NCERT Class X Chapter 11: Area Related To Circles Exercise 11.1 Question 2

NCERT Class X Chapter 11: Area Related To Circles

Question:

Find the area of a quadrant of a circle whose circumference is 22 cm. (Use \( \pi = \frac{22}{7} \))

Given:

Circumference of the circle = 22 cm

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

To Find:

Area of a quadrant of the circle

Formula:

  • Circumference of a circle: \( 2\pi r \)
  • Area of a circle: \( \pi r^2 \)
  • Area of a quadrant: \( \frac{1}{4} \pi r^2 \)

Solution:

Step 1: Let the radius of the circle be \( r \) cm. Write the formula for circumference and substitute values.

$$ 2\pi r = 22 \\ 2 \times \frac{22}{7} \times r = 22 $$

Step 2: Solve for \( r \).

$$ 2 \times \frac{22}{7} \times r = 22 \\ \frac{44}{7} r = 22 \\ r = \frac{22 \times 7}{44} \\ r = \frac{154}{44} \\ r = \frac{7}{2} \\ r = 3.5 \text{ cm} $$

Step 3: Find the area of the circle using the formula \( \pi r^2 \).

$$ \text{Area of circle} = \pi r^2 \\ = \frac{22}{7} \times (3.5)^2 \\ = \frac{22}{7} \times 12.25 \\ = 22 \times 1.75 \\ = 38.5 \text{ cm}^2 $$

Step 4: Find the area of a quadrant (one-fourth of the circle).

$$ \text{Area of quadrant} = \frac{1}{4} \times 38.5 \\ = 9.625 \text{ cm}^2 $$

Result:

The area of the quadrant is \( 9.625 \) cm\(^2\).

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