Se da:
ΔABC isoscel cu AB = BC
A se proiecteaza pe (MBC) in M
⇒ AM ⊥ (ABC) ⇒ AM ⊥ MB si AM ⊥ MC
ΔMBC este dreptunghic in M ⇒ MB ⊥ MC
MC = 6 cm
MB = 8 cm
Se cere:
AC = ?
Rezolvare:
[tex]\displaystyle\\
\text{Din triunghiul MBC dreptunghic in M calculam ipotenuza BC.}\\\\
BC= \sqrt{MB^2 + MC^2}= \sqrt{8^2 + 6^2}= \sqrt{64 + 36}= \sqrt{100}= \boxed{10~cm}\\\\
AB = BC = \boxed{10~cm}\\\\
\text{Din triunghiul MAB dreptunghic in M calculam cateta MA.}\\\\
MA = \sqrt{AB^2 - MB^2}= \sqrt{10^2 - 8^2}=\sqrt{100 - 64}=\sqrt{36}= \boxed{6~cm}\\\\
\text{In triunghiul MAC dreptunghic in M avem:}\\
m(\ \textless \ AMC) = 90^o\\
MA = 6~cm\\
MC = 6~cm
[/tex]
[tex]\displaystyle\\
\Longrightarrow~~ \Delta MAC \text{ este triunghi dreptung isoscel}\\\\
\text{Calculam ipotenuza AC.}\\\\
AC = \sqrt{MA^2 + MC^2}=\sqrt{6^2 + 6^2}=\sqrt{36 + 36}=\sqrt{72}= \boxed{\boxed{6 \sqrt{2}~cm} }[/tex]
