Question

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Answer 1

1. m(∡A)=m(∡D)=(360°-2*120°)/2=120/2⇒m(∡A)=60°

2.AD=AE+EF+FD, AE=FD iar EF=BC=20m

In ΔFCD m(∡FCD)=180°-(90°+60°)=180-150⇒m(∡FCD)=30°⇒FD=DC/2⇒FD=10m

Deci AD=2*10+20⇒AD=40m

3. BE=CF, ΔFCD (∡F=90°) ⇒CF²=CD²-FD²⇒CF²=20²-10²=400-100=300⇒ CF=√300⇒CF=BE=10√3m

4. Suprafata zonei B este aria dreptunghiului BCFE = BC*CF = 20*10√3 = 200√3m²

5. Ultimul rand BC are lungimea de 20 m=2000 cm

Daca un scaun are latimea de 50 cm, pe ultimul rand vor incapea 2000/50=40 scaune

6. Suprafata zonei A: ΔAEB

Suprafata zonei B: ΔFDC ⇒ AriaΔAEB=AriaΔFDC= FD*FC/2 = 10*10√3/2 =50√3m²

7.In ΔBED(∡E=90°)⇒BD²=BE²+ED²⇒BD²=(10√3)²+30²⇒BD²=300+900=1200 ⇒BD²=1200⇒BD=√1200⇒BD=20√3m

8. ΔABD are laturile: AB=20m, BD=20√3, AD=40m, observam ca AD²=AB²+BD²⇒ΔABD dreptunghic (conform reciprocei teoremei lui Pitagora)⇒m(∡ABD)=90°

9. m(∡EBD)=m(∡ABD)-m(∡ABE)=90°-30°⇒ m(∡EBD)=60°⇒ sin(∡EBD)= sin60°=√3/2

10.tg(∡ADB)=tg(∡EDB)=EB/ED=10√3/30=√3/3⇒tg(∡ADB)=30°