Question

Determine med DH.

segue anexo

Answer 1

Com pitagoras, calculamos AC
4²=2²+ AC²
16-4=AC²
AC=$\sqrt{12} =2 \sqrt{3}$

Agora calculamos AD
$(2 \sqrt{3} ) ^{2} = 2 ^{2} +AD ^{2}$
12-4=AD²
AD=$\sqrt{8} = 2 \sqrt{2}$

AD.DC=AC.DH
$2\sqrt{2} . 2 = 2 \sqrt{3} . DH$
$\frac{4 \sqrt{2} }{2 \sqrt{3} } =AD$
$\frac{2 \sqrt{2} }{ \sqrt{3} }$
$\frac{2 \sqrt{(2.3)} }{3}$
$\frac{2 \sqrt{6} }{3}$