W każdym zadaniu przykłady rozwiązuję od lewej do prawej, z góry na dół.
Przypomnijmy sobie Twierdzenie Pitagorasa: [tex]a^2+b^2=c^2[/tex]
Zad.1
[tex]a^2+a^2=17^2\\\\\sqrt{7}^2+6^2=x^2\\\\\sqrt{12}^2+x^2=\sqrt{17}^2\\\\\sqrt{28}^2+3^2=m^2\\\\f^2+11^2=21^2[/tex]
Zad.2
[tex]x^2+15^2=17^2\\x^2+225=289|-225\\x^2=289-225\\x^2=64|\sqrt{~}\\x=8\\\\k^2+10^2=12^2|-100\\k^2+100=144|-100\\k^2=144-100\\k^2=44|\sqrt{~}\\k=\sqrt{44}\\k=\sqrt{4*11}\\k=2\sqrt{11}\\\\5^2+5^2=a^2\\a^2=25+25\\a^2=50|\sqrt{~}\\a=\sqrt{50}\\a=\sqrt{25*2}\\a=5\sqrt{2}\\\\x^2+\sqrt{3}^2=\sqrt{19}^2\\x^2=3=19|-3\\x^2=19-3\\x^2=16|\sqrt{~}\\x=4\\\\9^2+8^2=m^2\\m^2=81+64\\m^2=145|\sqrt{~}\\m=\sqrt{145}[/tex]
Zad.3
[tex](\frac12d_1)^2+(\frac12d_2)^2=13^2\\(\frac12*24)^2+\frac14d_2$^2=169\\\frac14d_2$^2+12^2=169\\\frac14d_2$^2+144=169|-144\\\frac14d_2$^2=25|*4\\d_2^2=100|\sqrt{~}\\d_2=10[/tex]
Zad.4 + załącznik
[tex]27cm = 2x+11cm\\2x=16cm\\x=8cm\\(8cm)^2+h^2=(14cm)^2\\h^2+64cm^2=196cm^2|-64cm62\\h^2=196cm^2-64cm^2\\h^2=132|\sqrt{~}\\h=\sqrt{132}\\h=2\sqrt{33}[/tex]
