Question

Answer 1

Áp dụng HTL trong tam giác ABC vuông tại A:

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

\(\Rightarrow AH^2=\dfrac{1}{\dfrac{1}{AB^2}+\dfrac{1}{AC^2}}=\dfrac{1}{\dfrac{1}{3^2}+\dfrac{1}{4^2}}=\dfrac{144}{25}\Rightarrow AH=2,4\left(cm\right)\)

Xét tam giác ABC vuông tại A có:

\(BC^2=AB^2+AC^2\left(Pytago\right)\)

\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=5\left(cm\right)\)

\(\Rightarrow BM=\dfrac{1}{2}BC=2,5\left(cm\right)\)

\(S_{ABM}=\dfrac{1}{2}.AH.BM=\dfrac{1}{2}.2,4.2,5=3\left(cm^2\right)\)