A circle is inscribed in an equilateral triangle and square is inscribed in that circle. The ratio of the areas of the triangle and the square is

Options

A. `sqrt(3) :4`
B. `sqrt(3) :8`
C. `3sqrt(3) : 2`
D. `3sqrt(3) : 1`

Answer

To solve the problem of finding the ratio of the areas of an equilateral triangle and a square inscribed in a circle that is itself inscribed in the triangle, we can follow these steps:

### Step 1: Area of the Equilateral Triangle
The area \( A_T \) of an equilateral triangle with side length \( a \) is given by the formula:
\[ A_T = \frac{\sqrt{3}}{4} a^2 \]

### Step 2: Radius of the Inscribed Circle
The radius \( r \) of the inscribed circle (incircle) of an equilateral triangle can be calculated using the formula:
\[ r = \frac{a}{2\sqrt{3}} \]

### Step 3: Diameter of the Circle
The diameter \( D \) of the inscribed circle is twice the radius:
\[ D = 2r = 2 \left(\frac{a}{2\sqrt{3}}\right) = \frac{a}{\sqrt{3}} \]

### Step 4: Side Length of the Inscribed Square
For a square inscribed in a circle, the diagonal of the square is equal to the diameter of the circle. If \( s \) is the side length of the square, then the relationship between the side length and the diagonal (which is \( D \)) is given by:
\[ D = s\sqrt{2} \]
Thus, we can express \( s \) as:
\[ s = \frac{D}{\sqrt{2}} = \frac{a/\sqrt{3}}{\sqrt{2}} = \frac{a}{\sqrt{6}} \]

### Step 5: Area of the Inscribed Square
The area \( A_S \) of the square is given by:
\[ A_S = s^2 = \left(\frac{a}{\sqrt{6}}\right)^2 = \frac{a^2}{6} \]

### Step 6: Ratio of the Areas
Now, we can find the ratio of the area of the triangle to the area of the square:
\[ \text{Ratio} = \frac{A_T}{A_S} = \frac{\frac{\sqrt{3}}{4} a^2}{\frac{a^2}{6}} \]
Cancelling \( a^2 \) from the numerator and denominator, we have:
\[ \text{Ratio} = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{6}} = \frac{\sqrt{3}}{4} \times 6 = \frac{3\sqrt{3}}{2} \]

### Final Answer
Thus, the ratio of the areas of the triangle to the square is:
\[ \frac{3\sqrt{3}}{2} \]

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