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Find the lengths of the sides of a triangle with 20, 28, and 35 as the lengths of altitudes. Please help...

Note by Puneet Pinku 1 year, 3 months ago

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Area = 1/2 * a * 20 = 1/2 * b * 28 = 1/2 * c * 35 20a = 28b = 35c

a = 7k b = 5k c = 4k

Since a is longest side, then altitude of 20cm falls onto side a (if angle A is obtuse, then the other altitudes could fall outside of triangle a = a₁ + a₂

a₁² + 20² = b² ----> a₁ = √(b² − 400) = √(25k² − 400) = 5 √(k² − 16) a₂² + 20² = c² ----> a₂ = √(c² − 400) = √(16k² − 400) = 4 √(k² − 25)

a = a₁ + a₂ 7k = 5 √(k² − 16) + 4 √(k² − 25) -----> square both sides 49k² = 25 (k² − 16) + 16 √(k² − 25) + 40 √(k² − 16) √(k² − 25) 49k² = 25k² − 400 + 16k² − 400 + 40 √(k⁴ − 41k² + 400) 49k² = 41k² − 800 + 40 √(k⁴ − 41k² + 400) 8k² + 800 = 40 √(k⁴ − 41k² + 400) k² + 100 = 5 √(k⁴ − 41k² + 400) -----> square both sides again k⁴ + 200k² + 10000 = 25 (k⁴ − 41k² + 400) k⁴ + 200k² + 10000 = 25k⁴ − 1025k² + 10000 24k⁴ − 1225k² = 0 k² (24k² − 1225) = 0 k² = 1225/24

k = 35/(2√6)

a = 7k = 245/(2√6) b = 5k = 175/(2√6) c = 4k = 70/√6

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TopNewestArea = 1/2 * a * 20 = 1/2 * b * 28 = 1/2 * c * 35 20a = 28b = 35c

a = 7k b = 5k c = 4k

Since a is longest side, then altitude of 20cm falls onto side a (if angle A is obtuse, then the other altitudes could fall outside of triangle a = a₁ + a₂

a₁² + 20² = b² ----> a₁ = √(b² − 400) = √(25k² − 400) = 5 √(k² − 16)

a₂² + 20² = c² ----> a₂ = √(c² − 400) = √(16k² − 400) = 4 √(k² − 25)

a = a₁ + a₂

7k = 5 √(k² − 16) + 4 √(k² − 25) -----> square both sides

49k² = 25 (k² − 16) + 16 √(k² − 25) + 40 √(k² − 16) √(k² − 25)

49k² = 25k² − 400 + 16k² − 400 + 40 √(k⁴ − 41k² + 400)

49k² = 41k² − 800 + 40 √(k⁴ − 41k² + 400)

8k² + 800 = 40 √(k⁴ − 41k² + 400)

k² + 100 = 5 √(k⁴ − 41k² + 400) -----> square both sides again

k⁴ + 200k² + 10000 = 25 (k⁴ − 41k² + 400)

k⁴ + 200k² + 10000 = 25k⁴ − 1025k² + 10000

24k⁴ − 1225k² = 0

k² (24k² − 1225) = 0

k² = 1225/24

k = 35/(2√6)

a = 7k = 245/(2√6)

b = 5k = 175/(2√6)

c = 4k = 70/√6

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