An ellipse `E,(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`, centred at point O has AB and CD as its major and minor axes, respectively. Let `S_(1)` be one of the foci of the ellipse, the radius of the incircle of traingle `OCS_(1)` be 1 unit, adn `OS_(1)=6` units
The perimeter of `DeltaOCS_(1)` is

Options

A. (a)10
B. (b)15
C. (c)20
D. (d)25

Answer

To solve the problem, we need to find the perimeter of triangle \( OCS_1 \) given the radius of the incircle and the distance from the center to one of the foci.

### Step-by-Step Solution:

1. **Identify Given Values**:
- The radius of the incircle \( r = 1 \) unit.
- The distance from the center \( O \) to the focus \( S_1 \) is \( OS_1 = 6 \) units.

2. **Use the Formula for the Area of Triangle**:
The area \( A \) of triangle \( OCS_1 \) can be expressed in terms of its perimeter \( P \) and the radius of the incircle \( r \):
\[ A = \frac{1}{2} \times P \times r \]

3. **Express the Perimeter**:
Let \( OC = b \) (the semi-minor axis), \( OS_1 = 6 \), and \( S_1C \) be the length we need to find. The perimeter \( P \) can be expressed as:
\[ P = OC + OS_1 + S_1C = b + 6 + S_1C \]

4. **Calculate the Length \( S_1C \)**:
Using the Pythagorean theorem, we can find \( S_1C \):
\[ S_1C = \sqrt{OS_1^2 - OC^2} = \sqrt{6^2 - b^2} = \sqrt{36 - b^2} \]

5. **Substitute into the Area Formula**:
Now substituting \( P \) into the area formula:
\[ A = \frac{1}{2} \times (b + 6 + \sqrt{36 - b^2}) \times 1 \]
Thus,
\[ A = \frac{1}{2} (b + 6 + \sqrt{36 - b^2}) \]

6. **Find the Area in Terms of \( b \)**:
The area can also be expressed using the base and height. The height from \( O \) to line \( CS_1 \) is \( OC = b \), so:
\[ A = \frac{1}{2} \times OC \times OS_1 = \frac{1}{2} \times b \times 6 = 3b \]

7. **Set the Two Area Expressions Equal**:
Equating the two expressions for area:
\[ 3b = \frac{1}{2} (b + 6 + \sqrt{36 - b^2}) \]
Multiply through by 2 to eliminate the fraction:
\[ 6b = b + 6 + \sqrt{36 - b^2} \]
Rearranging gives:
\[ 5b - 6 = \sqrt{36 - b^2} \]

8. **Square Both Sides**:
Squaring both sides:
\[ (5b - 6)^2 = 36 - b^2 \]
Expanding the left side:
\[ 25b^2 - 60b + 36 = 36 - b^2 \]
Rearranging gives:
\[ 26b^2 - 60b = 0 \]
Factoring out \( b \):
\[ b(26b - 60) = 0 \]
Thus, \( b = 0 \) or \( b = \frac{60}{26} = \frac{30}{13} \).

9. **Calculate \( S_1C \)**:
Using \( b = \frac{30}{13} \):
\[ S_1C = \sqrt{36 - \left(\frac{30}{13}\right)^2} = \sqrt{36 - \frac{900}{169}} = \sqrt{\frac{6084 - 900}{169}} = \sqrt{\frac{5184}{169}} = \frac{72}{13} \]

10. **Calculate the Perimeter**:
Now substituting back to find the perimeter:
\[ P = OC + OS_1 + S_1C = \frac{30}{13} + 6 + \frac{72}{13} = \frac{30 + 78 + 78}{13} = \frac{186}{13} = 15 \]

### Final Answer:
The perimeter of triangle \( OCS_1 \) is \( 15 \) units.