A sphere of radius r is incribed in a right circular cone whose slant heights equals twice the radius of the base a. What is the relation between r and a ?

Options

A. `r = (a)/(sqrt(2))`
B. `r = (a)/(3)`
C. `r = (a)/(sqrt(3))`
D. 3r = 2a

Answer

To find the relation between the radius \( r \) of the inscribed sphere and the radius \( a \) of the base of the right circular cone, we can follow these steps:

### Step 1: Understand the Geometry
We have a right circular cone with a sphere inscribed in it. The slant height of the cone is given to be twice the radius of the base, which means if the radius of the base is \( a \), then the slant height \( AB = 2a \).

### Step 2: Identify the Elements
Let:
- \( O \) be the center of the sphere.
- \( D \) be the point where the sphere touches the base of the cone.
- \( A \) be the apex of the cone.
- \( B \) be the point on the circumference of the base directly below the apex.
- \( C \) be the point where the sphere touches the side of the cone.

### Step 3: Use Pythagorean Theorem
In triangle \( ADB \):
- \( AB = 2a \) (slant height)
- \( BD = a \) (radius of the base)

Using the Pythagorean theorem:
\[ AD^2 + BD^2 = AB^2 \]
Substituting the known values:
\[ AD^2 + a^2 = (2a)^2 \]
\[ AD^2 + a^2 = 4a^2 \]
\[ AD^2 = 4a^2 - a^2 = 3a^2 \]
Thus,
\[ AD = \sqrt{3}a \]

### Step 4: Relate the Dimensions
In triangle \( AOF \) (where \( F \) is the point where the sphere touches the side of the cone):
- \( AO = AD - OD \)
- \( OD = r \) (radius of the sphere)

Thus,
\[ AO = \sqrt{3}a - r \]

### Step 5: Use Sine Relation
In triangle \( AOB \):
- The angle \( \angle BAD \) can be determined using the sine function:
\[ \sin(\angle BAD) = \frac{BD}{AB} = \frac{a}{2a} = \frac{1}{2} \]
This implies that \( \angle BAD = 30^\circ \).

### Step 6: Apply Sine Rule
Using the sine rule in triangle \( AOF \):
\[ \sin(30^\circ) = \frac{OF}{AO} \]
Substituting the values:
\[ \frac{1}{2} = \frac{r}{\sqrt{3}a - r} \]
Cross-multiplying gives:
\[ \sqrt{3}a - r = 2r \]
Rearranging this equation:
\[ \sqrt{3}a = 3r \]
Thus, we find:
\[ r = \frac{\sqrt{3}}{3}a \]

### Step 7: Final Relation
To express \( r \) in terms of \( a \):
\[ r = \frac{a}{\sqrt{3}} \]

### Conclusion
The relation between the radius \( r \) of the inscribed sphere and the radius \( a \) of the base of the cone is:
\[ r = \frac{a}{\sqrt{3}} \]