Question

Cho tam giác \(ABC\) vuông tại \(A\left(AB< AC\right)\), đường cao \(AH.\)

\(a,\) Tính \(AB,AC\) nếu \(AH=6cm;HC-HB=9cm\)

\(b,\) Tính \(AB,AC\) nếu \(AH=6cm;S_{ABC}=37,5cm^2\)

\(c,\) Tính \(AH\) nếu \(AB=20cm;HC=9cm\)

\(d,\) Kẻ tia phân giác \(AD.\) Tính \(AH,BH,HD\) biết \(DB=7,5cm;DC=10cm\)

Answer 1

a: \(AH^2=HB\cdot HC\)

\(\Leftrightarrow HB\left(HB+9\right)=36\)

=>HB=3(cm)

=>HC=12cm

\(AC=\sqrt{12\cdot15}=5\sqrt{6}\left(cm\right)\)

\(AB=\sqrt{3\cdot15}=3\sqrt{5}\left(cm\right)\)

c: \(AB^2=BH\cdot BC\)

\(\Leftrightarrow BH\left(BH+9\right)-400=0\)

\(\Leftrightarrow\left(BH+25\right)\left(BH-16\right)=0\)

=>BH=16(cm)

\(AH=\sqrt{HB\cdot HC}=12\left(cm\right)\)

d: BC=BD+CD=17,5(cm)

Xét ΔABC có AD là phân giác

nên AB/DB=AC/DC

=>AB/3=AC/4=k

=>AB=3k; AC=4k

\(AC^2+AB^2=BC^2\)

\(\Leftrightarrow25k^2=17.5^2\)

\(\Leftrightarrow k^2=12.25\)

=>k=3,5

=>AB=10,5(cm); AC=14(cm)

\(AH=\dfrac{AB\cdot AC}{BC}=\dfrac{10.5\cdot14}{17.5}=8.4\left(cm\right)\)

\(BH=\dfrac{AB^2}{BC}=\dfrac{10.5^2}{17.5}=6.3\left(cm\right)\)