Question

ajutor...............​

Answer 1

Răspuns:

Explicație pas cu pas:

Ai desenul atasat

a. ΔAOM dreptunghic in A

∡O=60°⇒∡M=30°⇒ MO=2AO

MO=12 cm

b. AO=OB (raze)

m(∡AOB)=60°⇒ ΔAOB echilateral ⇒ $A_{AOB}=\frac{l^{2\sqrt{3} } }{4} =9\sqrt{3} cm^{2}$

In ΔAOM aplicam Pitagora

MO²=AO²+AM²

AM²=144-36=108

AM=6√3

$A_{AOM=\frac{6\cdot 6\sqrt{3} }{2} =18\sqrt{3}$

$A_{ABM}=A_{AOM}-A_{AOB}=18\sqrt{3}- 9\sqrt{3} =9\sqrt{3} cm^{2}$