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Q4(i):
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (Use $\pi = 3.14$)

Solution :

Given:

  • Radius of the circle ($r$) = $10\text{ cm}$
  • Angle subtended by the chord at the center ($\theta$) = $90^\circ$
  • Value of $\pi$ = $3.14$

To find:

The area of the minor segment.

O A B 10 cm 10 cm 90°

Step 1: Formula for the Area of a Minor Segment

The area of a minor segment is calculated by subtracting the area of the triangle formed by the chord and the radii from the area of the corresponding sector.

Formula: $\text{Area of minor segment} = \text{Area of sector} - \text{Area of } \triangle OAB$

Step 2: Calculate the Area of the Sector

The formula for the area of a sector is given by: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$

Substituting the given values:

$\text{Area of sector} = \frac{90}{360} \times 3.14 \times (10)^2$

$\text{Area of sector} = \frac{1}{4} \times 3.14 \times 100$

$\text{Area of sector} = 0.25 \times 314 = 78.5\text{ cm}^2$

Step 3: Calculate the Area of the Triangle ($\triangle OAB$)

Since the angle at the center is $90^\circ$, $\triangle OAB$ is a right-angled triangle with base $OA = 10\text{ cm}$ and height $OB = 10\text{ cm}$.

Formula: $\text{Area of } \triangle OAB = \frac{1}{2} \times \text{base} \times \text{height}$

$\text{Area of } \triangle OAB = \frac{1}{2} \times 10 \times 10$

$\text{Area of } \triangle OAB = \frac{1}{2} \times 100 = 50\text{ cm}^2$

Step 4: Calculate the Area of the Minor Segment

$\text{Area of minor segment} = \text{Area of sector} - \text{Area of } \triangle OAB$

$\text{Area of minor segment} = 78.5\text{ cm}^2 - 50\text{ cm}^2$

$\text{Area of minor segment} = 28.5\text{ cm}^2$

Final Answer: The area of the minor segment is 28.5 cm²


More Questions from Class 10 Mathematics Areas Related to Circles EXERCISE 11.1


CBSE Solutions for Class 10 Mathematics Areas Related to Circles


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