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.

Final Answer:

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

v_{1} = 4 cm v_{2} = 5 cm v_{3} = 6 cm h = \frac{1 / v _{1} + 1 / v _{2} + 1 / v _{3}}{2} = \frac{1 / 4 + 1 / 5 + 1 / 6}{2} = \frac{3 7}{1 2 0} ≐ 0.3083 t = h \cdot (h - 1 / v_{1}) \cdot (h - 1 / v_{2}) \cdot (h - 1 / v_{3}) = \frac{3 7}{1 2 0} \cdot (\frac{3 7}{1 2 0} - 1 / 4) \cdot (\frac{3 7}{1 2 0} - 1 / 5) \cdot (\frac{3 7}{1 2 0} - 1 / 6) ≐ 0.0003 s = 4 \cdot t = 4 \cdot 0.0003 ≐ 0.0665 S = \frac{1}{s} = \frac{1}{0 . 0 6 6 5} ≐ 15.0472 cm^{2} S = \frac{a \cdot v _{1}}{2} a = 2 \cdot S / v_{1} = 2 \cdot \frac{1}{0} / 4 = 7.5236 cm

Try calculation via our triangle calculator.

b = \frac{2 \cdot S}{v _{2}} = \frac{2 \cdot \frac{1}{0}}{5} = 6.0189 cm

c = \frac{2 \cdot S}{v _{3}} = \frac{2 \cdot \frac{1}{0}}{6} = 5.0157 cm

t_{1} = \frac{2 \cdot b ^{2} + 2 \cdot c ^{2} - a ^{2}}{2} = \frac{2 \cdot 6 . 0 1 8 9 ^{2} + 2 \cdot 5 . 0 1 5 7 ^{2} - 7 . 5 2 3 6 ^{2}}{2} = 4.0671 cm

t_{2} = \frac{2 \cdot a ^{2} + 2 \cdot c ^{2} - b ^{2}}{2} = \frac{2 \cdot 7 . 5 2 3 6 ^{2} + 2 \cdot 5 . 0 1 5 7 ^{2} - 6 . 0 1 8 9 ^{2}}{2} = 5.6413 cm

t_{3} = \frac{2 \cdot a ^{2} + 2 \cdot b ^{2} - c ^{2}}{2} = \frac{2 \cdot 7 . 5 2 3 6 ^{2} + 2 \cdot 6 . 0 1 8 9 ^{2} - 5 . 0 1 5 7 ^{2}}{2} = 6.3345 cm

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