Question

Answer 1

Bài 1:

Gọi tam giác đó là tam giác ABC vg tại A có đường cao AH:

\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\Rightarrow AH=\sqrt{\dfrac{1}{\dfrac{1}{AB^2}+\dfrac{1}{AC^2}}}=\sqrt{\dfrac{1}{\dfrac{1}{6^2}+\dfrac{1}{8^2}}}=4,8\left(cm\right)\)

\(BC^2=AB^2+AC^2\left(Pytago\right)\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)

\(\left\{{}\begin{matrix}AB^2=BH.BC\\AC^2=HC.BC\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{6^2}{10}=3,6\left(cm\right)\\HC=\dfrac{AC^2}{BC}=\dfrac{8^2}{10}=4,8\left(cm\right)\end{matrix}\right.\)