Question

cho tam giác ABC vuông tại A, đường cao AH,biết AB=24cm,\(\dfrac{HB}{HC}\)=\(\dfrac{9}{16}\).Tính AC,BC,AH

Answer 1

Ta có: \(\dfrac{HB}{HC}=\dfrac{9}{16}\Rightarrow HB=\dfrac{9}{16}HC\)

Ta có: \(AB^2=BH.BC=BH\left(BH+HC\right)=\dfrac{9}{16}HC\left(\dfrac{9}{16}HC+HC\right)\)

\(=\dfrac{9}{16}HC.\dfrac{25}{16}HC=\dfrac{225}{256}HC^2\)

\(\Rightarrow HC^2=\dfrac{256AB^2}{225}=\dfrac{16384}{25}\Rightarrow HC=\dfrac{128}{5}\left(cm\right)\)

\(\Rightarrow HB=\dfrac{72}{5}\Rightarrow BC=\dfrac{128+72}{5}=40\left(cm\right)\)

\(\Rightarrow AC=\sqrt{BC ^2-AB^2}=\sqrt{40^2-24^2}=32\)

Ta có: \(AB.AC=AH.BC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{24.32}{40}=\dfrac{96}{5}\left(cm\right)\)

Answer 2

\(\dfrac{HB}{HC}=\dfrac{9}{16}\Rightarrow HC=\dfrac{16}{9}HB\)

Áp dụng hệ thức lượng:

\(AB^2=HB.BC=HB\left(HB+HC\right)\)

\(\Leftrightarrow24^2=HB.\left(HB+\dfrac{16}{9}HB\right)\)

\(\Rightarrow HB^2=\dfrac{5184}{25}\Rightarrow HB=\dfrac{72}{5}\left(cm\right)\)

\(HC=\dfrac{16}{9}HB=\dfrac{128}{5}\) (cm)

\(BC=HB+HC=40\) (cm)

\(AC=\sqrt{BC^2-AB^2}=32\) (cm)

\(AH=\dfrac{AB.AC}{BC}=\dfrac{96}{5}\left(cm\right)\)