a) AB = BC = CF = 10cm
[tex]\it \mathcal{A}_{ACF} =\dfrac{CF\cdot AB}{2}=\dfrac{10\cdot10}{2}=50\ cm^2[/tex]
b) Cu teorema lui Pitagora în ΔABF, dreptunghic în B, ⇒ AF = 10√5 cm
[tex]\it \mathcal{A_{AFC}}=\dfrac{AF\cdot d(C,\ AF)}{2} \Rightarrow 50=\dfrac{10\sqrt5\cdot d}{2} \Rightarrow 100=10\sqrt5\cdot d \Rightarrow\\ \\ \\ \Rightarrow 10=\sqrt5\cdot d \Rightarrow 2\cdot5=\sqrt5\cdot d \Rightarrow 2\cdot\sqrt5\cdot\sqrt5=\sqrt5\cdot d \Rightarrow d=2\sqrt5cm[/tex]
c) AC = 10√2 cm (diagonala pătratului)
[tex]\it \mathcal{A}_{AFC} = \dfrac{AC\cdot AF\cdot sin(FAC)}{2} \Rightarrow 50=\dfrac{10\sqrt2\cdot10\sqrt5\cdot sin(FAC)}{2} \Rightarrow \\ \\ \\ \Rightarrow 100=100\sqrt{10}\cdot sin(FAC) \Rightarrow 1=\sqrt{10}\cdot sin(FAC) \Rightarrow \\ \\ \\ \Rightarrow sin(FAC)=\dfrac{^{\sqrt{10})}1}{\sqrt{10}}=\dfrac{\sqrt{10}}{10}[/tex]
