Question

vẽ tam giác ABC vuông tại A,đường cao AH. Hãy tính AB,AC,BC,CH trong các trường hợp sau

1.BH=3cm; AH=4cm;

Answer 1

\(AB=\sqrt{BH^2+AH^2}=\sqrt{3^2+4^2}=\sqrt{25}=5\left(cm\right)\)

\(\dfrac{1}{AH^2}=\dfrac{1}{BA^2}+\dfrac{1}{AC^2}\Leftrightarrow\dfrac{1}{4^2}=\dfrac{1}{5^2}+\dfrac{1}{AC^2}\Rightarrow\dfrac{1}{AC^2}=\dfrac{9}{400}\Rightarrow AC^2=\dfrac{400}{9}\Leftrightarrow AC=\dfrac{20}{3}\left(cm\right)\)

\(BC=\sqrt{AB^2+AC^2}=\sqrt{5^2+\left(\dfrac{20}{3}\right)^2}=\sqrt{\dfrac{625}{9}}=\dfrac{25}{3}\left(cm\right)\)

\(AC^2=HC.BC\Leftrightarrow\left(\dfrac{20}{3}\right)^2=HC.\dfrac{25}{3}\Rightarrow HC=\dfrac{16}{3}\left(cm\right)\)