A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are `m` and `n`, respectively, then `m + n^2` is equal to

Options

A. 408
B. 396
C. 312
D. 414

Answer

To solve the problem, we need to find the values of \( m \) (the area of the square inscribed in the circle) and \( n \) (the perimeter of the square), and then compute \( m + n^2 \).

### Step 1: Calculate the inradius of the equilateral triangle
The formula for the inradius \( r \) of an equilateral triangle with side length \( a \) is given by:

\[ r = \frac{a \sqrt{3}}{6} \]

Given that the side length \( a = 12 \):

\[ r = \frac{12 \sqrt{3}}{6} = 2 \sqrt{3} \]

### Step 2: Determine the diameter of the inscribed circle
The diameter \( D \) of the circle inscribed in the triangle is twice the inradius:

\[ D = 2r = 2 \times 2\sqrt{3} = 4\sqrt{3} \]

### Step 3: Relate the diameter to the side of the inscribed square
The diagonal \( d \) of the inscribed square is equal to the diameter of the circle:

\[ d = 4\sqrt{3} \]

### Step 4: Calculate the side length of the square
The relationship between the side length \( s \) of the square and its diagonal \( d \) is given by:

\[ d = s\sqrt{2} \]

Thus, we can find \( s \):

\[ s\sqrt{2} = 4\sqrt{3} \implies s = \frac{4\sqrt{3}}{\sqrt{2}} = 4\sqrt{\frac{3}{2}} = 4 \cdot \frac{\sqrt{6}}{2} = 2\sqrt{6} \]

### Step 5: Calculate the area \( m \) of the square
The area \( m \) of the square is given by:

\[ m = s^2 = (2\sqrt{6})^2 = 4 \cdot 6 = 24 \]

### Step 6: Calculate the perimeter \( n \) of the square
The perimeter \( n \) of the square is given by:

\[ n = 4s = 4 \cdot 2\sqrt{6} = 8\sqrt{6} \]

### Step 7: Calculate \( n^2 \)
Now we compute \( n^2 \):

\[ n^2 = (8\sqrt{6})^2 = 64 \cdot 6 = 384 \]

### Step 8: Calculate \( m + n^2 \)
Finally, we find \( m + n^2 \):

\[ m + n^2 = 24 + 384 = 408 \]

### Final Answer
Thus, the value of \( m + n^2 \) is:

\[ \boxed{408} \]