Understanding Shapes Inscribed in a Circle

This question asks for the ratio of the areas of two different shapes, a square and a regular hexagon, when both are drawn inside the same circle such that all their vertices touch the circle's circumference. This is what it means for a shape to be 'inscribed' in a circle.

Calculating the Area of an Inscribed Square

Let's consider a circle with radius \(R\). When a square is inscribed in this circle, the diagonal of the square is equal to the diameter of the circle, which is \(2R\).

Let the side length of the square be \(s\). Using the Pythagorean theorem for a right-angled triangle formed by two sides and a diagonal of the square:

\[s^2 + s^2 = (2R)^2\]

\[2s^2 = 4R^2\]

\[s^2 = 2R^2\]

The area of the square is given by \(s^2\). So, the area of the inscribed square is \(2R^2\).

Calculating the Area of an Inscribed Regular Hexagon

A regular hexagon inscribed in a circle can be divided into 6 congruent equilateral triangles, where each vertex of the triangles is at the center of the circle and two vertices are on the circumference. The side length of each equilateral triangle is equal to the radius \(R\) of the circle.

The area of one equilateral triangle with side length \(a\) is given by the formula \(\frac{\sqrt{3}}{4}a^2\). In our case, the side length is \(R\), so the area of one equilateral triangle is \(\frac{\sqrt{3}}{4}R^2\).

Since the regular hexagon is made up of 6 such equilateral triangles, the total area of the hexagon is 6 times the area of one triangle:

\[\text{Area of Hexagon} = 6 \times \frac{\sqrt{3}}{4}R^2\]

\[\text{Area of Hexagon} = \frac{6\sqrt{3}}{4}R^2\]

\[\text{Area of Hexagon} = \frac{3\sqrt{3}}{2}R^2\]

Finding the Ratio of Areas

We need to find the ratio of the area of the square to the area of the regular hexagon. Both shapes are inscribed in the same circle, so they share the same radius \(R\).

\[\text{Ratio} = \frac{\text{Area of Square}}{\text{Area of Regular Hexagon}}\]

\[\text{Ratio} = \frac{2R^2}{\frac{3\sqrt{3}}{2}R^2}\]

We can cancel out the \(R^2\) terms, as \(R\) is a common factor and not zero:

\[\text{Ratio} = \frac{2}{\frac{3\sqrt{3}}{2}}\]

To simplify the fraction, multiply the numerator by the reciprocal of the denominator:

\[\text{Ratio} = 2 \times \frac{2}{3\sqrt{3}}\]

\[\text{Ratio} = \frac{4}{3\sqrt{3}}\]

So, the ratio of the areas of the square and the regular hexagon is \(4 : 3\sqrt{3}\).

Final Ratio Calculation

Area of Square : Area of Hexagon \( = 2R^2 : \frac{3\sqrt{3}}{2}R^2 \)

\( = 2 : \frac{3\sqrt{3}}{2} \)

Multiply by 2:

\( = 4 : 3\sqrt{3} \)

Revision Table: Inscribed Shape Areas

Additional Information: Polygons Inscribed in a Circle

When any regular polygon is inscribed in a circle, its vertices lie on the circumference. The center of the circle is also the center of the regular polygon.

• For an inscribed regular n-sided polygon, the polygon can be divided into \(n\) congruent isosceles triangles. The two equal sides of each triangle are the radii of the circle (\(R\)), and the third side is a side of the polygon.
• For specific regular polygons like squares and hexagons, there are simpler ways to find the area in terms of the circle's radius.
• Understanding how to relate the polygon's dimensions (side length, apothem) to the circle's radius is key to solving such problems. For a regular n-gon with side length \(s\) and apothem \(a\), inscribed in a circle of radius \(R\), we have relationships like \(s = 2R \sin(\pi/n)\) and \(a = R \cos(\pi/n)\). The area is also \(\frac{1}{2} \times \text{perimeter} \times \text{apothem}\) or \(\frac{1}{2} n R^2 \sin(2\pi/n)\).
• For a square (n=4), \(s = 2R \sin(\pi/4) = 2R (\sqrt{2}/2) = R\sqrt{2}\). Area = \(s^2 = (R\sqrt{2})^2 = 2R^2\).
• For a regular hexagon (n=6), \(s = 2R \sin(\pi/6) = 2R (1/2) = R\). Area = \(\frac{1}{2} \times 6s \times R \cos(\pi/6) = 3R \times R (\sqrt{3}/2) = \frac{3\sqrt{3}}{2}R^2\). These results match the geometric methods used above.