Question

va rog ajutor la acest ex geometrie​

Answer 1

Explicație pas cu pas:

a)

${\bf m(\angle BEC)} = \dfrac{m(BC)}{2} = \dfrac{60^{0}}{2} =\bf 30^{0}$

m(CDE) = m(CD) + m(DE) = 90°+120° = 210°

${\bf m(\angle CAE)} = \dfrac{m(CDE)}{2} = \dfrac{210^{0}}{2} =\bf 105^{0}$

m(BCD) = m(BC) + m(CD) = 60°+90° = 150°

${\bf m(\angle BED)} = \dfrac{m(BCD)}{2} = \dfrac{150^{0}}{2} =\bf 75^{0}$

${\bf m(\angle BOC)} = m(BC) =\bf 60^{0}$

b)

m(AE) = 180°-m(DE) = 180°-120° = 60°

${\bf m(CE;DA)} = \dfrac{m(CD) + m(AE)}{2} = \dfrac{90^{0} + 60^{0}}{2} = \bf 75^{0}$

c)

OB≡OC și ∢BOC = 60° ⇒ ΔBOC este echilateral

r = OB = OC = 4 cm

${\bf \mathcal{P}_{\Delta ABC}} = 3 \cdot OB = 3 \cdot 4 =\bf 12 \ cm$

${\bf \mathcal{A}_{\Delta BOC}} = \dfrac{OB^{2} \cdot \sqrt{3} }{4} = \dfrac{4^{2} \cdot \sqrt{3} }{4} = \bf 4 \sqrt{3} \ {cm}^{2}$

d)

m(AB)+m(BC)+m(CD) = 30°+60°+90° = 180° ⇒ AD este diametru ⇒ ΔADE este dreptunghic

r = OA = 4 cm ⇒ AD = 2×OA = 2×4 = 8 cm

m(∢DAE) = ½×m(DE) = 60° ⇒ m(∢ADE) = 30° ⇒ AE este cateta opusă unghiului de 30°

⇒ AE = 4 cm

T.P: DE² = AD²-AE² = 8²-4² = 32

⇒ DE = 4√3 cm

${\bf \mathcal{P}_{\Delta ADE}} = AD+AE+DE = 8+4+4 \sqrt{3} = \bf 4(3 + \sqrt{3}) \ cm$

${\bf \mathcal{A}_{\Delta ADE}} = \dfrac{AE \cdot DE}{2} = \dfrac{4 \cdot 4 \sqrt{3} }{2} = \bf8 \sqrt{3} \ {cm}^{2}$

e)

r = OD = OE = 4 cm

${\bf \mathcal{P}_{\Delta DOE}} = OD+OE+DE = 4+4+4 \sqrt{3} = \bf 4(2 + \sqrt{3}) \ cm$

m(∢DOE) = m(DE) = 120°

${\bf \mathcal{A}_{\Delta DOE}} = \dfrac{OD \cdot OE \cdot \sin DOE}{2} = \dfrac{4^{2} \cdot \sin 120^{0}}{2} = \dfrac{4^{2} \cdot \dfrac{\sqrt{3}}{2}}{2} = \bf4 \sqrt{3} \ cm^{2}$