Respuesta:
Explicación paso a paso:
1. Ao = R² = (2[tex]\sqrt{2}[/tex]u)²
Ao = 8 u²
2. AB = diámetro = 2R = 12 cm
radio R = 6 cm
Ao = R² = (6 cm)²
Ao = 36 cm²
3. área del sector circular
Asector = [tex]\frac{\pi. R^{2}.\alpha }{360\º}[/tex] α = 120º
As = π(6 cm)² 120/360
As = 12π cm²
5. A = π(R² - r²)
A = π( (10 cm)² - (6 cm)²)
A = 54π cm²
6. sector AOB = 90º
Asector = [tex]\frac{\pi. R^{2}.\alpha }{360\º}[/tex] sector AOB = 90º
As = π(12 cm)² 90º/360º
As = 36π cm²
7. Ao = πR² = 100π m²
R² = 100 m²
R = 10 m
D = 2R = 2*10 m
D = 20 m
8. AC hipotenusa, aplica Teorema de Pitágoras
AC² = AB² + BC²
AC² = (6 cm)² + (8 cm)²
AC² = 36 cm² + 64 cm² = 100 cm²
AC = 10 cm = diámetro
2R = 10 cm
R = 5 cm
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Verified answer
Respuesta:
Explicación paso a paso:
1. Ao = R² = (2[tex]\sqrt{2}[/tex]u)²
Ao = 8 u²
2. AB = diámetro = 2R = 12 cm
radio R = 6 cm
Ao = R² = (6 cm)²
Ao = 36 cm²
3. área del sector circular
Asector = [tex]\frac{\pi. R^{2}.\alpha }{360\º}[/tex] α = 120º
As = π(6 cm)² 120/360
As = 12π cm²
5. A = π(R² - r²)
A = π( (10 cm)² - (6 cm)²)
A = 54π cm²
6. sector AOB = 90º
Asector = [tex]\frac{\pi. R^{2}.\alpha }{360\º}[/tex] sector AOB = 90º
As = π(12 cm)² 90º/360º
As = 36π cm²
7. Ao = πR² = 100π m²
R² = 100 m²
R = 10 m
D = 2R = 2*10 m
D = 20 m
8. AC hipotenusa, aplica Teorema de Pitágoras
AC² = AB² + BC²
AC² = (6 cm)² + (8 cm)²
AC² = 36 cm² + 64 cm² = 100 cm²
AC = 10 cm = diámetro
2R = 10 cm
R = 5 cm