Question

Ofer coroană!! Vă rog ajutați-mă!​

Answer 1

a)

$\it AB=BC=CD=AD=12\ cm;\ \ AE=EB=12:2=6cm;\\ \\ BF=2CF\Big|_{+CF} \Rightarrow BF+CF=3CF \Rightarrow BC=3CF\Big|_{:3} \Rightarrow CF=BC:3=4cm\\ \\ \\ BF=2CF=2\cdot4=8cm$

$\it \mathcal{A}_{D EF}=\mathcal{A}_{ABCD}-(\mathcal{A}_{ADE}+\mathcal{A}_{BEF}+\mathcal{A}_{CFD})=\\ \\ \\ =12^2-\Big(\dfrac{6\cdot12}{2}+\dfrac{6\cdot8}{2}+\dfrac{4\cdot12}{2}\Big)=144-(36+24+24)=144-84=60cm^2$

b)

$\it Cu \ \ teorema \ \ lui\ \ Pitagora \Rightarrow DE=6\sqrt5\ cm;\ \ DF=4\sqrt10\ cm.\\ \\ \\\mathcal{A}_{D EF}=\dfrac{DE\cdot DF\cdot sin (EDF)}{2} \Rightarrow 60=\dfrac{6\sqrt5\cdot4\sqrt{10}\cdot sin(EDF)}{2} \Rightarrow \\ \\ \\ \Rightarrow 60=12\sqrt{50}\cdot sin(EDF) \Rightarrow sin(EDF)=\dfrac{60}{12\sqrt{50}}=\dfrac{5}{5\sqrt2}=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}\\ \\ \\ Dar,\ \ sin45^o=\dfrac{\sqrt2}{2},\ \ deci\ \ \widehat{EDF}=45^o\ .$

Answer 2

Răspuns:

a) 60 cm²; b) 45°

Explicație pas cu pas:

ABCD este pătrat => AB=BC=CD=AD = 12 cm

AE ≡ EB

AE = EB = ½×AB = ½× 12

=> AE = 6 cm, EB = 6 cm

BF = 2CF => CF = ⅓×BC = ⅓×12

=> CF = 4 cm

BF = 12 - 4 => BF = 8 cm

Aria(DEF) = Aria(ABCD) - Aria(AED) - Aria(BEF) - Aria(CDF) = AB² - ½×AE×AD - ½×BE×BF - ½×DC×CF = 12² - ½×(6×12 + 6×8 + 12×4) = 144 - ½×(72+48+48) = 144 - ½×168 = 144 - 84 = 60

=> Aria(DEF) = 60 cm²

b)

T.P. în ΔADE:

DE² = AD² + AE² = 12² + 6² = 180

=> DE = 6√5 cm

T.P. în ΔCDF:

DF² = DC² + CF² = 12² + 4² = 160

=> DF = 4√10 cm

Aria(DEF) = ½×DE×DF×sin(∢EDF)

60 = ½×6√5×4√10×sin(∢EDF)

60 = 60√2×sin(∢EDF)

=> sin(∢EDF) = 1/√2 => ∢EDF = 45°

q.e.d.