Răspuns :
Rezolvi folosind Teorema lui Pitagora.
[tex] {ip}^{2} = {c1}^{2} + {c2}^{2} \\ {(2 \sqrt{7}) }^{2} = {2}^{2} + {c2}^{2} \\ = > {c2}^{2} = {(2 \sqrt{7} )}^{2} - 4 \\ {c2}^{2} = 28 - 4 \\ {c2}^{2} = 24 \: \: | \: \sqrt{} \\ c2 = 2 \sqrt{6} [/tex]
[tex] {ip}^{2} = {c1}^{2} + {c2}^{2} \\ {(2 \sqrt{7}) }^{2} = {2}^{2} + {c2}^{2} \\ = > {c2}^{2} = {(2 \sqrt{7} )}^{2} - 4 \\ {c2}^{2} = 28 - 4 \\ {c2}^{2} = 24 \: \: | \: \sqrt{} \\ c2 = 2 \sqrt{6} [/tex]
[tex] In \: \: \triangle \: ABC, \: m( \angle~A) = 90^{o}~ \overset{\text{T.P. }}{\Longrightarrow} \: \mathbf{BC^{2} = AC^{2} + AB^{2}} \\ \mathbf{ (2 \sqrt{7} ) ^{2} = {2}^{2} + AB^{2}} \\ \mathbf{4 \cdot 7 = 4 + AB^{2}} \\ \mathbf{ 28 = 4 + AB^{2}} \\ \mathbf{4 + AB^{2} = 28} \\ \mathbf{AB^{2} = 28 - 4} \\ \mathbf{AB^{2} = 24} \\ \mathbf{ AB = \sqrt{24}} \\ \mathbf{AB = 2 \sqrt{6} \: cm}[/tex]