Sea
r r r el radio de la circunferencia.
Para hallar el área del triángulo que queda fuera de la circunferencia, podemos hallar el área del triángulo que queda dentro y restarla.
Como
∠ A B O \angle ABO ∠ A B O y
∠ A C O \angle ACO ∠ A C O son ángulos rectos, se tiene
∠ B O C = 120 ∘ \angle BOC = 120^{\circ} ∠ B O C = 12 0 ∘ .
Por lo tanto, el área del sector
O B C OBC O B C es
120 360 ⋅ π r 2 = π r 2 3 . \dfrac{120}{360} \cdot \pi r^2 = \dfrac{\pi r^2}{3}. 360 120 ⋅ π r 2 = 3 π r 2 . Ahora hallamos el área de
△ B O C . \triangle BOC. △ B O C . Usando la fórmula del área de un triángulo con el seno, obtenemos
1 2 sin ( 120 ∘ ) ⋅ r 2 = r 2 3 4 . \dfrac{1}{2} \sin (120^{\circ}) \cdot r^2 = \dfrac{r^2\sqrt{3}}{4}. 2 1 sin ( 12 0 ∘ ) ⋅ r 2 = 4 r 2 3 . Entonces el área del triángulo que queda dentro de la circunferencia es
π r 2 3 − r 2 3 4 = r 2 ( 4 π − 3 3 ) 12 . \dfrac{\pi r^2}{3} - \dfrac{r^2\sqrt{3}}{4} = \dfrac{r^2(4\pi - 3\sqrt{3})}{12}. 3 π r 2 − 4 r 2 3 = 12 r 2 ( 4 π − 3 3 ) . La fracción del triángulo que queda dentro de la circunferencia es
r 2 ( 4 π − 3 3 ) 12 ⋅ 4 3 r 2 3 , \dfrac{r^2(4\pi-3\sqrt3)}{12}\cdot \dfrac{4}{3r^2\sqrt3}, 12 r 2 ( 4 π − 3 3 ) ⋅ 3 r 2 3 4 , que se simplifica a
4 π − 3 3 9 3 . \dfrac{4\pi-3\sqrt3}{9\sqrt3}. 9 3 4 π − 3 3 . La fracción buscada es entonces
1 − 4 π − 3 3 9 3 = 1 − 4 π 3 27 + 1 3 = 4 3 − 4 π 3 27 . \begin{aligned} &1-\dfrac{4\pi-3\sqrt3}{9\sqrt3}\\ &=1-\dfrac{4\pi\sqrt3}{27}+\dfrac13\\ &=\dfrac43-\dfrac{4\pi\sqrt3}{27}. \end{aligned} 1 − 9 3 4 π − 3 3 = 1 − 27 4 π 3 + 3 1 = 3 4 − 27 4 π 3 . Por lo tanto, E es la respuesta correcta.
Let the radius of the circle be
r . r. r . To find the area of the triangle outside of the circle, we can find the area of the triangle inside the circle and subtract it.
We get that
∠ B O C = 120 ∘ \angle BOC = 120^{\circ} ∠ B O C = 12 0 ∘ since
∠ A B O \angle ABO ∠ A B O and
∠ A C O \angle ACO ∠ A C O are right angles.
This means that the area of sector
O B C OBC O B C is
120 360 ⋅ π r 2 = π r 2 3 . \dfrac{120}{360} \cdot \pi r^2 = \dfrac{\pi r^2}{3}. 360 120 ⋅ π r 2 = 3 π r 2 . Now, we need to find the area of
△ B O C . \triangle BOC. △ B O C . Using the formula for the area of a triangle with sine, we get the area to be
1 2 sin ( 120 ∘ ) ⋅ r 2 = r 2 3 4 . \dfrac{1}{2} \sin (120^{\circ}) \cdot r^2 = \dfrac{r^2\sqrt{3}}{4}. 2 1 sin ( 12 0 ∘ ) ⋅ r 2 = 4 r 2 3 . Then the area of the triangle inside the circle is
π r 2 3 − r 2 3 4 = r 2 ( 4 π − 3 3 ) 12 . \dfrac{\pi r^2}{3} - \dfrac{r^2\sqrt{3}}{4} = \dfrac{r^2(4\pi - 3\sqrt{3})}{12}. 3 π r 2 − 4 r 2 3 = 12 r 2 ( 4 π − 3 3 ) . The area of
△ A B C \triangle ABC △ A B C is
( r 3 ) 2 3 4 = 3 r 2 3 4 . \dfrac{(r\sqrt{3})^2\sqrt{3}}{4} = \dfrac{3r^2\sqrt{3}}{4}. 4 ( r 3 ) 2 3 = 4 3 r 2 3 . The fraction of the triangle lying inside the circle is
r 2 ( 4 π − 3 3 ) 12 ⋅ 4 3 r 2 3 , \dfrac{r^2(4\pi-3\sqrt3)}{12}\cdot \dfrac{4}{3r^2\sqrt3}, 12 r 2 ( 4 π − 3 3 ) ⋅ 3 r 2 3 4 , which simplifies to
4 π − 3 3 9 3 . \dfrac{4\pi-3\sqrt3}{9\sqrt3}. 9 3 4 π − 3 3 . Hence the desired fraction is
1 − 4 π − 3 3 9 3 = 1 − 4 π 3 27 + 1 3 = 4 3 − 4 π 3 27 . \begin{aligned} &1-\dfrac{4\pi-3\sqrt3}{9\sqrt3}\\ &=1-\dfrac{4\pi\sqrt3}{27}+\dfrac13\\ &=\dfrac43-\dfrac{4\pi\sqrt3}{27}. \end{aligned} 1 − 9 3 4 π − 3 3 = 1 − 27 4 π 3 + 3 1 = 3 4 − 27 4 π 3 . Thus, E is the correct answer.
