Question

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

1)AC/AB=2cm;AC=4cm

2)AC/AB=√3cm;AC=2cm

3)AB=2√3cm;AB/AC=1cm

Answer 1

1: AB=AC/2=2cm

\(BC=\sqrt{2^2+4^2}=2\sqrt{5}\left(cm\right)\)

\(AH=\dfrac{AB\cdot AC}{BC}=\dfrac{8}{2\sqrt{5}}=\dfrac{4\sqrt{5}}{5}\left(cm\right)\)

\(BH=\dfrac{AB^2}{BC}=\dfrac{2^2}{2\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\left(cm\right)\)

\(CH=BC-BH=2\sqrt{5}-\dfrac{1}{5}\cdot2\sqrt{5}=\dfrac{8}{5}\sqrt{5}\left(cm\right)\)

2: \(AB=\dfrac{AC}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\left(cm\right)\)

\(BC=\sqrt{AB^2+AC^2}=\dfrac{4\sqrt{3}}{3}\left(cm\right)\)

\(AH=\dfrac{AB\cdot AC}{BC}=\sqrt{3}\left(cm\right)\)

\(BH=\dfrac{AB^2}{BC}=\dfrac{\sqrt{3}}{3}\left(cm\right)\)