### Teorija

"Ja kvadrāta malas garums ir $$1\ \mathrm{m}$$, tad tāda kvadrāta laukums ir $$1$$ kvadrātmetrs ($$1\ \mathrm{m}^2$$)."
Svarīgi!
$$100\ \mathrm{m}^2 = 1\ \mathrm{a}$$ (ārs)
$$1\ \mathrm{ha}$$ (hektārs) $$= 100\ \mathrm{a} = 10\ 000\ \mathrm{m}^2$$
$1\phantom{\rule{0.147em}{0ex}}\mathrm{ha}\phantom{\rule{0.147em}{0ex}}=100\cdot 100\phantom{\rule{0.147em}{0ex}}{m}^{2}$ (tas ir kvadrāts ar malas garumu $$100\ \mathrm{m}$$)

Piemērs:
 Pārveidojums Risinājums Pamatojums $$2\ \mathrm{ha} = 200\ \mathrm{a}$$ $2\cdot 100\phantom{\rule{0.147em}{0ex}}a$ jo $$1\ \mathrm{ha} = 100\ \mathrm{a}$$ $$3\ \mathrm{m}^2 = 30\ 000\ \mathrm{cm}^2$$ $3\cdot 10\phantom{\rule{0.147em}{0ex}}000\phantom{\rule{0.147em}{0ex}}{\mathrm{cm}}^{2}$ jo $$1\ \mathrm{m} = 100\ \mathrm{cm}$$,$1\phantom{\rule{0.147em}{0ex}}{m}^{2}=100\cdot 100\phantom{\rule{0.147em}{0ex}}{\mathrm{cm}}^{2}$ $$4\$$${m}^{2}$$$= 400\$$${\mathrm{dm}}^{2}$ $4\cdot 100\phantom{\rule{0.147em}{0ex}}{\mathrm{dm}}^{2}$ jo $$1\ \mathrm{m} = 10\ \mathrm{dm}$$,$1\phantom{\rule{0.147em}{0ex}}{m}^{2}=10\cdot 10\phantom{\rule{0.147em}{0ex}}{\mathrm{dm}}^{2}$ $$5\$$${\mathrm{dm}}^{2}$$$= 500\$$${\mathrm{cm}}^{2}$ $5\cdot 100\phantom{\rule{0.147em}{0ex}}{\mathrm{cm}}^{2}$ jo $$1\ \mathrm{dm} = 10\ \mathrm{cm}$$,$1\phantom{\rule{0.147em}{0ex}}{\mathrm{dm}}^{2}=10\cdot 10\phantom{\rule{0.147em}{0ex}}{\mathrm{cm}}^{2}$ $$6\$$${\mathrm{dm}}^{2}$$$= 60\ 000\$$${\mathrm{mm}}^{2}$ $6\cdot 10\phantom{\rule{0.147em}{0ex}}000\phantom{\rule{0.147em}{0ex}}{\mathrm{mm}}^{2}$ jo $$1\ \mathrm{dm} = 100\ \mathrm{mm}$$,$1\phantom{\rule{0.147em}{0ex}}{\mathrm{dm}}^{2}=100\cdot 100\phantom{\rule{0.147em}{0ex}}{\mathrm{mm}}^{2}$

Atsauce:
Matemātika 5.klasei/ Jānis Mencis (sen.), Jānis Mencis (jun.). Rīga: Zvaigzne ABC, 2008.- 288 lpp.- izmantotā literatūra: 68. lpp.