Câu hỏi của Hoàng Phúc

Question

Cho tam giác ABC vuông tại A.Biết $\frac{AB}{AC}=\frac{5}{7}$,đường cao AH=15cma)Tính HB,HCb)Tính chu vi tam giác ABC

Lê Hà Phương · Accepted Answer

a) Ta thấy: $AB.AC=BC.AH$$\Leftrightarrow AB^2.AC^2=BC^2.AH^2$$\Leftrightarrow AH^2=\frac{AB^2.AC^2}{BC^2}$$\Leftrightarrow AH^2=\frac{AB^2.AC^2}{AB^2+AC^2}$$\Leftrightarrow\frac{1}{AH^2}=\frac{AB^2+AC^2}{AB^2.AC^2}=\frac{1}{AB^2}+\frac{1}{AC^2}$Ta có: $\frac{AB}{AC}=\frac{5}{7}\Rightarrow AB:AC=\frac{5}{7}\Rightarrow AB=\frac{5}{7}AC$Áp dụng công thức trên: $\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{15^2}=\frac{1}{\frac{25}{49}AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{49}{25}.\frac{1}{AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{1}{AC^2}\left(\frac{49}{25}+1\right)$$\Rightarrow\frac{1}{225}=\frac{1}{AC^2}.\frac{74}{25}\Rightarrow\frac{1}{AC^2}=\frac{1}{225}.\frac{25}{74}=\frac{1}{666}\Rightarrow AC^2=666\Rightarrow AC=\sqrt{666}=3\sqrt{74}cm$Do đó: $AB=\frac{5}{7}.3\sqrt{74}=\frac{15\sqrt{74}}{7}cm$Xét tam giác ABH có: $AH^2+BH^2=AB^2\Leftrightarrow15^2+BH^2=\left(\frac{15\sqrt{74}}{7}\right)^2\Leftrightarrow BH^2=\frac{16650}{49}-225=\frac{5625}{49}$$\Rightarrow BH=\frac{\sqrt{5625}}{\sqrt{49}}=\frac{75}{7}cm$Xét tam giác ACH có: $AH^2+HC^2=AC^2\Leftrightarrow15^2+HC^2=666\Leftrightarrow HC^2=666-225=441$$\Rightarrow HC=\sqrt{441}=21cm$Vậy: $BH=\frac{75}{7}cm$ và $HC=21cm$b) Chu vi tam giác ABC là: $AB+AC+BC=\frac{15\sqrt{74}}{7}+3\sqrt{74}+21+\frac{75}{7}\approx76cm$Câu trả lời: a) Ta thấy: $AB.AC=BC.AH$$\Leftrightarrow AB^2.AC^2=BC^2.AH^2$$\Leftrightarrow AH^2=\frac{AB^2.AC^2}{BC^2}$$\Leftrightarrow AH^2=\frac{AB^2.AC^2}{AB^2+AC^2}$$\Leftrightarrow\frac{1}{AH^2}=\frac{AB^2+AC^2}{AB^2.AC^2}=\frac{1}{AB^2}+\frac{1}{AC^2}$Ta có: $\frac{AB}{AC}=\frac{5}{7}\Rightarrow AB:AC=\frac{5}{7}\Rightarrow AB=\frac{5}{7}AC$Áp dụng công thức trên: $\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{15^2}=\frac{1}{\frac{25}{49}AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{49}{25}.\frac{1}{AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{1}{AC^2}\left(\frac{49}{25}+1\right)$$\Rightarrow\frac{1}{225}=\frac{1}{AC^2}.\frac{74}{25}\Rightarrow\frac{1}{AC^2}=\frac{1}{225}.\frac{25}{74}=\frac{1}{666}\Rightarrow AC^2=666\Rightarrow AC=\sqrt{666}=3\sqrt{74}cm$Do đó: $AB=\frac{5}{7}.3\sqrt{74}=\frac{15\sqrt{74}}{7}cm$Xét tam giác ABH có: $AH^2+BH^2=AB^2\Leftrightarrow15^2+BH^2=\left(\frac{15\sqrt{74}}{7}\right)^2\Leftrightarrow BH^2=\frac{16650}{49}-225=\frac{5625}{49}$$\Rightarrow BH=\frac{\sqrt{5625}}{\sqrt{49}}=\frac{75}{7}cm$Xét tam giác ACH có: $AH^2+HC^2=AC^2\Leftrightarrow15^2+HC^2=666\Leftrightarrow HC^2=666-225=441$$\Rightarrow HC=\sqrt{441}=21cm$Vậy: $BH=\frac{75}{7}cm$ và $HC=21cm$b) Chu vi tam giác ABC là: $AB+AC+BC=\frac{15\sqrt{74}}{7}+3\sqrt{74}+21+\frac{75}{7}\approx76cm$

