Areas Related to Circles - Areas of Sector and Segment of a Circle

Find the area of a sector of a circle with radius $6 \text{ cm}$ if the angle of the sector is $60^\circ$. (Use $\pi = \frac{22}{7}$)

Given: $r = 6 \text{ cm}$, $\theta = 60^\circ$. Using the formula for the area of a sector: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$ $\text{Area} = \frac{60}{360} \times \frac{22}{7} \times 6 \times 6$ $\text{Area} = \frac{1}{6} \times \frac{22}{7} \times 36$ $\text{Area} = \frac{22 \times 6}{7} = \frac{132}{7} \approx 18.86 \text{ cm}^2$

We apply the sector area formula by substituting the given radius and central angle. The fraction $\frac{60}{360}$ simplifies to $\frac{1}{6}$, which then cancels one factor of the $r^2$ term.

A chord of a circle of radius $10 \text{ cm}$ subtends a right angle at the center. Find the area of the corresponding minor segment. (Use $\pi = 3.14$)

1. Area of minor sector $OAPB$ with $\theta = 90^\circ$: $\text{Area of sector} = \frac{90}{360} \times 3.14 \times 10 \times 10 = \frac{1}{4} \times 314 = 78.5 \text{ cm}^2$ 2. Area of $\triangle OAB$ (Right-angled triangle): $\text{Area of } \triangle = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2$ 3. Area of minor segment: $\begin{array}{r} 78.5 \\ - 50.0 \\ \hline 28.5 \end{array}$ $\text{Area of segment} = 28.5 \text{ cm}^2$

To find the segment area, we first find the area of the quarter-circle (sector with $90^\circ$) and then subtract the area of the right-angled triangle formed by the radii and the chord.

Given: $r = 21 \text{ cm}$, $\theta = 60^\circ$

(i) Length of arc $(l) = \frac{\theta}{360^\circ} \times 2\pi r$ $l = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21$ $l = \frac{1}{6} \times 2 \times 22 \times 3$ $l = 22 \text{ cm}$

(ii) Area of sector $= \frac{\theta}{360^\circ} \times \pi r^2$ $\text{Area} = \frac{60}{360} \times \frac{22}{7} \times 21 \times 21$ $\text{Area} = \frac{1}{6} \times 22 \times 3 \times 21$ $\text{Area} = 11 \times 21 = 231 \text{ cm}^2$

To find the arc length, we use the fraction of the circumference corresponding to the central angle. For the sector area, we take the same fraction of the total area of the circle.

The horse is tied at the corner of a square, so it can graze in the shape of a quadrant (sector with $\theta = 90^\circ$). Given: $r = 5 \text{ m}$, $\theta = 90^\circ$

$\text{Area of Grazing} = \frac{\theta}{360^\circ} \times \pi r^2$ $\text{Area} = \frac{90}{360} \times 3.14 \times 5 \times 5$ $\text{Area} = \frac{1}{4} \times 3.14 \times 25$ $\text{Area} = \frac{78.5}{4} = 19.625 \text{ m}^2$

Since the field is square, the angle at the corner is $90^\circ$. The length of the rope acts as the radius of the circular sector the horse can reach.