Giải thích các bước giải:
Bài 1:
Ta có: $\Delta ABC$ vuông tại $A$
$\to BC^2=AB^2+AC^2$
$\to AC^2=BC^2-AB^2=256$
$\to AC=16$
Mà $AH\perp BC$
$\to S_{ABC}=\dfrac12AH\cdot BC=\dfrac12AB\cdot AC$
$\to AH=\dfrac{AB\cdot AC}{BC}=\dfrac{48}5$
$\to BH=\sqrt{AB^2-AH^2}=\dfrac{36}5, CH=BC-BH=\dfrac{64}5$
Bài 2:
Xét $\Delta MNK,\Delta MKP$ có:
$\widehat{MKN}=\widehat{MKP}(=90^o)$
$\widehat{NMK}=90^o-\widehat{KMP}=\widehat{MPK}$
$\to\Delta MNK\sim\Delta PMK(g.g)$
$\to \dfrac{MK}{PK}=\dfrac{NK}{MK}$
$\to MK^2=KN\cdot KP$
$\to MK^2=4$
$\to MK=2$
$\to MN=\sqrt{MK^2+KN^2}=\sqrt5, MP=\sqrt{MK^2+KP^2}=2\sqrt{5}$