Question

Intr-un trapez dreptunghic, baza mare=12,baza mica=8 si unghiul ABC=60°. Aflati perimetrul si aria.​

Answer 1

ducem CE ⊥ AB

EB = AB - DC = 12 - 8 = 4 cm

cosABC = EB/BC

cos60 = 4/BC

1/2 = 4/BC

BC = 8cm

Conform Pitagora

CE = √(CB² - EB²) = √(64 - 16) = √48 = 4√3 cm

CE = DA = 4√3 cm

Perimetrul = AB+BC+CD+DA = 12+8+8+4√3 = 28+4√3 = 4(7 + √3) cm

Aria = (AB+CD)/2 × DA = (12+8)/2 × 4√3 = 40√3 cm²

Answer 2

$\it ABCD-tapez\ dreptunghic,\ AB||CD,\ AB>CD,\ DA\perp AB,\ m(\hat B)=60^o\\ \\ Ducem\ CF\perpAB,\ F\in AB\\ \\ AFCD-dreptunghi\ \Rightarrow AF=CD=8cm \Rightarrow FB=12-8=4cm\\ \\ Din\ \Delta CFB \Rightarrow m(\widehat{ BCF}) =30^o\ (complementul\ lui\ 60^o) \stackrel{T.\angle30^o}{\Longrightarrow} \\ \\ \Rightarrow BC=2BF=2\cdot4=8cm$

Cu teorema lui Pitagora, în triunghiul FBC, obținem:

$\it CF = 4\sqrt3\Rightarrow DA=CF=4\sqrt3cm\\ \\ \mathcal{P}=AB+BC+CD+DA=12+8+8+4\sqrt3=28+4\sqrt3cm\\ \\ \mathcal{A}=\dfrac{AB+CD}{2}\cdot DA= \dfrac{12+8}{2}\cdot4\sqrt3=10\cdot4\sqrt3=40\sqrt3cm^2$