Let x be the area of a square inscribed in a circle of radius r and y be the area of an equilateral triangle inscribed in the same circle. Which one of the following is correct ?

Question

Prepp Team · Accepted Answer

Understanding the Problem: Inscribed Shapes in a Circle

The question asks us to find a relationship between the area of a square and the area of an equilateral triangle, both of which are inscribed in the same circle with radius \(r\). We are given that \(x\) is the area of the square and \(y\) is the area of the equilateral triangle.

Calculating the Area of the Inscribed Square

When a square is inscribed in a circle, its vertices lie on the circle's circumference. The diagonal of the square is equal to the diameter of the circle.

Let the side length of the square be \(s\).
The diagonal of the square is \(s\sqrt{2}\).
The diameter of the circle is \(2r\).
So, \(s\sqrt{2} = 2r\).
Squaring both sides, \((s\sqrt{2})^2 = (2r)^2\), which gives \(2s^2 = 4r^2\).
Dividing by 2, we get \(s^2 = 2r^2\).
The area of the square, \(x\), is \(s^2\).
Therefore, \(x = 2r^2\).

Calculating the Area of the Inscribed Equilateral Triangle

When an equilateral triangle is inscribed in a circle, its vertices lie on the circle. The circle is the circumcircle of the triangle.

Let the side length of the equilateral triangle be \(a\).
For an equilateral triangle with side \(a\), the radius of its circumcircle is given by the formula \(R = \frac{a}{\sqrt{3}}\).
In this case, the circumcircle radius is \(r\). So, \(r = \frac{a}{\sqrt{3}}\).
Solving for \(a\), we get \(a = r\sqrt{3}\).
The area of an equilateral triangle with side \(a\) is given by the formula \(A = \frac{\sqrt{3}}{4}a^2\).
The area of the triangle, \(y\), is \(y = \frac{\sqrt{3}}{4}a^2\).
Substitute the value of \(a\): \(y = \frac{\sqrt{3}}{4}(r\sqrt{3})^2\).
\(y = \frac{\sqrt{3}}{4}(r^2 \times 3) = \frac{3\sqrt{3}}{4}r^2\).

Finding the Relationship Between the Areas

We have the areas in terms of \(r\):

Area of square: \(x = 2r^2\)
Area of equilateral triangle: \(y = \frac{3\sqrt{3}}{4}r^2\)

We want to find a relationship between \(x\) and \(y\) that does not involve \(r\). From the equation for \(x\), we can express \(r^2\) in terms of \(x\):

\(r^2 = \frac{x}{2}\)

Now substitute this expression for \(r^2\) into the equation for \(y\):

\(y = \frac{3\sqrt{3}}{4}\left(\frac{x}{2}\right)\)

\(y = \frac{3\sqrt{3}x}{8}\)

To remove the square root and find a cleaner relationship like the options provided, we can square both sides:

\(y^2 = \left(\frac{3\sqrt{3}x}{8}\right)^2\)

\(y^2 = \frac{(3\sqrt{3})^2 x^2}{8^2}\)

\(y^2 = \frac{(9 \times 3) x^2}{64}\)

\(y^2 = \frac{27 x^2}{64}\)

Multiplying both sides by 64 gives:

\(64y^2 = 27x^2\)

This can also be written as \(27x^2 = 64y^2\).

Comparing with the Given Options

Let's check our derived relationship \(27x^2 = 64y^2\) against the given options:

\(9x^2 = 16y^2\) (Incorrect)
\(27x^2 = 64y^2\) (Correct)
\(36x^2 = 49y^2\) (Incorrect)
\(16x^2 = 21y^2\) (Incorrect)

Our derived relationship matches option 2.

Revision Table: Key Formulas

Shape Inscribed in Circle (Radius \(r\))	Side Length	Area
Square	\(s\) where \(s = r\sqrt{2}\)	\(x = s^2 = 2r^2\)
Equilateral Triangle	\(a\) where \(a = r\sqrt{3}\)	\(y = \frac{\sqrt{3}}{4}a^2 = \frac{3\sqrt{3}}{4}r^2\)

Additional Information: Inscribed Polygons

Understanding how to calculate the area of regular polygons inscribed in a circle is a common geometry topic. Here are some key points:

For any regular \(n\)-sided polygon inscribed in a circle of radius \(r\), the vertices lie on the circle.
Each side of the polygon subtends an angle of \(\frac{360^\circ}{n}\) at the center of the circle.
The polygon can be divided into \(n\) congruent isosceles triangles with two sides equal to the radius \(r\).
The area of each triangle is \(\frac{1}{2} \times r \times r \times \sin\left(\frac{360^\circ}{n}\right) = \frac{1}{2}r^2\sin\left(\frac{360^\circ}{n}\right)\).
The total area of the polygon is \(n \times \frac{1}{2}r^2\sin\left(\frac{360^\circ}{n}\right) = \frac{n}{2}r^2\sin\left(\frac{360^\circ}{n}\right)\).
Let's check this formula for the square (n=4): Area = \(\frac{4}{2}r^2\sin\left(\frac{360^\circ}{4}\right) = 2r^2\sin(90^\circ) = 2r^2 \times 1 = 2r^2\). This matches our result for \(x\).
Let's check this formula for the equilateral triangle (n=3): Area = \(\frac{3}{2}r^2\sin\left(\frac{360^\circ}{3}\right) = \frac{3}{2}r^2\sin(120^\circ) = \frac{3}{2}r^2 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{4}r^2\). This matches our result for \(y\).

These formulas can be useful for calculating the areas of other regular polygons inscribed in a circle.

Prepp Team · Answer

9x2 = 16y2

Prepp Team · Answer

36x2 = 49y2

Prepp Team · Answer

16x2 = 21y2

Understanding the Problem: Inscribed Shapes in a Circle

Calculating the Area of the Inscribed Square

Calculating the Area of the Inscribed Equilateral Triangle

Finding the Relationship Between the Areas

Comparing with the Given Options

Revision Table: Key Formulas

Additional Information: Inscribed Polygons

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