Xét tam giác ABC vuông A, đcao AH:
`@`\(AC=\sqrt{\dfrac{AB^2.AH^2}{AB^2-AH^2}}=\sqrt{\dfrac{5^2.4^2}{5^2-4^2}}=\dfrac{20}{3}\left(cm\right)\)
`@`\(BC=\sqrt{AB^2+AC^2}=\sqrt{5^2+\left(\dfrac{20}{3}\right)^2}=\dfrac{25}{3}\left(cm\right)\)
`@`\(CH=\dfrac{AC^2}{BC}=\dfrac{\left(\dfrac{20}{3}\right)^2}{\dfrac{25}{3}}=\dfrac{16}{3}\left(cm\right)\)
`@`\(HB=BC-CH=\dfrac{25}{3}-\dfrac{16}{3}=3\left(cm\right)\)
`AH` là đường cao hay sao vậy bạn?
Áp dụng Pytago: `BH = sqrt(AB^2 - AH^2) = sqrt(5^2 - 4^2) = sqrt 9 = 3 cm`.
`@AH^2 = BH . CH <=> 4^2 = 3 . CH -> CH = 16/3 cm`.
`@BC = BH + CH = 16/3 + 3 = 25/3 cm`.
Áp dụng Pytago: `AC= sqrt(BC^2 - AB^2) = sqrt(625/9 - 25) = sqrt(400/9) = 20/3 cm`
