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Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24% B. 30% C. 36% D. 42% E. 48%

If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):

Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.

This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24% B. 30% C. 36% D. 42% E. 48%

If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):

Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.

This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)

The closest answer is 42%.

Answer: D

Bunuel,

I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24% B. 30% C. 36% D. 42% E. 48%

If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):

Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.

This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)

The closest answer is 42%.

Answer: D

Bunuel,

I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24% B. 30% C. 36% D. 42% E. 48%

If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):

Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.

This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)

The closest answer is 42%.

Answer: D

Bunuel,

I am confused at the step in which you derive the height side of the first 30-60-90 triangle. the "x" side is r/2, because the "2x" side is r, but why is "x sqrt 3" side not = to (r sqrt 3)/2? instead you have it divided by 4? I must be missing something. You also say that side AX = XB? That can't be because we just found it to be a 30-60-90 right triangle. You must've meant to say that AX = AX, thus we know AC?

that link was helpful, but now I'm just trying to tie it back in with your slightly alternative explanation above. Still, i cannot see why you say that AX = \(\frac{\sqrt{3}r}{4}\) because we have an equilateral triangle (AOB), with sides R, and height AX. The formula states that the height AX would be \(\frac{\sqrt{3}r}{2}\). Please advise.

Approximately what percent of the area of the circle shown is shaded, if polygon ABCDEF is a regular hexagon?

A. 24% B. 30% C. 36% D. 42% E. 48%

If polygon \(ABCDEF\) is a regular hexagon, then the figure is very symmetrical and actually looks like the drawing. We should avoid rigorous proofs and instead make quick arguments "from symmetry" - that is, recognizing that many parts of the diagram are equivalent.

We know that for the hexagon, each side is the same length and each interior angle is the same. Since the interior angles of a hexagon sum up to \(180(n - 2)^\circ = 180(6 - 2)^\circ = 720^\circ\), and there are 6 interior, equal angles, then each of those angles must measure \(\frac{720^\circ}{6} = 120^\circ\).

Moreover, each side of the hexagon is equal in length to the radius, since any regular hexagon can be chopped up into 6 smaller equilateral triangles, as shown by the lines in blue through the circle's center \(O\):

Consider small triangle \(AXB\). The hypotenuse of this right triangle, \(AB\), has length \(r\). Angle \(ABO\) is \(60^\circ\), so \(AXB\) is a 30-60-90 triangle. This means that \(XB\) is \(\frac{r}{2}\) in length, and \(AX\) is \(\frac{\sqrt{3}r}{4}\) in length. Since \(AX\) and \(XB\) are the same length (by symmetry), \(AC\) is \(\sqrt{3}r\) in length.

This means that the area of triangle \(ABC\) is \(\frac{1}{2} \leftarrow( \sqrt{3}r \rightarrow) \leftarrow( \frac{r}{4} \rightarrow) = \frac{\sqrt{3} r^2}{4}\). Since there are three shaded triangles in all (including \(ABC\)), the total shaded area is \(\frac{3 \sqrt{3} r^2}{4}\). The area of the circle is , so the ratio of the shaded area to the area of the circle is given by \(\frac{3 \sqrt{3} r^2}{4} \div \pi r^2 = \frac{3 \sqrt{3}}{4 \pi} \approx \frac{\sqrt{3}}{4} \approx \frac{1.7}{4} \approx 0.43\)

The closest answer is 42%.

Answer: D

Hey Bunuel, I think the highlighted part above should be ax=xc. and that is why it is creating a bit of confusion.

I think this is a high-quality question and I agree with explanation. There is a typo. AX equals (√3*r)/2, not (√3*r)/4 as it mentioned in the second line below the picture in the explanation. The rest is correct.