Question

Answer 1

\(1,\\ a,AC=\sqrt{BC^2-AB^2}=8\left(cm\right)\left(pytago\right)\)

Áp dụng HTL:

\(\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=3,6\left(cm\right)\\HC=BC-BH=6,4\left(cm\right)\\AH=\sqrt{BH\cdot HC}=4,8\left(cm\right)\end{matrix}\right.\)

\(b,BC=\sqrt{AB^2+AC^2}=2\left(cm\right)\left(pytago\right)\)

Áp dụng HTL:

\(\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{3}{4}\left(cm\right)\\CH=BC-BH=\dfrac{13}{4}\left(cm\right)\\AH=\sqrt{\dfrac{3}{4}\cdot\dfrac{13}{4}}=\dfrac{\sqrt{39}}{2}\left(cm\right)\end{matrix}\right.\)

\(c,BC=BH+HC=25a\)

Áp dụng HTL:

\(\left\{{}\begin{matrix}AH=\sqrt{BH\cdot CH}=12a\\AB=\sqrt{BC\cdot BH}=20a\\AC=\sqrt{BC\cdot CH}=15a\end{matrix}\right.\)

\(d,BC=\sqrt{AB^2+AC^2}=25a\left(pytago\right)\)

Áp dụng HTL: \(AH=\dfrac{AB\cdot AC}{BC}=\dfrac{300a^2}{625a}=\dfrac{2\sqrt{3}a}{5}\)

 

Answer 2

d: Xét ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC

nên \(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)

\(\Leftrightarrow\dfrac{1}{AH^2}=\dfrac{1}{225a^2}+\dfrac{1}{400a^2}\)

\(\Leftrightarrow AH=12a\)