Areas Related to Circles

Given: Radius $r = 6$ cm, angle of sector $\theta = 60°$

Formula: Area of sector $= \dfrac{\theta}{360} \times \pi r^2$

Solution:
$$\text{Area of sector} = \frac{60}{360} \times \frac{22}{7} \times 6 \times 6$$
$$= \frac{1}{6} \times \frac{22}{7} \times 36$$
$$= \frac{22 \times 36}{6 \times 7} = \frac{792}{42} = \frac{132}{7}$$
$$= 18\frac{6}{7} \text{ cm}^2$$

Answer: Area of the sector $= \dfrac{132}{7}$ cm² $= 18\dfrac{6}{7}$ cm²

Given: Circumference $= 22$ cm

Step 1: Find the radius.
$$2\pi r = 22 \implies r = \frac{22}{2\pi} = \frac{22 \times 7}{2 \times 22} = \frac{7}{2} \text{ cm}$$

Step 2: Find the area of the quadrant.
A quadrant corresponds to $\theta = 90°$.
$$\text{Area of quadrant} = \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$$
$$= \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{1}{4} \times \frac{22 \times 49}{28} = \frac{1}{4} \times \frac{1078}{28} = \frac{1078}{112} = \frac{77}{8}$$

Answer: Area of the quadrant $= \dfrac{77}{8}$ cm² $= 9\dfrac{5}{8}$ cm²

Given: Length of minute hand (radius) $r = 14$ cm

Step 1: Find the angle swept in 5 minutes.
The minute hand completes $360°$ in 60 minutes.
$$\text{Angle in 5 minutes} = \frac{360°}{60} \times 5 = 30°$$

Step 2: Find the area swept.
$$\text{Area} = \frac{\theta}{360} \times \pi r^2 = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14$$
$$= \frac{1}{12} \times \frac{22}{7} \times 196$$
$$= \frac{1}{12} \times \frac{22 \times 196}{7} = \frac{1}{12} \times 22 \times 28 = \frac{616}{12} = \frac{154}{3}$$

Answer: Area swept by the minute hand in 5 minutes $= \dfrac{154}{3}$ cm² $= 51\dfrac{1}{3}$ cm²

Given: Radius $r = 10$ cm, $\theta = 90°$, $\pi = 3.14$

(i) Area of minor segment:

Area of minor sector (with $\theta = 90°$):
$$= \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 10 \times 10 = \frac{314}{4} = 78.5 \text{ cm}^2$$

Area of triangle OAB (right-angled at O, with OA = OB = 10 cm):
$$= \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2$$

Area of minor segment:
$$= \text{Area of minor sector} - \text{Area of } \triangle OAB$$
$$= 78.5 - 50 = 28.5 \text{ cm}^2$$

(ii) Area of major sector:
Angle of major sector $= 360° - 90° = 270°$
$$= \frac{270}{360} \times \pi r^2 = \frac{3}{4} \times 3.14 \times 100 = \frac{942}{4} = 235.5 \text{ cm}^2$$

Answers:
- Area of minor segment $= 28.5$ cm²
- Area of major sector $= 235.5$ cm²

Given: Radius $r = 21$ cm, $\theta = 60°$

(i) Length of the arc:
$$l = \frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21$$
$$= \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22 \text{ cm}$$

(ii) Area of the sector:
$$= \frac{\theta}{360} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 21 \times 21$$
$$= \frac{1}{6} \times \frac{22}{7} \times 441 = \frac{1}{6} \times 1386 = 231 \text{ cm}^2$$

(iii) Area of the segment:

Since $\theta = 60°$ and OA = OB = 21 cm, triangle OAB is equilateral (all sides = 21 cm).

$$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times (21)^2 = \frac{\sqrt{3}}{4} \times 441 = \frac{441\sqrt{3}}{4} \text{ cm}^2$$

$$\text{Area of segment} = 231 - \frac{441\sqrt{3}}{4} = \left(231 - \frac{441\sqrt{3}}{4}\right) \text{ cm}^2$$

Answer: Area of segment $= \left(231 - \dfrac{441\sqrt{3}}{4}\right)$ cm²

Given: Radius $r = 15$ cm, $\theta = 60°$, $\pi = 3.14$, $\sqrt{3} = 1.73$

Area of minor sector:
$$= \frac{60}{360} \times 3.14 \times 15^2 = \frac{1}{6} \times 3.14 \times 225 = \frac{706.5}{6} = 117.75 \text{ cm}^2$$

