b,
$\Delta ABC$ vuông tại $A$ có:
$BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=5(cm)$
Xét $\Delta AHB$ và $\Delta CAB$ có:
$\widehat{AHB}=\widehat{CAB}=90^o$
$\widehat{ABC}$ chung
$\Rightarrow \Delta AHB\backsim \Delta CAB$ (g.g)
$\Rightarrow \dfrac{AH}{AB}=\dfrac{AC}{BC}$
$\Leftrightarrow \dfrac{AH}{3}=\dfrac{4}{5}$
$\Leftrightarrow AH=2,4(cm)$
Theo câu $a$, $BD=2,1(cm)$
$\Delta AHB\backsim\Delta CAB$ (cmt)
$\Rightarrow \dfrac{BH}{AB}=\dfrac{AB}{BC}$
$\Leftrightarrow BH=\dfrac{AB^2}{BC}=1,8(cm)$
$\to HD=BD-BH=0,3(cm)$
$\Delta AHD$ vuông tại $H$ có:
$AD=\sqrt{AH^2+HD^2}\approx 2,42(cm)$
