Question

[Nguyễn Minh Vũ]

Vì \(AD\) là đường trung tuyến ứng vs cạnh huyền \(BC\)

=> \(AD=\frac{BC}{2}=\frac{\sqrt{41}}{2}\left(cm\right)\)

Vì  \(AH\div AD=40\div41\)

=> \(AH=\frac{40}{41}AD\)

=> \(AH=\frac{40}{41}.\frac{\sqrt{41}}{2}\)

=> \(AH=\frac{20}{\sqrt{41}}\left(cm\right)\)

AD ĐL Pytago vào \(\Delta AHD\left(\widehat{A}=90^o\right)\)

\(AH^2+HD^2=AD^2\)

=> \(HD^2=AD^2-AH^2\)

=> \(HD^2=\left(\frac{\sqrt{41}}{2}\right)^2-\left(\frac{20}{\sqrt{41}}\right)^2\)

=>  \(HD^2=\frac{81}{164}\)

=> \(HD=\frac{9}{2\sqrt{41}}\)

=> \(HB=\frac{\sqrt{41}}{2}-\frac{9}{2\sqrt{41}}\)

=> \(HB=\frac{16}{\sqrt{41}}\)

AD HTL trong \(\Delta ABC\left(\widehat{A}=90^o\right)\)

\(AB^2=HB.BC\)

=> \(AB^2=\frac{16}{\sqrt{41}}.\sqrt{41}\)

=> \(AB=\sqrt{16}=4\)

AD ĐL Pytago vào \(\Delta ABC\left(\widehat{A}=90^o\right)\)

\(AB^2+AC^2=BC^2\)

=> \(16+AC^2=41\)

=> \(AC=\sqrt{25}=5\)