Minato Namikaze · Answer

A B C H 15 cm   Vì tam giác ABC vuông tại A => góc B + góc C = 90oVì tam giác HAC vuông tại H => góc HAC + góc C = 90o=> góc HAC = góc BXét tam giác HAC và tam giác HBA có:     góc HAC = góc B (cmt)     góc AHC = góc AHB (=90o)=> tam giác HAC đồng dạng với tam giác HBA (TH3)=> $\frac{AC}{AB}=\frac{AH}{BH}=\frac{HC}{AH}=\frac{7}{5}$=> $HC=15.\frac{7}{5}=21\left(cm\right);HB=15.\frac{5}{7}=\frac{75}{7}\left(cm\right)$Sau đó tính AB; AC; BC. Ngại là lắm, làm nốt nhá ._.Câu trả lời: A B C H 15 cm   Vì tam giác ABC vuông tại A => góc B + góc C = 90oVì tam giác HAC vuông tại H => góc HAC + góc C = 90o=> góc HAC = góc BXét tam giác HAC và tam giác HBA có:     góc HAC = góc B (cmt)     góc AHC = góc AHB (=90o)=> tam giác HAC đồng dạng với tam giác HBA (TH3)=> $\frac{AC}{AB}=\frac{AH}{BH}=\frac{HC}{AH}=\frac{7}{5}$=> $HC=15.\frac{7}{5}=21\left(cm\right);HB=15.\frac{5}{7}=\frac{75}{7}\left(cm\right)$Sau đó tính AB; AC; BC. Ngại là lắm, làm nốt nhá ._.

Lê Hà Phương · Answer

a) Ta có: $\frac{AB}{AC}=\frac{5}{7}\Rightarrow AB:AC=\frac{5}{7}\Rightarrow AB=\frac{5}{7}AC$Áp dụng công thức: $\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{15^2}=\frac{1}{\frac{25}{49}AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{49}{25}.\frac{1}{AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{1}{AC^2}\left(\frac{49}{25}+1\right)$$\Rightarrow\frac{1}{225}=\frac{1}{AC^2}.\frac{74}{25}\Rightarrow\frac{1}{AC^2}=\frac{1}{225}.\frac{25}{74}=\frac{1}{666}\Rightarrow AC^2=666\Rightarrow AC=\sqrt{666}=3\sqrt{74}cm$Do đó: $AB=\frac{5}{7}.3\sqrt{74}=\frac{15\sqrt{74}}{7}cm$Suy ra: $CH=21cm$ và $BH=\frac{75}{7}cm$ ( Áp đụng định lý Pi - ta go )b) Chu vi tam giác ABC là: AB + AC + BC = $\frac{15\sqrt{74}}{7}+3\sqrt{74}+21+\frac{75}{7}\approx76cm$Cũng k biết đúng k nữa =.=Câu trả lời: a) Ta có: $\frac{AB}{AC}=\frac{5}{7}\Rightarrow AB:AC=\frac{5}{7}\Rightarrow AB=\frac{5}{7}AC$Áp dụng công thức: $\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{15^2}=\frac{1}{\frac{25}{49}AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{49}{25}.\frac{1}{AC^2}+\frac{1}{AC^2}\Leftrightarrow\frac{1}{225}=\frac{1}{AC^2}\left(\frac{49}{25}+1\right)$$\Rightarrow\frac{1}{225}=\frac{1}{AC^2}.\frac{74}{25}\Rightarrow\frac{1}{AC^2}=\frac{1}{225}.\frac{25}{74}=\frac{1}{666}\Rightarrow AC^2=666\Rightarrow AC=\sqrt{666}=3\sqrt{74}cm$Do đó: $AB=\frac{5}{7}.3\sqrt{74}=\frac{15\sqrt{74}}{7}cm$Suy ra: $CH=21cm$ và $BH=\frac{75}{7}cm$ ( Áp đụng định lý Pi - ta go )b) Chu vi tam giác ABC là: AB + AC + BC = $\frac{15\sqrt{74}}{7}+3\sqrt{74}+21+\frac{75}{7}\approx76cm$Cũng k biết đúng k nữa =.=

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