Question

[Trần Hữu Ngọc Minh]

Cho tam giác \(ABC\)nhọn,trực tâm \(H\)có \(HA=8cm,HB=\sqrt{14}cm,HC=\sqrt{44}cm.\)Tính diện tích tam giác \(ABC\)

Answer 1

[Tuyển Trần Thị]

dat HI=x, HF=y, HG=z

ta co \(\frac{SBHC}{SABC}=\frac{\frac{1}{2}.HI.BC}{\frac{1}{2}AI.BC}=\frac{HI}{AI}=\) \(\frac{x}{x+8}\)

ttu \(\frac{SAHC}{SABC}=\frac{y}{y+\sqrt{14}}\) \(\frac{SHAB}{SABC}=\frac{z}{z+\sqrt{44}}\)

cộng vế vs vế  \(\frac{x}{x+8}+\frac{y}{y+\sqrt{14}}+\frac{z}{z+\sqrt{44}}=\frac{SHBC+SHAC+SHAB}{SABC}=1\) (1)

do \(\Delta AHF\simeq\Delta BHI\rightarrow\frac{HF}{HI}=\frac{y}{x}=\frac{AH}{BH}=\frac{8}{\sqrt{14}}\Rightarrow y=\frac{8}{\sqrt{14}}x\)

ttu \(\Delta AHG\simeq\Delta CHI\Rightarrow z=\frac{8}{\sqrt{44}}x\)

the vao 1 ta co \(\frac{x}{x+8}+\frac{\frac{8}{\sqrt{14}}x}{\frac{8}{\sqrt{14}}x+\sqrt{14}}+\frac{\frac{8x}{\sqrt{44}}}{\frac{8x}{\sqrt{44}}+\sqrt{44}}=1\)

\(\Leftrightarrow\frac{x}{x+8}+\frac{8x}{8x+14}+\frac{8x}{8x+44}=1\)

 giải ra bn có  x=2

ap dung dl pitago vao tam giac vuong BHI \(BI^2=14-x^2=14-4=10\Rightarrow BI=\sqrt{10}\)

                             . ............................HIC \(IC=\sqrt{40}\)

\(\Rightarrow BC=BI+IC=\sqrt{10}+\sqrt{40}\)

MA AI=\(AH+HI=8+2=10\)

\(\Rightarrow SABC=\frac{10.\left(\sqrt{10}+\sqrt{40}\right)}{2}=15\sqrt{10}\)

Answer 2

[Strike Eagle]

\frac{x}{x+8}+\frac{\frac{8}{\sqrt{14}}x}{\frac{8}{\sqrt{14}}x+\sqrt{14}}+\frac{\frac{8x}{\sqrt{44}}}{\frac{8x}{\sqrt{44}}+\sqrt{44}}=1x+8x​+14​8​x+14​14​8​x​+44​8x​+44​44​8x​​=