Areas Related to Circles - Exercise 11.1

Radius of circle, $r = 7\text{ cm}$
Sector angle, $\theta = 45^\circ$
The formula for the area of a sector of angle $\theta$ is: $$\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2$$

Substituting the given values (taking $\pi = \frac{22}{7}$): $$\text{Area} = \frac{45^\circ}{360^\circ} \times \frac{22}{7} \times (7)^2$$ $$\text{Area} = \frac{1}{8} \times \frac{22}{7} \times 49 = \frac{1}{8} \times 154 = \frac{77}{4} = 19.25\text{ cm}^2$$

Given circumference of circle $= 44\text{ cm}$.
$$2\pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44$$ $$\frac{44}{7} \times r = 44 \implies r = \frac{44 \times 7}{44} = 7\text{ cm}$$

A quadrant represents one-fourth of a circle (central angle $\theta = 90^\circ$): $$\text{Area of Quadrant} = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (7)^2$$ $$= \frac{1}{4} \times \frac{22}{7} \times 49 = \frac{154}{4} = 38.5\text{ cm}^2 \text{ (or } \frac{77}{2}\text{ cm}^2\text{)}$$

In 60 minutes, the minute hand completes a full rotation of $360^\circ$.
Angle swept in 1 minute $= \frac{360^\circ}{60} = 6^\circ$.
Therefore, angle swept in 10 minutes: $$\theta = 10 \times 6^\circ = 60^\circ$$

Radius $r = 21\text{ cm}$, $\theta = 60^\circ$: $$\text{Area Swept} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21$$ $$= \frac{1}{6} \times \frac{22}{7} \times 441 = \frac{1}{6} \times 22 \times 3 \times 21$$ $$= \frac{1}{6} \times 1386 = 231\text{ cm}^2$$

Radius $r = 14\text{ cm}$, $\theta = 90^\circ$.
$$\text{Area of Minor Sector } OAB = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 616 = 154\text{ cm}^2$$ Since $\triangle OAB$ is a right-angled triangle at $O$ with sides $OA = OB = 14\text{ cm}$: $$\text{Area of } \triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 14 = 98\text{ cm}^2$$ $$\text{Area of Minor Segment} = \text{Area of Sector } OAB - \text{Area of } \triangle OAB$$ $$= 154 - 98 = 56\text{ cm}^2$$

Angle of major sector $= 360^\circ - 90^\circ = 270^\circ$.
$$\text{Area of Major Sector} = \frac{270^\circ}{360^\circ} \times \pi r^2 = \frac{3}{4} \times 616 = 462\text{ cm}^2$$ Alternatively: $\text{Total Area} - \text{Minor Sector Area} = 616 - 154 = 462\text{ cm}^2$.

Radius $r = 35\text{ cm}$, $\theta = 60^\circ$: $$l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 35$$ $$= \frac{1}{6} \times 220 = \frac{110}{3} \approx 36.67\text{ cm}$$

$$\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 35 \times 35$$ $$= \frac{1}{6} \times 3850 = \frac{1925}{3} \approx 641.67\text{ cm}^2$$

Since $\theta = 60^\circ$ and $OA = OB = 35\text{ cm}$, $\triangle OAB$ is an equilateral triangle.
$$\text{Area of Equilateral } \triangle OAB = \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times (35)^2 = \frac{1.73}{4} \times 1225 \approx 529.81\text{ cm}^2$$ $$\text{Area of Segment} = \text{Area of Sector} - \text{Area of } \triangle OAB$$ $$= 641.67 - 529.81 = 111.86\text{ cm}^2$$

Given $r = 10\text{ cm}$ and $\theta = 60^\circ$.
$$\text{Area of Sector} = \frac{60^\circ}{360^\circ} \times 3.14 \times (10)^2 = \frac{1}{6} \times 314 \approx 52.33\text{ cm}^2$$ Since $\theta = 60^\circ$ and $OA = OB$, $\triangle OAB$ is equilateral: $$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times (10)^2 = \frac{1.73}{4} \times 100 = 43.25\text{ cm}^2$$

$$\text{Area of Minor Segment} = 52.33 - 43.25 = 9.08\text{ cm}^2$$ $$\text{Total Area of Circle} = \pi r^2 = 3.14 \times 100 = 314\text{ cm}^2$$ $$\text{Area of Major Segment} = \text{Area of Circle} - \text{Area of Minor Segment}$$ $$= 314 - 9.08 = 304.92\text{ cm}^2$$

Radius $r = 10\text{ cm}$, $\theta = 120^\circ$: $$\text{Area of Sector} = \frac{120^\circ}{360^\circ} \times 3.14 \times (10)^2 = \frac{1}{3} \times 314 \approx 104.67\text{ cm}^2$$

