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.

Correct result:

a =  7.5236 cm
b =  6.0189 cm
c =  5.0157 cm
t₁ =  4.0671 cm
t₂ =  5.6413 cm
t₃ =  6.3345 cm

Solution:

$$\begin{gathered} v_1=4\text{ cm } \\ {v}_2=5\text{ cm } \\ {v}_3=6\text{ cm } \\ h={\displaystyle \frac{1\mathrm{/}{v}_1+1\mathrm{/}{v}_2+1\mathrm{/}{v}_3}{2}}={\displaystyle \frac{1\mathrm{/}4+1\mathrm{/}5+1\mathrm{/}6}{2}}={\displaystyle \frac{37}{120}}\doteq 0.3083 \\ t=h\cdot (h-1\mathrm{/}{v}_1)\cdot (h-1\mathrm{/}{v}_2)\cdot (h-1\mathrm{/}{v}_3)=0.3083\cdot (0.3083-1\mathrm{/}4)\cdot (0.3083-1\mathrm{/}5)\cdot (0.3083-1\mathrm{/}6)\doteq 0.0003 \\ s=4\cdot \sqrt{t}=4\cdot \sqrt{0.0003}\doteq 0.0665 \\ S={\displaystyle \frac{1}{s}}={\displaystyle \frac{1}{0.0665}}\doteq 15.0472{\text{cm}}^2 \\ S={\displaystyle \frac{a\cdot v_1}{2}} \\ a=2\cdot S\mathrm{/}{v}_1=2\cdot 15.0472\mathrm{/}4=7.5236\text{ cm} \end{gathered}$$

Try calculation via our triangle calculator.

$b={\displaystyle \frac{2\cdot S}{v_2}}={\displaystyle \frac{2\cdot 15.0472}{5}}=6.0189\text{ cm}$

$c={\displaystyle \frac{2\cdot S}{v_3}}={\displaystyle \frac{2\cdot 15.0472}{6}}=5.0157\text{ cm}$

$t_1={\displaystyle \frac{\sqrt{2\cdot b^2+2\cdot c^2-a^2}}{2}}={\displaystyle \frac{\sqrt{2\cdot 6.0189^2+2\cdot 5.0157^2-7.5236^2}}{2}}=4.0671\text{ cm}$

$t_2={\displaystyle \frac{\sqrt{2\cdot a^2+2\cdot c^2-b^2}}{2}}={\displaystyle \frac{\sqrt{2\cdot 7.5236^2+2\cdot 5.0157^2-6.0189^2}}{2}}=5.6413\text{ cm}$

$t_3={\displaystyle \frac{\sqrt{2\cdot a^2+2\cdot b^2-c^2}}{2}}={\displaystyle \frac{\sqrt{2\cdot 7.5236^2+2\cdot 6.0189^2-5.0157^2}}{2}}=6.3345\text{ cm}$

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