \ \text{cm}$
$c=\dfrac{ 2 \cdot \ S }{ v_{3} }=\dfrac{ 2 \cdot \ 15.0472 }{ 6 } \doteq 5.0157 \doteq 5.016 \ \text{cm}$
$t_{1}=\dfrac{ \sqrt{ 2 \cdot \ b^2+2 \cdot \ c^2-a^2 } }{ 2 }=\dfrac{ \sqrt{ 2 \cdot \ 6.0189^2+2 \cdot \ 5.0157^2-7.5236^2 } }{ 2 } \doteq 4.0671 \doteq 4.067 \ \text{cm}$
$t_{2}=\dfrac{ \sqrt{ 2 \cdot \ a^2+2 \cdot \ c^2-b^2 } }{ 2 }=\dfrac{ \sqrt{ 2 \cdot \ 7.5236^2+2 \cdot \ 5.0157^2-6.0189^2 } }{ 2 } \doteq 5.6417 \doteq 5.642 \ \text{cm}$
$t_{3}=\dfrac{ \sqrt{ 2 \cdot \ a^2+2 \cdot \ b^2-c^2 } }{ 2 }=\dfrac{ \sqrt{ 2 \cdot \ 7.5236^2+2 \cdot \ 6.0189^2-5.0157^2 } }{ 2 } \doteq 6.3348 \doteq 6.335 \ \text{cm}$

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