In $\triangle OAB$, $OA = OB = 10\text{ cm}$ and $\angle AOB = 120^\circ$.
Using trigonometry or formula $\text{Area} = \frac{1}{2} r^2 \sin 120^\circ$: $$\sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$$ $$\text{Area of } \triangle OAB = \frac{1}{2} \times (10)^2 \times \frac{\sqrt{3}}{2} = 25\sqrt{3} = 25 \times 1.73 = 43.25\text{ cm}^2$$

$$\text{Area of Segment} = \text{Area of Sector} - \text{Area of } \triangle OAB$$ $$= 104.67 - 43.25 = 61.42\text{ cm}^2$$

The peg is at the corner of a square field, so the central angle is $\theta = 90^\circ$. The grazing region forms a quadrant of radius $r = 7\text{ m}$: $$\text{Grazing Area} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (7)^2 = \frac{154}{4} = 38.5\text{ m}^2$$

When the rope length increases to $R = 14\text{ m}$: $$\text{New Grazing Area} = \frac{1}{4} \times \frac{22}{7} \times (14)^2 = \frac{1}{4} \times \frac{22}{7} \times 196 = \frac{1}{4} \times 616 = 154\text{ m}^2$$ $$\text{Increase in Area} = 154 - 38.5 = 115.5\text{ m}^2$$

Diameter $d = 42\text{ mm}$, therefore radius $r = \frac{42}{2} = 21\text{ mm}$.
The wire is used along the circular boundary and along 5 diameters: $$\text{Circumference of Circle} = \pi d = \frac{22}{7} \times 42 = 132\text{ mm}$$ $$\text{Length of 5 Diameters} = 5 \times 42 = 210\text{ mm}$$ $$\text{Total Wire Required} = 132 + 210 = 342\text{ mm}$$

Total area of the circular brooch: $$\text{Total Area} = \pi r^2 = \frac{22}{7} \times 21 \times 21 = 22 \times 3 \times 21 = 1386\text{ mm}^2$$ Since the brooch is divided into 10 equal sectors: $$\text{Area of Each Sector} = \frac{1386}{10} = 138.6\text{ mm}^2$$

Radius of flat circular umbrella, $r = 50\text{ cm}$.
Number of equally spaced sectors formed by 10 ribs $= 10$.
Angle between two consecutive ribs: $$\theta = \frac{360^\circ}{10} = 36^\circ$$ $$\text{Area between consecutive ribs} = \frac{1}{10} \times \pi r^2 = \frac{1}{10} \times \frac{22}{7} \times (50)^2$$ $$= \frac{1}{10} \times \frac{22}{7} \times 2500 = \frac{55000}{70} = \frac{5500}{7} \approx 785.71\text{ cm}^2$$

Length of wiper blade (radius), $r = 21\text{ cm}$.
Sweep angle for each wiper, $\theta = 120^\circ$.
Area swept by one wiper: $$\text{Area}_1 = \frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times (21)^2 = \frac{1}{3} \times 22 \times 3 \times 21 = 462\text{ cm}^2$$ Since there are two non-overlapping blades: $$\text{Total Area Cleaned} = 2 \times 462 = 924\text{ cm}^2$$

Distance light spreads (radius), $r = 15\text{ km}$.
Sector angle, $\theta = 72^\circ$.
Area of the sector over which warning is spread: $$\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{72^\circ}{360^\circ} \times 3.14 \times (15)^2$$ $$= \frac{1}{5} \times 3.14 \times 225 = 3.14 \times 45 = 141.3\text{ km}^2$$

Radius $r = 30\text{ cm}$. Since there are 6 equal designs, central angle for each sector: $$\theta = \frac{360^\circ}{6} = 60^\circ$$ $$\text{Area of Sector} = \frac{60^\circ}{360^\circ} \times 3.14 \times (30)^2 = \frac{1}{6} \times 3.14 \times 900 = 471\text{ cm}^2$$ Each triangle is equilateral with side $a = 30\text{ cm}$: $$\text{Area of Equilateral } \triangle = \frac{\sqrt{3}}{4} a^2 = \frac{1.73}{4} \times 900 = 389.25\text{ cm}^2$$ $$\text{Area of 1 Design Segment} = 471 - 389.25 = 81.75\text{ cm}^2$$

$$\text{Total Area of 6 Designs} = 6 \times 81.75 = 490.5\text{ cm}^2$$ Rate of making designs $= ₹0.50\text{ per cm}^2$: $$\text{Total Cost} = 490.5 \times 0.50 = ₹245.25$$

The standard formula for the area of a sector of angle $p^\circ$ and radius $R$ is: $$\text{Area} = \frac{p^\circ}{360^\circ} \times \pi R^2$$ Multiplying both numerator and denominator by 2: $$\text{Area} = \frac{p^\circ}{2 \times 360^\circ} \times 2\pi R^2 = \frac{p^\circ}{720^\circ} \times 2\pi R^2$$ This matches Option (D).