Area of triangle OAB:
Since $\theta = 60°$ and OA = OB = 15 cm, $\triangle OAB$ is equilateral.
$$= \frac{\sqrt{3}}{4} \times 15^2 = \frac{1.73}{4} \times 225 = \frac{389.25}{4} = 97.3125 \approx 97.31 \text{ cm}^2$$

Area of minor segment:
$$= 117.75 - 97.31 = 20.44 \text{ cm}^2$$

Area of circle:
$$= \pi r^2 = 3.14 \times 225 = 706.5 \text{ cm}^2$$

Area of major segment:
$$= 706.5 - 20.44 = 686.06 \text{ cm}^2$$

Answers:
- Area of minor segment $\approx 20.44$ cm²
- Area of major segment $\approx 686.06$ cm²

Given: Radius $r = 12$ cm, $\theta = 120°$, $\pi = 3.14$, $\sqrt{3} = 1.73$

Area of sector:
$$= \frac{120}{360} \times 3.14 \times 12^2 = \frac{1}{3} \times 3.14 \times 144 = \frac{452.16}{3} = 150.72 \text{ cm}^2$$

Area of triangle OAB:
Draw OM $\perp$ AB. Since $\theta = 120°$, $\angle AOM = 60°$.
$$OM = r\cos 60° = 12 \times \frac{1}{2} = 6 \text{ cm}$$
$$AM = r\sin 60° = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \text{ cm}$$
$$AB = 2 \times AM = 12\sqrt{3} \text{ cm}$$
$$\text{Area of } \triangle OAB = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 12\sqrt{3} \times 6 = 36\sqrt{3}$$
$$= 36 \times 1.73 = 62.28 \text{ cm}^2$$

Area of segment:
$$= 150.72 - 62.28 = 88.44 \text{ cm}^2$$

Answer: Area of the corresponding segment $= 88.44$ cm²

Given: Side of square field $= 15$ m, length of rope $= 5$ m (radius), $\pi = 3.14$

The horse is tied at a corner of the square. The angle at each corner of a square is $90°$.

(i) Grazing area with rope = 5 m:
The horse can graze in a sector of radius 5 m and angle 90°.
$$\text{Area} = \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 5^2 = \frac{1}{4} \times 3.14 \times 25 = \frac{78.5}{4} = 19.625 \text{ m}^2$$

(ii) Grazing area with rope = 10 m:
$$\text{Area} = \frac{1}{4} \times 3.14 \times 10^2 = \frac{1}{4} \times 3.14 \times 100 = \frac{314}{4} = 78.5 \text{ m}^2$$

Increase in grazing area:
$$= 78.5 - 19.625 = 58.875 \text{ m}^2$$

Answers:
- (i) Area of grazing with 5 m rope $= 19.625$ m²
- (ii) Increase in grazing area $= 58.875$ m²

Given: Diameter $= 35$ mm, so radius $r = 17.5$ mm; 5 diameters divide the circle into 10 equal sectors.

(i) Total length of silver wire:
- Circumference of circle $= 2\pi r = 2 \times \dfrac{22}{7} \times 17.5 = 2 \times \dfrac{22}{7} \times \dfrac{35}{2} = 2 \times 55 = 110$ mm
- Length of 5 diameters $= 5 \times 35 = 175$ mm
- Total length $= 110 + 175 = 285$ mm

(ii) Area of each sector:
The circle is divided into 10 equal sectors, so each sector has angle $= \dfrac{360°}{10} = 36°$.
$$\text{Area of each sector} = \frac{36}{360} \times \pi r^2 = \frac{1}{10} \times \frac{22}{7} \times (17.5)^2$$
$$= \frac{1}{10} \times \frac{22}{7} \times 306.25 = \frac{1}{10} \times \frac{6737.5}{7} = \frac{6737.5}{70} = \frac{962.5}{10} = 96.25 \text{ mm}^2$$

Answers:
- (i) Total length of silver wire $= 285$ mm
- (ii) Area of each sector $= 96.25$ mm²

