MP Board · Class 10 · Mathematics · Areas Related to CirclesA chord of a circle of radius 15 cm subtends an angle of $60^{\circ}$ at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Given Data:

• Radius of the circle ($r$) = $15$ cm
• Angle subtended by the chord at the centre ($\theta$) = $60^{\circ}$
• Value of $\pi = 3.14$
• Value of $\sqrt{3} = 1.73$

Step 1: Area of the Minor Sector\nThe formula for the area of a sector is:

$$\text{Area of sector} = \frac{\theta}{360^{\circ}} \times \pi r^2$$\nSubstitute the given values: $$\text{Area of minor sector} = \frac{60^{\circ}}{360^{\circ}} \times 3.14 \times (15)^2$$ $$\text{Area of minor sector} = \frac{1}{6} \times 3.14 \times 225$$ $$\text{Area of minor sector} = \frac{706.5}{6} = 117.75\text{ cm}^2$$

Step 2: Area of $\triangle OAB$\nSince the angle $\theta = 60^{\circ}$ and two sides (radii) are equal ($OA = OB = 15$ cm), $\triangle OAB$ is an equilateral triangle. \nThe formula for the area of an equilateral triangle is:

$$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times \text{side}^2$$ $$\text{Area} = \frac{1.73}{4} \times (15)^2 = \frac{1.73}{4} \times 225$$ $$\text{Area} = \frac{389.25}{4} = 97.3125\text{ cm}^2$$

Step 3: Area of the Minor Segment

$$\text{Area of minor segment} = \text{Area of minor sector} - \text{Area of } \triangle OAB$$ $$\text{Area of minor segment} = 117.75 - 97.3125 = 20.4375\text{ cm}^2$$

Step 4: Area of the Major Segment\nTotal area of the circle = $\pi r^2 = 3.14 \times (15)^2 = 3.14 \times 225 = 706.5\text{ cm}^2$

$$\text{Area of major segment} = \text{Total area of circle} - \text{Area of minor segment}$$ $$\text{Area of major segment} = 706.5 - 20.4375 = 686.0625\text{ cm}^2$$

Final Answer:

• Area of the minor segment = $20.44\text{ cm}^2$
• Area of the major segment = $686.06\text{ cm}^2$