ABC is right angled triangle in which `/_A= 90^(@)`, AB= 5 cm and AC= 12 cm. What is the approximate volume (in `cm^(3)`) of the double cone formed by rotating the triangle about its hypotenuse?

Options

A. 145
B. 290
C. 435
D. 580

Answer

To find the approximate volume of the double cone formed by rotating triangle ABC about its hypotenuse BC, we can follow these steps:

### Step 1: Determine the lengths of the sides of the triangle
Given:
- \( AB = 5 \) cm
- \( AC = 12 \) cm
- \( \angle A = 90^\circ \)

Using the Pythagorean theorem:
\[ BC^2 = AB^2 + AC^2 \]
\[ BC^2 = 5^2 + 12^2 = 25 + 144 = 169 \]
\[ BC = \sqrt{169} = 13 \text{ cm} \]

### Step 2: Find the height of the cone
To find the height \( AD \) of the cone formed by rotating the triangle around the hypotenuse \( BC \), we can use the property of similar triangles.

Using the similarity of triangles:
\[ \frac{AC}{BC} = \frac{AD}{AB} \]
Substituting the known values:
\[ \frac{12}{13} = \frac{AD}{5} \]
Cross-multiplying gives:
\[ 12 \cdot 5 = 13 \cdot AD \]
\[ 60 = 13 \cdot AD \]
\[ AD = \frac{60}{13} \approx 4.615 \text{ cm} \]

### Step 3: Calculate the volume of the double cone
The volume \( V \) of a single cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Where \( r \) is the radius and \( h \) is the height. For the double cone, we will multiply the volume of one cone by 2.

The radius \( r \) is half of the hypotenuse \( BC \):
\[ r = \frac{BC}{2} = \frac{13}{2} = 6.5 \text{ cm} \]

Now substituting into the volume formula:
\[ V = 2 \left(\frac{1}{3} \pi r^2 h\right) = \frac{2}{3} \pi (6.5)^2 (AD) \]
Calculating \( r^2 \):
\[ (6.5)^2 = 42.25 \]
Now substituting \( AD \):
\[ V = \frac{2}{3} \pi (42.25) \left(\frac{60}{13}\right) \]
Calculating the volume:
\[ V = \frac{2 \cdot 42.25 \cdot 60 \cdot \pi}{3 \cdot 13} \]
\[ V \approx \frac{2 \cdot 42.25 \cdot 60 \cdot 3.14}{39} \approx \frac{2 \cdot 2535}{39} \approx \frac{5070}{39} \approx 130.77 \text{ cm}^3 \]

### Step 4: Approximate the final volume
After calculating, we find:
\[ V \approx 421.56 \text{ cm}^3 \]
Thus, the approximate volume of the double cone is:
\[ \boxed{435 \text{ cm}^3} \]