Naloge

Morda opaziš pitagorejsko trojico $3$, $4$, $5$ ali pa višino izračunaš.

$v_c=\sqrt{a^2-(\frac{c}{2})^2}$

$v_c=\sqrt{(5\,\rm{cm})^2-(\frac{8}{2}\,\rm{cm})^2}$

$v_c=\sqrt{25\,\rm{cm}^2-(4\,\rm{cm})^2}$

$v_c=\sqrt{25\,\rm{cm}^2-16\,\rm{cm}^2}$

$v_c=\sqrt{9\,\rm{cm}^2}$

$v_c=3 \,\rm{cm}$

Tudi natančna meritev ob natančnem načrtovanju da dolžino višine na osnovnico $3 \,\rm{cm}$.

$a=\sqrt{v^2+(\frac{c}{2})^2}$

$a=\sqrt{(24\,\rm{dm})^2+(\frac{20}{2}\,\rm{dm})^2}$

$a=\sqrt{576\,\rm{dm}^2+100\,\rm{dm}^2}$

$a=\sqrt{676\,\rm{dm}^2}$

$a=26 \,\rm{dm}$

$\frac{c}{2}=\sqrt{a^2-v^2}$

$\frac{c}{2}=\sqrt{(29\,\rm{dm})^2-(20\,\rm{dm})^2}$

$\frac{c}{2}=\sqrt{441\,\rm{dm}^2}$

$\frac{c}{2}=21 \,\rm{dm}$

$c=2\cdot 21 \,\rm{dm}=42 \,\rm{dm}$