Odpowiedź:
[tex]\huge\boxed{\begin{array}{l|l}a)&Ob=20\sqrt2mm\\b)&Ob=60cm\\c)&Ob=28\sqrt3dm\\d)&Ob=4\sqrt6cm\end{array}}[/tex]
Przekątna kwadratu
Przekątna kwadratu o boku a ma długość a√2
[tex]\huge\boxed{d=a\sqrt2}[/tex]
Z Twierdzenia Pitagorasa:
[tex]\boxed{\begin{array}{l}a^2+a^2=d^2\\d^2=2a^2 |\sqrt{}\\d=\sqrt{2a^2}\\d=a\sqrt2\end{array}}[/tex]
Rozwiązanie:
a)
[tex]d=10mm\\a\sqrt2=10mm |:\sqrt2\\a=\dfrac{10}{\sqrt2}mm\\a=\dfrac{10\sqrt2}2mm\\a=5\sqrt2mm\\\\Ob=4\cdot 5\sqrt2mm\\\boxed{Ob=20\sqrt2mm}[/tex]
b)
[tex]d=15\sqrt2cm\\a\sqrt{2}\!\!\!\!\!\diagup=15\sqrt2\!\!\!\!\!\diagup cm\\a=15cm\\\\Ob=4\cdot 15cm\\\boxed{Ob=60cm}[/tex]
c)
[tex]d=7\sqrt6dm\\a\sqrt2=7\sqrt6dm |:\sqrt2\\a=\dfrac{7\sqrt6}{\sqrt2}dm\\\\a=7\sqrt3dm\\\\Ob=4\cdot 7\sqrt3dm\\\boxed{Ob=28\sqrt3dm}[/tex]
d)
[tex]d=2\sqrt3cm\\a\sqrt2=2\sqrt3cm |:\sqrt2\\a=\dfrac{2\sqrt3}{\sqrt2}cm\\\\a=\dfrac{2\sqrt3\cdot \sqrt2}2cm\\a=\sqrt6cm\\\\Ob=4\cdot \sqrt6cm\\\boxed{Ob=4\sqrt6cm}[/tex]
