Câu 1
Có $AB//EF(\bot BE) $
$\to \frac {AB}{EF}=\frac{BC}{CE}$ (hệ quả định lí Ta-lét)
$\to \frac {AB}{18,6}=\frac{79,6}{34,2}$
$\to AB \approx 43,29 (m) $
Câu 2
Có $\widehat{DEB}=\widehat{DEB}$
$\to DE//AC $
$\to \frac{BE}{BC}=\frac{BD}{BA}$
$\to \frac{BE}{BE+EC}=\frac{BD}{BA}$
$\to \frac{2,1}{2,1+1,4}=\frac{1,5}{BA}$
$\to AB=2,5 (m)$
