A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed.

Answer

To solve the problem of finding the volume and surface area of the double cone formed by revolving a right triangle with sides 3 cm and 4 cm about its hypotenuse, we will follow these steps:

### Step 1: Calculate the length of the hypotenuse
Using the Pythagorean theorem, we can find the hypotenuse \( c \) of the triangle.

\[ c = \sqrt{a^2 + b^2} \]
where \( a = 3 \, \text{cm} \) and \( b = 4 \, \text{cm} \).

\[ c = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{cm} \]

### Step 2: Identify the radius of the cones formed
When the triangle revolves around the hypotenuse, it forms two cones. The radius \( r_1 \) of the cone formed by the side 3 cm is:

\[ r_1 = \frac{3}{5} \times \text{hypotenuse} = \frac{3}{5} \times 5 = 3 \, \text{cm} \]

The radius \( r_2 \) of the cone formed by the side 4 cm is:

\[ r_2 = \frac{4}{5} \times \text{hypotenuse} = \frac{4}{5} \times 5 = 4 \, \text{cm} \]

### Step 3: Calculate the volume of the double cone
The volume \( V \) of a cone is given by the formula:

\[ V = \frac{1}{3} \pi r^2 h \]

Since we have two cones, we will calculate the volume of each cone and add them together.

For the cone with radius \( r_1 = 3 \, \text{cm} \) and height \( h = 4 \, \text{cm} \):

\[ V_1 = \frac{1}{3} \pi (3^2)(4) = \frac{1}{3} \pi (9)(4) = \frac{36}{3} \pi = 12\pi \, \text{cm}^3 \]

For the cone with radius \( r_2 = 4 \, \text{cm} \) and height \( h = 3 \, \text{cm} \):

\[ V_2 = \frac{1}{3} \pi (4^2)(3) = \frac{1}{3} \pi (16)(3) = \frac{48}{3} \pi = 16\pi \, \text{cm}^3 \]

Thus, the total volume \( V \) of the double cone is:

\[ V = V_1 + V_2 = 12\pi + 16\pi = 28\pi \, \text{cm}^3 \]

### Step 4: Calculate the surface area of the double cone
The surface area \( S \) of a cone is given by the formula:

\[ S = \pi r \sqrt{r^2 + h^2} \]

Calculating the surface area for both cones:

For the cone with radius \( r_1 = 3 \, \text{cm} \) and height \( h = 4 \, \text{cm} \):

\[ S_1 = \pi (3) \sqrt{3^2 + 4^2} = \pi (3) \sqrt{9 + 16} = \pi (3) \sqrt{25} = \pi (3)(5) = 15\pi \, \text{cm}^2 \]

For the cone with radius \( r_2 = 4 \, \text{cm} \) and height \( h = 3 \, \text{cm} \):

\[ S_2 = \pi (4) \sqrt{4^2 + 3^2} = \pi (4) \sqrt{16 + 9} = \pi (4) \sqrt{25} = \pi (4)(5) = 20\pi \, \text{cm}^2 \]

Thus, the total surface area \( S \) of the double cone is:

\[ S = S_1 + S_2 = 15\pi + 20\pi = 35\pi \, \text{cm}^2 \]

### Final Results
- Volume of the double cone: \( 28\pi \, \text{cm}^3 \)
- Surface area of the double cone: \( 35\pi \, \text{cm}^2 \)