Question

Answer 1

Bài 38:

Ta có: \(AH^2=BH\cdot CH\)

\(\Rightarrow CH=\dfrac{AH^2}{BH}=\dfrac{12^2}{9}=16\left(cm\right)\)

Mà: \(BC=BH+CH=9+16=25\left(cm\right)\)

\(\Rightarrow S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC=\dfrac{1}{2}\cdot12\cdot25=150\left(cm^2\right)\)

Answer 2

Bài 39: 

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

\(\Rightarrow AH^2=\dfrac{AB^2\cdot AC^2}{AB^2+AC^2}\Leftrightarrow12^2=\dfrac{AB^2\cdot20^2}{AB^2+20^2}\)

\(\Leftrightarrow AB=15\left(cm\right)\)

Áp dụng định lý Py-ta-go ta có:

\(BC=\sqrt{AC^2+AB^2}=\sqrt{20^2+15^2}=25\left(cm\right)\)

\(\Rightarrow S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC=\dfrac{1}{2}\cdot12\cdot25=150\left(cm^2\right)\)

Answer 3

Bài 40:

Ta có:

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

\(\Rightarrow AH^2=\dfrac{AC^2AB^2}{AC^2+AB^2}\Rightarrow4,8^2=\dfrac{AC^2\cdot6^2}{AC^2+6^2}\)

\(\Leftrightarrow AC=8\left(cm\right)\)

Áp dụng định lý Py-ta-go ta có:

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

\(\Rightarrow S_{ABC}=\dfrac{1}{2}\cdot AH\cdot BC=\dfrac{1}{2}\cdot4,8\cdot10=24\left(cm^2\right)\)