What is the height of the cone which is formed by joining the two ends of a sector of circle with radius r and angle `60^(@)` ?

Options

A. a. `(sqrt(35))/(6)r`
B. b. `(sqrt(25))/(6) r `
C. c. `(r^(2))/(sqrt(3))`
D. d. none of these

Answer

To find the height of the cone formed by joining the two ends of a sector of a circle with radius \( r \) and angle \( 60^\circ \), we can follow these steps:

### Step 1: Determine the length of the arc of the sector
The length of the arc \( L \) of a sector is given by the formula:
\[ L = \frac{\theta}{360^\circ} \times 2\pi r \]
where \( \theta \) is the angle of the sector in degrees.

For our case, \( \theta = 60^\circ \):
\[ L = \frac{60}{360} \times 2\pi r = \frac{1}{6} \times 2\pi r = \frac{\pi r}{3} \]

### Step 2: Relate the arc length to the circumference of the base of the cone
When the sector is rolled to form a cone, the length of the arc becomes the circumference \( C \) of the base of the cone:
\[ C = 2\pi R \]
where \( R \) is the radius of the base of the cone.

Setting the two equations equal gives:
\[ \frac{\pi r}{3} = 2\pi R \]

### Step 3: Solve for \( R \)
We can simplify the equation:
\[ \frac{r}{3} = 2R \implies R = \frac{r}{6} \]

### Step 4: Use the Pythagorean theorem to find the height of the cone
The height \( h \) of the cone, the radius of the base \( R \), and the slant height (which is the radius of the sector) \( r \) form a right triangle. According to the Pythagorean theorem:
\[ r^2 = h^2 + R^2 \]

Substituting \( R = \frac{r}{6} \):
\[ r^2 = h^2 + \left(\frac{r}{6}\right)^2 \]
\[ r^2 = h^2 + \frac{r^2}{36} \]

### Step 5: Rearrange to solve for \( h^2 \)
Subtract \( \frac{r^2}{36} \) from both sides:
\[ r^2 - \frac{r^2}{36} = h^2 \]
\[ \frac{36r^2 - r^2}{36} = h^2 \]
\[ \frac{35r^2}{36} = h^2 \]

### Step 6: Take the square root to find \( h \)
\[ h = \sqrt{\frac{35r^2}{36}} = \frac{r\sqrt{35}}{6} \]

### Final Answer
Thus, the height of the cone is:
\[ h = \frac{r\sqrt{35}}{6} \]

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