Rectangle of maximum area that can be inscribed in an equilateral triangle of side awill have area=

Options

A. `(a^2sqrt3)/2`
B. `(a^2sqrt3)/4`
C. `(a^2 sqrt3)/8`
D. None

Answer

To find the area of the rectangle of maximum area that can be inscribed in an equilateral triangle of side length \( A \), we will follow these steps:

### Step 1: Understand the Geometry
We start with an equilateral triangle \( ABC \) with each side of length \( A \). We will inscribe a rectangle \( A'B'C'D' \) such that its base lies on side \( BC \) and its top vertices touch the sides \( AB \) and \( AC \).

### Step 2: Set Up the Coordinates
Let’s place the triangle in the coordinate system:
- Point \( A(0, \frac{\sqrt{3}}{2}A) \)
- Point \( B(-\frac{A}{2}, 0) \)
- Point \( C(\frac{A}{2}, 0) \)

### Step 3: Define the Rectangle
Let the height of the rectangle be \( h \) and the width be \( w \). The rectangle's vertices touching \( AB \) and \( AC \) will have coordinates based on the height \( h \).

### Step 4: Relate Height and Width
Using the properties of the triangle, the height \( h \) of the rectangle can be expressed in terms of \( w \):
- The height of the triangle from point \( A \) to line \( BC \) is \( \frac{\sqrt{3}}{2}A \).
- The height of the rectangle will be \( h = \frac{\sqrt{3}}{2}A - \frac{\sqrt{3}}{3}w \) (using the slope of the sides).

### Step 5: Area of the Rectangle
The area \( A_r \) of the rectangle can be expressed as:
\[ A_r = w \cdot h = w \left( \frac{\sqrt{3}}{2}A - \frac{\sqrt{3}}{3}w \right) \]
\[ A_r = \frac{\sqrt{3}}{2}Aw - \frac{\sqrt{3}}{3}w^2 \]

### Step 6: Maximize the Area
To find the maximum area, we take the derivative of \( A_r \) with respect to \( w \) and set it to zero:
\[ \frac{dA_r}{dw} = \frac{\sqrt{3}}{2}A - \frac{2\sqrt{3}}{3}w = 0 \]
Solving for \( w \):
\[ \frac{\sqrt{3}}{2}A = \frac{2\sqrt{3}}{3}w \]
\[ w = \frac{3A}{4} \]

### Step 7: Find the Corresponding Height
Substituting \( w \) back into the height equation:
\[ h = \frac{\sqrt{3}}{2}A - \frac{\sqrt{3}}{3}\left(\frac{3A}{4}\right) \]
\[ h = \frac{\sqrt{3}}{2}A - \frac{\sqrt{3}}{4}A = \frac{\sqrt{3}}{4}A \]

### Step 8: Calculate the Maximum Area
Now substituting \( w \) and \( h \) into the area formula:
\[ A_r = w \cdot h = \left(\frac{3A}{4}\right) \left(\frac{\sqrt{3}}{4}A\right) \]
\[ A_r = \frac{3\sqrt{3}}{16}A^2 \]

### Final Result
Thus, the area of the rectangle of maximum area that can be inscribed in an equilateral triangle of side \( A \) is:
\[ \boxed{\frac{3\sqrt{3}}{16}A^2} \]

---