Question

Answer 1

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

\(\cos C=\dfrac{AC}{BC}=\dfrac{4}{5}\approx\cos37^0\\ \Rightarrow\widehat{C}\approx37^0\)

b, Áp dụng HTL: \(\left\{{}\begin{matrix}AH^2=HB\cdot HC\\AH^2=AE\cdot AC\end{matrix}\right.\Leftrightarrow HB\cdot HC=AE\cdot AC\)

c, Áp dụng HTL: \(AH=\dfrac{AB\cdot AC}{BC}=12\left(cm\right)\)

Suy ra \(AE=\dfrac{AH^2}{AC}=7,2\left(cm\right)\)

Suy ra \(S_{ABE}=\dfrac{1}{2}AE\cdot AB=54\left(cm^2\right)\)

Mà \(S_{ABC}=\dfrac{1}{2}AB\cdot AC=150\left(cm^2\right)\)

Vậy \(S_{BEC}=S_{ABC}-S_{ABE}=96\left(cm^2\right)\)