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Question

Find the area (in cm2) of the sector whose perimeter is \[ \frac{64}{3} \, \text{cm} \] and central angle is \( 60^\circ \). (Use \( \pi = \frac{22}{7} \).)

This question was previously asked in
SSC CGL 2024 (Tier-I) Previous Year Paper (17-Sep-2024) (Shift 3)
The correct answer is

77/3

Solution

We are tasked to find the area of a sector whose perimeter is \( \frac{64}{3} \, \text{cm} \) and central angle is \( 60^\circ \). We use \( \pi = \frac{22}{7} \).

Step 1: Formula for the perimeter of a sector

The perimeter of a sector is given by:

\[ \text{Perimeter} = 2r + \text{Arc Length}. \]

The arc length for a sector with central angle \( \theta \) is:

\[ \text{Arc Length} = \frac{\theta}{360} \times 2\pi r. \]

Step 2: Substituting values

The perimeter is \( \frac{64}{3} \), and the central angle is \( 60^\circ \). Substituting into the formula:

\[ \frac{64}{3} = 2r + \frac{60}{360} \times 2\pi r. \]

Simplify \( \frac{60}{360} \) to \( \frac{1}{6} \):

\[ \frac{64}{3} = 2r + \frac{\pi r}{3}. \]

Substitute \( \pi = \frac{22}{7} \):

\[ \frac{64}{3} = 2r + \frac{\left( \frac{22}{7} \right) r}{3}. \]

Step 3: Simplify and solve for \( r \)

Combine terms with a common denominator:

\[ 2r = \frac{42r}{21}, \quad \frac{\pi r}{3} = \frac{22r}{21}. \]

Thus, the equation becomes:

\[ \frac{64}{3} = \frac{64r}{21}. \]

Multiplying through by \( 21 \):

\[ 21 \times \frac{64}{3} = 64r \implies r = 7 \, \text{cm}. \]

Step 4: Calculate the area of the sector

The formula for the area of a sector is:

\[ \text{Area} = \frac{\theta}{360} \times \pi r^2. \]

Substitute \( \theta = 60^\circ \), \( r = 7 \), and \( \pi = \frac{22}{7} \):

\[ \text{Area} = \frac{60}{360} \times \frac{22}{7} \times (7)^2. \]

Simplify step by step:

  • \( \frac{60}{360} = \frac{1}{6} \).
  • \( (7)^2 = 49 \).
  • \( \frac{22}{7} \times 49 = 154 \).

Thus:

\[ \text{Area} = \frac{1}{6} \times 154 = \frac{154}{6} = 25.67 \, \text{cm}^2. \]

Final Answer

The area of the sector is approximately \( \boxed{25.67 \, \text{cm}^2} \).

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