Potrzebuję na jutro! :-).... Question from @Bllankaaa

Cześć!

$a^2=9^2+12^2\\\\a^2=81+144\\\\a^2=225\\\\a=\sqrt{225}\\\\\huge\boxed{a=15}$

$b^2=5^2+12^2\\\\b^2=25+144\\\\b^2=169\\\\b=\sqrt{169}\\\\\huge\boxed{b=13}$

$c^2=5^2+6^2\\\\c^2=25+36\\\\c^2=61\\\\\huge\boxed{c=\sqrt{61}}$

$d^2=10^2-6^2\\\\d^2=100-36\\\\d^2=64\\\\d=\sqrt{64}\\\\\huge\boxed{d=8}$

$e^2=5^2-(\sqrt{11})^2\\\\e^2=25-11\\\\e^2=14\\\\\huge\boxed{e=\sqrt{14}}$

$f^2=(\sqrt{10})^2-(\sqrt6)^2\\\\f^2=10-6\\\\f^2=4\\\\f=\sqrt4\\\\\huge\boxed{f=2}$

2 votes Thanks 1

LoczeeQ

Zastosujmy twierdzenie Pitagorasa

${9}^{2} + {12}^{2} = {a}^{2} \\ 81 + 144 = {a}^{2} \\ 225 = {a}^{2} \\ a = \sqrt{225} \\ a = 15$

${12}^{2} + {5}^{2} = {b}^{2} \\ 144 + 25 = {b}^{2} \\ 169 = {b}^{2} \\ b = \sqrt{169} \\ b = 13$

${6}^{2} + {5}^{2} = {c}^{2} \\ 36 + 25 = {c}^{2} \\ 61 = {c}^{2} \\ c = \sqrt{61}$

${6}^{2} + {d}^{2} = {10}^{2} \\ 36 + {d}^{2} = 100 \\ {d}^{2} = 64 \\ d = \sqrt{64} \\ d = 8$

${e}^{2} + ( \sqrt{11} )^{2} = {5}^{2} \\ {e}^{2} + 11 = 25 \\ {e}^{2} = 14 \\ e = \sqrt{14}$

${f}^{2} + ( \sqrt{6} )^{2} = ( \sqrt{10} )^{2} \\ {f}^{2} + 6 = 10 \\ {f}^{2} = 4 \\ f = \sqrt{4} \\ f = 2$

Myślę że pomogłem ;)

0 votes Thanks 0