Given: Radius $r = 45$ cm, number of ribs $= 8$

The 8 equally spaced ribs divide the circle into 8 equal sectors.
$$\text{Angle of each sector} = \frac{360°}{8} = 45°$$

Area between two consecutive ribs (one sector):
$$= \frac{45}{360} \times \pi r^2 = \frac{1}{8} \times \frac{22}{7} \times 45 \times 45$$
$$= \frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{22 \times 2025}{56} = \frac{44550}{56} = \frac{22275}{28} = 795.535... \approx 795.54 \text{ cm}^2$$

Answer: Area between two consecutive ribs $= \dfrac{22275}{28}$ cm² $\approx 795.54$ cm²

Given: Length of blade (radius) $r = 25$ cm, angle $\theta = 115°$, number of wipers $= 2$

Area cleaned by one wiper in one sweep:
$$= \frac{\theta}{360} \times \pi r^2 = \frac{115}{360} \times \frac{22}{7} \times 25 \times 25$$
$$= \frac{115}{360} \times \frac{22}{7} \times 625$$
$$= \frac{115 \times 22 \times 625}{360 \times 7} = \frac{1581250}{2520} = \frac{158125}{252} \approx 627.48 \text{ cm}^2$$

Total area cleaned by two wipers:
$$= 2 \times \frac{158125}{252} = \frac{158125}{126} = \frac{158125}{126} \approx 1254.96 \text{ cm}^2$$

More precisely:
$$= \frac{115}{360} \times \frac{22}{7} \times 625 \times 2 = \frac{23 \times 22 \times 625}{72 \times 7} = \frac{316250}{504} = \frac{158125}{252}$$

Answer: Total area cleaned $= \dfrac{158125}{126}$ cm² $\approx 1254.96$ cm²

Given: Radius $r = 16.5$ km, angle $\theta = 80°$, $\pi = 3.14$

Area of the sector:
$$= \frac{\theta}{360} \times \pi r^2 = \frac{80}{360} \times 3.14 \times (16.5)^2$$
$$= \frac{2}{9} \times 3.14 \times 272.25$$
$$= \frac{2}{9} \times 854.865 = \frac{1709.73}{9} = 189.97 \text{ km}^2$$

Answer: Area of the sea over which ships are warned $\approx 189.97$ km²

Given: Radius $r = 28$ cm, 6 equal designs, rate $= ₹0.35$ per cm², $\sqrt{3} = 1.7$

The 6 equal designs are 6 segments formed by 6 equal chords (each chord subtends $60°$ at the centre, since $360°/6 = 60°$).

Area of one segment:

Area of one sector (angle $= 60°$):
$$= \frac{60}{360} \times \frac{22}{7} \times 28^2 = \frac{1}{6} \times \frac{22}{7} \times 784 = \frac{1}{6} \times 2464 = \frac{2464}{6} = \frac{1232}{3} \text{ cm}^2$$

Area of equilateral triangle (since $\theta = 60°$, OA = OB = 28 cm, triangle is equilateral with side 28 cm):
$$= \frac{\sqrt{3}}{4} \times 28^2 = \frac{1.7}{4} \times 784 = \frac{1332.8}{4} = 333.2 \text{ cm}^2$$

Area of one segment:
$$= \frac{1232}{3} - 333.2 = 410.67 - 333.2 = 77.47 \text{ cm}^2$$

Total area of 6 designs:
$$= 6 \times 77.47 = 464.82 \text{ cm}^2$$

Cost of making the designs:
$$= 464.82 \times 0.35 = ₹162.69$$

Answer: Cost of making the designs $= ₹162.68$ (approximately)

Correct Answer: (D) $\dfrac{p}{720} \times 2\pi R^2$

Justification:
The standard formula for area of a sector is:
$$\text{Area} = \frac{p}{360} \times \pi R^2$$

This can be rewritten as:
$$= \frac{p}{360} \times \pi R^2 = \frac{2p}{720} \times \pi R^2 = \frac{p}{720} \times 2\pi R^2$$

Option (D) $\dfrac{p}{720} \times 2\pi R^2 = \dfrac{p}{360} \times \pi R^2$, which matches the standard formula.

Options (A) and (C) give arc length (not area), and option (B) has the wrong denominator (180 instead of 360).