Cho tam giác ABC vuông tại A, đường cao AH . Biết BH= 4cm

* Hướng dẫn giải

$$\begin{gathered} a, \Delta ABC, \stackrel{⏜}{A}={90}^0,AH\perp BC\left(gt\right)\Rightarrow AH=\sqrt{BH.CH}=\sqrt{4.9}=6cm \\ \Delta ABH, \stackrel{⏜}{H}={90}^0\left(gt\right)\Rightarrow \tan B=\frac{AH}{BH}=\frac{6}{4}\Rightarrow \stackrel{⏜}{B}\approx 56,3^0 \\ b, \Delta ABC, \stackrel{⏜}{A}={90}^0,MB=MC\left(gt\right)\Rightarrow AM=\frac{1}{2}BC=\frac{1}{2}.13=6,5cm \\ {S}_{\Delta AHM}=\frac{1}{2}MH.AH=\frac{1}{2}.2,5.6=7,5c{m}^2 \end{gathered}$$