Statement I perimeter of a regular pentagon inscribed in a circle with centre O and radius a cm equals `10a sin 36^(@)` cm.
Statement II Perimeter of a regular polygon inscribed in a circle with centre O and radius a cm equals `(3n-5)sin ((360^(@))/(2n))cm,` then it is n sided, where `n ge3.`

Options

A. (a)Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
B. (b)Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I
C. (c)Statement I is correct but Statement II is incorrect
D. (d)Statement I is correct but Statement I is incorrect

Answer

To solve the problem, we need to verify the two statements regarding the perimeter of a regular pentagon and a regular polygon inscribed in a circle.

### Step-by-Step Solution

**Step 1: Analyze Statement I**

The first statement claims that the perimeter \( P \) of a regular pentagon inscribed in a circle with center \( O \) and radius \( a \) is given by:

\[ P = 10a \sin(36^\circ) \]

To find the perimeter of a regular pentagon, we can use the formula for the side length \( s \) of a regular polygon inscribed in a circle:

\[ s = 2a \sin\left(\frac{180^\circ}{n}\right) \]

For a pentagon, \( n = 5 \):

\[ s = 2a \sin\left(\frac{180^\circ}{5}\right) = 2a \sin(36^\circ) \]

The perimeter \( P \) of the pentagon is:

\[ P = 5s = 5 \times 2a \sin(36^\circ) = 10a \sin(36^\circ) \]

Thus, Statement I is **true**.

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**Step 2: Analyze Statement II**

The second statement claims that the perimeter \( P \) of a regular polygon inscribed in a circle with center \( O \) and radius \( a \) is given by:

\[ P = (3n - 5) \sin\left(\frac{360^\circ}{2n}\right) \]

Using the same formula for the side length \( s \):

\[ s = 2a \sin\left(\frac{180^\circ}{n}\right) \]

The perimeter \( P \) for \( n \) sides is:

\[ P = n \cdot s = n \cdot 2a \sin\left(\frac{180^\circ}{n}\right) \]

This can be rewritten as:

\[ P = 2na \sin\left(\frac{180^\circ}{n}\right) \]

Now, we need to check if this matches the given expression \( (3n - 5) \sin\left(\frac{360^\circ}{2n}\right) \).

Substituting \( \frac{360^\circ}{2n} = \frac{180^\circ}{n} \):

\[ P = (3n - 5) \sin\left(\frac{180^\circ}{n}\right) \]

This implies:

\[ 2na \sin\left(\frac{180^\circ}{n}\right) = (3n - 5) \sin\left(\frac{180^\circ}{n}\right) \]

For this to hold true, we need:

\[ 2a = 3 - \frac{5}{n} \]

This is not generally true for all \( n \geq 3 \). Therefore, Statement II is **false**.

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### Conclusion

- **Statement I** is **true**.
- **Statement II** is **false**.

### Final Answer

The correct option is **C** (Statement I is true, Statement II is false).

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