NCERT Class X Chapter 11: Area Related To Circles Example 1
NCERT Class X Chapter 11: Area Related To Circles Example 1
Question:
Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector (Use π = 3.14).
Given:
Radius, \( r = 4 \) cm
Angle, \( \theta = 30^\circ \)
\( \pi = 3.14 \)
To Find:
1. Area of the sector (minor sector)
2. Area of the corresponding major sector
Formula:
Area of a sector: $$ \text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 $$ Area of major sector: $$ \text{Area of major sector} = \pi r^2 - \text{Area of minor sector} $$
Solution:
Step 1: Write the formula for the area of the sector.
$$ \text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2 $$Step 2: Substitute the given values: \( \theta = 30^\circ \), \( r = 4 \) cm, \( \pi = 3.14 \).
$$ \text{Area of sector} = \frac{30^\circ}{360^\circ} \times 3.14 \times (4)^2 $$Step 3: Simplify the fraction and calculate \( 4^2 \).
$$ \frac{30}{360} = \frac{1}{12}, \quad 4^2 = 16 $$ $$ \text{Area of sector} = \frac{1}{12} \times 3.14 \times 16 $$Step 4: Multiply the values to get the area of the sector.
$$ \text{Area of sector} = \frac{1}{12} \times 50.24 = 4.1867 \text{ cm}^2 $$Step 5: Write the formula for the area of the major sector.
$$ \text{Area of major sector} = \pi r^2 - \text{Area of minor sector} $$Step 6: Substitute the values: \( \pi = 3.14 \), \( r = 4 \) cm, and area of minor sector \( = 4.1867 \) cm\(^2\).
$$ \text{Area of major sector} = 3.14 \times (4)^2 - 4.1867 $$Step 7: Calculate \( 3.14 \times 16 \) and subtract the area of the minor sector.
$$ 3.14 \times 16 = 50.24 $$ $$ \text{Area of major sector} = 50.24 - 4.1867 = 47.1133 \text{ cm}^2 $$Result:
Area of the sector of angle \( 30^\circ = 4.19 \text{ cm}^2 \)
Area of the corresponding major sector \( = 46.05 \text{ cm}^2 \)
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