The volume of the greatest cone obtained by rotating a right - angled triangle of hypotenuse 2 units about a side is `(kpi)/(9sqrt3)` cubic units, then the value of k is equal to

Answer

To solve the problem, we need to find the volume of the greatest cone obtained by rotating a right-angled triangle with a hypotenuse of 2 units about one of its sides. We will follow these steps:

### Step 1: Define the triangle
Let the right-angled triangle have vertices A, B, and C, where:
- A is the right angle,
- AB is one leg (height, h),
- AC is the other leg (radius, r),
- BC is the hypotenuse.

Given that the hypotenuse BC = 2 units, we can denote:
- AO = h (height),
- OC = r (radius).

### Step 2: Use the Pythagorean theorem
According to the Pythagorean theorem:
\[ AB^2 + AC^2 = BC^2 \]
Substituting the values:
\[ h^2 + r^2 = 2^2 \]
\[ h^2 + r^2 = 4 \]

### Step 3: Express r in terms of h
From the equation \( h^2 + r^2 = 4 \), we can express \( r^2 \) as:
\[ r^2 = 4 - h^2 \]

### Step 4: Write the volume of the cone
The volume \( V \) of the cone formed by rotating the triangle about side AB is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( r^2 \):
\[ V = \frac{1}{3} \pi (4 - h^2) h \]
\[ V = \frac{1}{3} \pi (4h - h^3) \]

### Step 5: Differentiate the volume with respect to h
To find the maximum volume, we differentiate \( V \) with respect to \( h \):
\[ \frac{dV}{dh} = \frac{1}{3} \pi (4 - 3h^2) \]

### Step 6: Set the derivative to zero
Setting the derivative equal to zero to find critical points:
\[ 4 - 3h^2 = 0 \]
\[ 3h^2 = 4 \]
\[ h^2 = \frac{4}{3} \]
\[ h = \frac{2}{\sqrt{3}} \]

### Step 7: Find r using h
Now, substitute \( h \) back to find \( r \):
\[ r^2 = 4 - h^2 = 4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3} \]
\[ r = \frac{2\sqrt{2}}{\sqrt{3}} \]

### Step 8: Calculate the volume
Now substitute \( r \) and \( h \) back into the volume formula:
\[ V = \frac{1}{3} \pi \left(\frac{8}{3}\right) \left(\frac{2}{\sqrt{3}}\right) \]
\[ V = \frac{16\pi}{9\sqrt{3}} \]

### Step 9: Identify k
The volume is given as \( \frac{k \pi}{9 \sqrt{3}} \). Comparing:
\[ \frac{16\pi}{9\sqrt{3}} = \frac{k \pi}{9\sqrt{3}} \]
Thus, \( k = 16 \).

### Final Answer
The value of \( k \) is **16**.