Problema cu roșiile
[tex]\it a)\ 120\cdot4=480\ \ell ei\ (\hat\imath \d s i\ propusese\ s\breve a\ ob\c{\it t}in\breve a)\\ \\ \\ b)\ 120-\dfrac{20}{100}\cdot120=120-2\cdot12=120-24=96\ kg\ (au\ r\breve amas\ pentru\ v\hat anzare)\\ \\ \\ 480:96=5\ \ell ei\ (noul\ pre\c{\it t}\ al\ unui\ kg\ de\ ro\d s ii)[/tex]
Ultima problemă
a)
[tex]\it\mathcal{A}_{D EF}=\mathcal{A}_{ABCD}-(\mathcal{A}_{AD E}+\mathcal{A}_{B EF}+\mathcal{A}_{FDC})=12^2-\Big(\dfrac{6\cdot12}{2}+\dfrac{6\cdot8}{2}+\dfrac{4\cdot12}{2}\Big)=\\ \\ \\ =144-(36+24+24)=144-84=60\ cm^2[/tex]
b)
[tex]\it Cu \ teorema \ lui \ Pitagora, \ rezult\breve a:\ ED=6\sqrt5\ cm,\ \ FD=4\sqrt{10}\ cm\\ \\ \\ \mathcal{A}_{D EF}=\dfrac{ED\cdot FD\cdot sin(EDF)}{2} \Rightarrow 60=\dfrac{6\sqrt5\cdot4\sqrt{10}\cdot sin(EDF)}{2} \Rightarrow\\ \\ \\ \Rightarrow60=3\sqrt5\cdot4\sqrt5\cdot\sqrt2\cdot sin(EDF) \Rightarrow 60=60\sqrt2\cdot sin(EDF)\Big|_{:60} \Rightarrow\\ \\ \Rightarrow 1=\sqrt2\cdot sin(EDF) \Rightarrow sin(EDF)=\dfrac{^{\sqrt2)}1}{\ \ \sqrt2}=\dfrac{\sqrt2}{2} \Rightarrow \widehat{EDF}=45^o[/tex]
