[tex]\it 1.\ \ \mathcal{P}=4\cdot \ell=4\cdot10=40\ cm \\ \\ \mathcal{A}=\ell^2=10^2=100\ cm^2\\ \\ Diagonala=\ell\sqrt2=10\sqrt2\ cm[/tex]
2.
[tex]\it \mathcal{P}=2(L+\ell)=2(20+16)=2\cdot36=72\ cm\\ \\ \mathcal{A}=L\cdot\ell=20\cdot16=320\ cm^2[/tex]
Notăm dreptunghiul ABCD, cu AB = 20cm, BC=16cm.
Cu teorema lui Pitagora vom determina diagonala AC, din
triunghiul ABC, dreptunghic în B.
[tex]\it AC^2=AB^2+BC^2=20^2+16^2=400+256=656 \Rightarrow AC=\sqrt{656}=\\ \\ =\sqrt{16\cdot41}=4\sqrt{41}\ cm[/tex]
[tex]\it 4.\ \ \mathcal{A} =\ell\cdot h=10\cdot6=60\ cm^2[/tex]
[tex]\it 5.\ \ \mathcal{A}=\dfrac{\ell^3\sqrt3}{4}=\dfrac{8^2\sqrt3}{4}=\dfrac{8\cdot8\sqrt3}{4}=2\cdot8\sqrt3=16\sqrt3\ cm^2[/tex]
[tex]\it\ h=\dfrac{\ell\sqrt3}{2}=\dfrac{8\sqrt3}{2}=4\sqrt3\ cm[/tex]
