a)
Pole rombu obliczamy korzystając ze wzoru
[tex]P=\frac{1}{2}\cdot e\cdot f[/tex] e,f - przekątne rombu
[tex]e=2\sqrt{2}\\\\f=4\sqrt{3}\\\\P=\frac{1}{\not2}\cdot\not2\sqrt{2}\cdot4\sqrt{3}=\sqrt{2}\cdot4\sqrt{3}=4\sqrt{2\cdot3}=4\sqrt{6}\\\\Odp. Pole\ \ rombu\ \ wynosi\ \ 4\sqrt{6}\\\\\\b)\\\\Obliczamy\ \ ile\ \ wynosi\ \ dlugo\'s\'c\ \ boku\ \ tego\ \ kwadratu\\\\P=a^2\\\\a^2=20\\\\a=\sqrt{20}=\sqrt{4\cdot5}=2\sqrt{5}\ \ cm\\\\Obliczamy\ \ obw\'od\ \ kwadratu\\\\Obw=4a\\\\Obw=4\cdot2\sqrt{5}=8\sqrt{5}cm\\\\Odp.\ \ Obw\'o\ \ kwadratu\ \ wynosi\ \ 8\sqrt{5}cm[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
Verified answer
a)
Pole rombu obliczamy korzystając ze wzoru
[tex]P=\frac{1}{2}\cdot e\cdot f[/tex] e,f - przekątne rombu
[tex]e=2\sqrt{2}\\\\f=4\sqrt{3}\\\\P=\frac{1}{\not2}\cdot\not2\sqrt{2}\cdot4\sqrt{3}=\sqrt{2}\cdot4\sqrt{3}=4\sqrt{2\cdot3}=4\sqrt{6}\\\\Odp. Pole\ \ rombu\ \ wynosi\ \ 4\sqrt{6}\\\\\\b)\\\\Obliczamy\ \ ile\ \ wynosi\ \ dlugo\'s\'c\ \ boku\ \ tego\ \ kwadratu\\\\P=a^2\\\\a^2=20\\\\a=\sqrt{20}=\sqrt{4\cdot5}=2\sqrt{5}\ \ cm\\\\Obliczamy\ \ obw\'od\ \ kwadratu\\\\Obw=4a\\\\Obw=4\cdot2\sqrt{5}=8\sqrt{5}cm\\\\Odp.\ \ Obw\'o\ \ kwadratu\ \ wynosi\ \ 8\sqrt{5}cm[/tex]