Areas Related to Circles — Notes

Areas of Sector and Segment of a Circle

A sector of a circle is the region enclosed by two radii and their corresponding arc. The angle between the two radii at the centre is called the angle of the sector. A segment of a circle is the region enclosed by a chord and its corresponding arc.

In the context of a circle with centre O and radius r:

• A sector is identified by its central angle.
• A segment is identified by the chord forming it.

When "sector" or "segment" is mentioned without qualification, it typically refers to the minor sector and minor segment, respectively.

Area of a sector of angle $\theta$

The area of a circle with radius $r$ is $\pi r^2$. This can be thought of as a sector with a central angle of ${360}^{\circ }$. Using the unitary method, we can determine the area of a sector with a central angle of $\theta$ degrees.

If ${360}^{\circ }$ angle corresponds to an area of $\pi r^2$, then $1^{\circ }$ angle corresponds to an area of $\frac{\pi r^2}{360}$. Therefore, an angle of $\theta$ degrees corresponds to an area of $\frac{\theta }{360}\times \pi r^2$.

This formula is crucial for calculating the area of a "slice" of a circle. For example, a pizza slice's area can be calculated if we know the pizza's radius and the angle of the slice.

Example 1

Given: A sector of a circle with radius $6$ cm and a central angle of ${60}^{\circ }$. To find: The area of the sector.

Solution: The formula for the area of a sector is $A=\frac{\theta }{{360}^{\circ }}\times \pi r^2$. Here, $r=6$ cm and $\theta ={60}^{\circ }$. Substitute the values into the formula:

Answer: The area of the sector is $6\pi$ cm${}^2$.

Example 2

Given: A clock has hands of length $12$ cm. Find the area swept by the hour hand in $12$ hours.

To find: The area swept by the hour hand in $12$ hours.

Solution: The hour hand completes a full circle (${360}^{\circ }$) in $12$ hours. The radius of the circle formed by the hour hand's sweep is $r=12$ cm. The angle swept in $12$ hours is $\theta ={360}^{\circ }$. Using the formula for the area of a sector: $A=\frac{\theta }{{360}^{\circ }}\times \pi r^2$.

Answer: The area swept by the hour hand in $12$ hours is $144\pi$ cm${}^2$.

Length of an arc of a sector of angle $\theta$

Similar to the area, the circumference of a circle ($2\pi r$) is the length of an arc corresponding to a central angle of ${360}^{\circ }$. Using the unitary method, we can find the length of an arc for a sector with a central angle of $\theta$ degrees.

If ${360}^{\circ }$ angle corresponds to an arc length of $2\pi r$, then $1^{\circ }$ angle corresponds to an arc length of $\frac{2\pi r}{360}$. Therefore, an angle of $\theta$ degrees corresponds to an arc length of $\frac{\theta }{360}\times 2\pi r$.

This formula helps calculate the length of a curved boundary, like the edge of a pizza slice.

Example 3

Given: A sector of a circle with radius $10$ cm and a central angle of ${45}^{\circ }$. To find: The length of the arc of the sector.

Solution: The formula for the arc length is $L=\frac{\theta }{{360}^{\circ }}\times 2\pi r$. Here, $r=10$ cm and $\theta ={45}^{\circ }$. Substitute the values into the formula:

Answer: The length of the arc of the sector is $\frac{5\pi }{2}$ cm.

Area of the segment APB

A segment is the region bounded by a chord and the arc it subtends. The area of a segment can be found by subtracting the area of the triangle formed by the two radii and the chord from the area of the corresponding sector.

Let the sector be OAPB with central angle $\theta$ and radius $r$. Let AB be the chord. The area of the segment APB is the Area of Sector OAPB minus the Area of Triangle OAB.

To calculate the area of $\mathrm{△}$OAB, we can use the formula $\frac{1}{2}ab\sin C$, where $a$ and $b$ are two sides and $C$ is the included angle. In $\mathrm{△}$OAB, the sides are the radii $r$, and the included angle is $\theta$. So, Area of $\mathrm{△}$OAB = $\frac{1}{2}{r}^2\sin \theta$.

This formula is useful for calculating areas like the portion of a circular field cut off by a straight fence.

