Question

Biết AC=6, BH=32/5. Tính AB?

Answer 1

AC^2 = BC.CH = (BH+CH)CH

\(\Leftrightarrow CH^2+\frac{32}{5}CH-36=0\Leftrightarrow\left[{}\begin{matrix}CH=\frac{18}{5}\\CH=-10\left(loai\right)\end{matrix}\right.\)

=> BC = 10

=> \(AB^2=BC.BH=10\cdot\frac{32}{5}=64\Rightarrow AB=8\)

Answer 2

\(\widehat{CAB}=90^o???\)

\(CH=HB???\)