A piece of paper is in the form of a right angle triangle in which the ratio of base and perpendicular is 3:4 and hypotenuse is 20 cm. What is the volume of the beiggest cone that can be taking right angle vertex of the paper as the vertex of the cone ?

Options

A. a. `45.8 cm^(3) `
B. b. `56.1 cm^(3) `
C. c. `61.5 cm^(3) `
D. d. `48 cm^(3) `

Answer

To solve the problem step by step, we will follow the given information about the right angle triangle and the cone.

### Step 1: Understand the triangle dimensions
The triangle has a base and a height in the ratio of 3:4. Let the base be \(3x\) and the height (perpendicular) be \(4x\). The hypotenuse is given as 20 cm.

### Step 2: Apply the Pythagorean theorem
According to the Pythagorean theorem:
\[ \text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2 \]
Substituting the values we have:
\[ 20^2 = (3x)^2 + (4x)^2 \]
This simplifies to:
\[ 400 = 9x^2 + 16x^2 \]
\[ 400 = 25x^2 \]

### Step 3: Solve for \(x\)
Now, we can solve for \(x\):
\[ x^2 = \frac{400}{25} = 16 \]
Taking the square root:
\[ x = 4 \]

### Step 4: Calculate base and height
Now, we can find the base and height:
- Base = \(3x = 3 \times 4 = 12 \, \text{cm}\)
- Height = \(4x = 4 \times 4 = 16 \, \text{cm}\)

### Step 5: Determine the radius of the cone
The largest cone that can be formed will have its vertex at the right angle vertex of the triangle. The radius of the base of the cone will be equal to the base of the triangle:
\[ \text{Radius} (r) = \frac{\text{Base}}{2} = \frac{12}{2} = 6 \, \text{cm} \]

### Step 6: Calculate the height of the cone
The height of the cone will be equal to the height of the triangle:
\[ \text{Height} (h) = 16 \, \text{cm} \]

### Step 7: Calculate the volume of the cone
The volume \(V\) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting the values we have:
\[ V = \frac{1}{3} \pi (6)^2 (16) \]
\[ V = \frac{1}{3} \pi (36)(16) \]
\[ V = \frac{1}{3} \pi (576) \]
\[ V = 192\pi \, \text{cm}^3 \]

### Step 8: Final answer
Using \(\pi \approx \frac{22}{7}\) for calculation:
\[ V \approx 192 \times \frac{22}{7} \approx 56.57 \, \text{cm}^3 \]

Thus, the volume of the biggest cone that can be formed is approximately \(56.57 \, \text{cm}^3\).