twierdzenie Pitagorasa
a^2+b^2=c^2
-------------------
I
x^2=6^2+8^2
x^2=36+64
x=\sqrt{100}=10
II
y^2=17^2-15^2
y^2=289-225
y=\sqrt{64}=8
III
z^2=2^2+6^2
z^2=4+36
z=\sqrt{40}=\sqrt{4*10}
z=2\sqrt{10}
IV
a^2=(3\sqrt2)^2-4^2
a^2=9*2-16
a=\sqrt3
V
b^2=(2\sqrt5)^2+(3\sqrt3)^2
b^2=4*5+9*3=20+27
b=\sqrt{47}
VI
c^2=(\sqrt{13})^2-3^2=13-9
c=\sqrt{4}=2
zadanie 8
a)
|AB|^2=6^2+5^2
|AB|^2=36+25
|AB|=\sqrt{59}
b)
|AB|^2=12,5^2-7,5^2
|AB|^2=156,25-56,25
|AB|=\sqrt{100}=10
c)
|AB|^2=5^2-(\sqrt{11})^2
|AB|^2=25-11
|AB|=\sqrt{14}