Đáp án+Giải thích các bước giải:
4,
Ta có:
`\frac{AB}{AC}=\frac{5}{7}⇒AB=\frac{5AC}{7}`
Lại có:
`\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}`
`⇒\frac{1}{225}=\frac{1}{\frac{25AC^2}{49}}+\frac{1}{AC^2}`
`⇔\frac{1}{225}=\frac{49}{25AC^2}+\frac{1}{AC^2}`
`⇔\frac{1}{225}=\frac{49+25}{25AC^2}`
`⇒AC^2=666`
`⇒AC=3\sqrt{74}(cm)`
`⇒AB=3\sqrt{74}.\frac{5}{7}=\frac{15\sqrt{74}}{7}(cm)`
`⇒BC=\sqrt{AB^2+AC^2}=\sqrt{(3\sqrt{74})^2+(\frac{15\sqrt{74}}{7})^2}=\frac{222}{7}(cm)`
`HB.BC=AB^2`
`⇒HB=\frac{AB^2}{BC}=\frac{(\frac{15\sqrt{74}}{7})^2}{\frac{222}{7}}=\frac{75}{7}(cm)`
`⇒HC=BC-BH=\frac{222}{7}-\frac{75}{7}=21(cm)`
Vậy `HB=\frac{75}{7}cm;HC=21cm`
