W zadaniu sprawdzana jest znajomość Twierdzenia Pitagorasa. Przypomnijmy sobie wzór:
[tex]a^{2} + b^{2} = c^{2}[/tex]
a, b - przyprostokątne
c - przeciwprostokątna
a) a = 1
b = 1
[tex]1^{2} + 1^{2} = c^2\\1 + 1 = c^2\\c^2 = 2\\c = \sqrt{2}[/tex]
AC = [tex]\sqrt{2}[/tex]
a = [tex]\sqrt{2}[/tex]
b = 1
[tex]\sqrt{2} ^{2} + 1^{2} = c^{2} \\[/tex]
[tex]2 + 1 = c^{2}[/tex]
[tex]c^{2} = 3[/tex]
[tex]c = \sqrt{3}[/tex]
AD = [tex]\sqrt{3}[/tex]
[tex]\sqrt{3}^{2} + 1^{2} = c^{2} \\[/tex]
[tex]3 + 1 = c^{2}\\c^2 = 4\\c = 2[/tex]
AE = 2
[tex]2^2 + 1^2 = c^2\\4 + 1 = c^2\\c^2 = 5\\c = \sqrt{5}[/tex]
AF = [tex]\sqrt{5}[/tex]
[tex]\sqrt{5}^2 + 1^2 = c^2\\5 + 1 = c^2\\c^2 = 6\\c = \sqrt{6[/tex]
AG = [tex]\sqrt{6}[/tex]
[tex]\sqrt{6}^2 + 1^2 = c^2\\6 + 1 = c^2\\c^2 = 7\\c = \sqrt{7} \\AH = \sqrt{7}[/tex]
[tex]\sqrt{7}^2 + 1^2 = c^2\\7 + 1 = c^2\\c^2 = 8\\c = \sqrt{8} = 2\sqrt{2} \\AI = 2\sqrt{2}[/tex]
[tex]2\sqrt{2}^2 + 1^2 = c^2\\4 * 2 + 1 = c^2\\c^2 = 9\\c = \sqrt{9} = 3[/tex]
[tex]AK = 3[/tex]
[tex]3^2 + 1^2 = c^2\\9 + 1 = c^2\\10 = c^2\\c = \sqrt{10} \\AL = \sqrt{10}[/tex]
[tex]\sqrt{10}^2 + 1^2 = c^2\\10 + 1 = c^2\\c^2 = 11\\c = \sqrt{11\\} \\AM = \sqrt{11}[/tex]
[tex]\sqrt{11}^2 + 1^2 = c^2\\11 + 1 = c^2\\c^2 = 12\\c = \sqrt{12} = 2\sqrt{3} \\AN = 2\sqrt{3}[/tex]
[tex]2\sqrt{3}^2 * 1^2 = c^2\\4 * 3 + 1^2 = c^2\\c^2 = 13\\c = \sqrt{13} \\AO = \sqrt{13}[/tex]
