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Explicație pas cu pas:

1) ABC triunghi echilateral:

[tex]\ell = 4 \ \ cm \\ \implies P = 3 \cdot \ell = 3 \cdot 4 = \bf 12 \ \ cm[/tex]

2) OC bisectoare:

[tex]m(∢COB) = \frac{1}{2} \cdot m(∢AOB) = \frac{140}{2} \\ \implies \bf m(∢COB) = 70 \degree[/tex]

3) AB = 6 cm, m(∢B) = 30°, Aria = 18 cm²

[tex]Aria = \frac{AB \cdot BC \cdot sin \angle B}{2} \iff \\ 18 = \frac{6 \cdot BC \cdot \frac{1}{2} }{2} \implies BC = \bf 12 \: cm[/tex]

4) ABCDEF hexagon regulat, AB = 12 cm

AB ≡ OF => OF = 12 cm

FC = 2×OF = 2×12 = 24 => FC = 24 cm

5) ABCD dreptunghi, AC = 2BC

AC = 2×OC => OC ≡ BC ≡ OB

=> ΔBOC echilateral

=> m(∢BOC) = 60°

6) ABCD pătrat, aria = 72 cm²

[tex]aria = {\ell}^{2} \implies \ell = 6 \sqrt{2} \ \ cm[/tex]

[tex]d = \ell \sqrt{2} = 6 \sqrt{2} \cdot \sqrt{2} = \bf 12 \: cm[/tex]

1.∆echilateral P=3×latura=3×4=12cm d)

2.AOB=140° AOC=BOC

COB=140/2=70°. c)

3.∆ABC AB=6cm <B=30°

Aria∆ABC=18cm²=BC×h/2

h=AB/2=6/2=3cm conf T sin

BC=18×2/3=12 cm. b)

4.AB=12cm la hexagon raza =latura=12cm

diamentrul =24cm b)

5.AC=2BC ∆ABC dreptunghic =>T sin

A=30°

BOC echilateral =>BOC=60°. c)

6 .aria 4=72cm

latura√72=6√2cm

diagonala=latura √2=6√2×√2=12cm. a)