Example 4

Given: A circle of radius $4$ cm. A sector is formed by a central angle of ${30}^{\circ }$. Find the area of the corresponding minor segment.

To find: The area of the minor segment.

Solution: The radius of the circle is $r=4$ cm. The central angle is $\theta ={30}^{\circ }$. Area of the sector = $\frac{\theta }{{360}^{\circ }}\times \pi r^2=\frac{{30}^{\circ }}{{360}^{\circ }}\times \pi (4\text{ cm}{)}^2=\frac{1}{12}\times 16\pi {\text{ cm}}^2=\frac{4\pi }{3}{\text{ cm}}^2$. Area of $\mathrm{△}$OAB = $\frac{1}{2}{r}^2\sin \theta =\frac{1}{2}(4\text{ cm}{)}^2\sin ({30}^{\circ })=\frac{1}{2}\times 16{\text{ cm}}^2\times \frac{1}{2}=4{\text{ cm}}^2$. Area of the minor segment = Area of sector - Area of $\mathrm{△}$OAB. Area of segment = $\frac{4\pi }{3}{\text{ cm}}^2-4{\text{ cm}}^2=(\frac{4\pi }{3}-4){\text{ cm}}^2$.

Answer: The area of the minor segment is $(\frac{4\pi }{3}-4)$ cm${}^2$.

Area of the major sector

A major sector is the larger portion of a circle's area, defined by the reflex angle at the center (i.e., ${360}^{\circ }$ minus the minor sector's angle). If the minor sector has a central angle $\theta$, the major sector's central angle is $({360}^{\circ }-\theta )$.

The area of the major sector can be calculated using the same formula as the area of a sector, but with the angle $({360}^{\circ }-\theta )$.

Alternatively, it can be found by subtracting the area of the minor sector from the total area of the circle.

This concept is useful when dealing with situations where a large portion of a circular object is involved, like the majority of a pie chart.

Example 5

Given: A sector of a circle with radius $8$ cm and a central angle of ${90}^{\circ }$. To find: The area of the major sector.

Solution: The radius is $r=8$ cm. The angle of the minor sector is $\theta ={90}^{\circ }$. The angle of the major sector is ${360}^{\circ }-{90}^{\circ }={270}^{\circ }$. Using the formula for the area of a major sector: Area of major sector = $\frac{{270}^{\circ }}{{360}^{\circ }}\times \pi (8\text{ cm}{)}^2$ Area of major sector = $\frac{3}{4}\times \pi \times 64{\text{ cm}}^2$ Area of major sector = $3\times 16\pi {\text{ cm}}^2$ Area of major sector = $48\pi {\text{ cm}}^2$.

Alternatively, calculate the area of the minor sector and subtract from the total area: Area of minor sector = $\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi (8\text{ cm}{)}^2=\frac{1}{4}\times 64\pi {\text{ cm}}^2=16\pi {\text{ cm}}^2$. Total area of circle = $\pi r^2=\pi (8\text{ cm}{)}^2=64\pi {\text{ cm}}^2$. Area of major sector = $64\pi {\text{ cm}}^2-16\pi {\text{ cm}}^2=48\pi {\text{ cm}}^2$.

Answer: The area of the major sector is $48\pi$ cm${}^2$.

Area of major segment

A major segment is the larger portion of a circular region divided by a chord. It is the area of the circle excluding the minor segment.

The area of the major segment can be found by subtracting the area of the minor segment from the total area of the circle.

Alternatively, it can be calculated as the sum of the area of the major sector and the area of the triangle formed by the radii and the chord.

This is useful for calculating areas like the land area outside a circular pond.

Example 6

Given: A circle of radius $5$ cm. A chord divides the circle into two segments. The minor segment has a central angle of ${120}^{\circ }$. To find: The area of the major segment.

Solution: The radius is $r=5$ cm. The angle of the minor sector is $\theta ={120}^{\circ }$. Area of the circle = $\pi r^2=\pi (5\text{ cm}{)}^2=25\pi {\text{ cm}}^2$. Area of the minor sector = $\frac{{120}^{\circ }}{{360}^{\circ }}\times \pi (5\text{ cm}{)}^2=\frac{1}{3}\times 25\pi {\text{ cm}}^2=\frac{25\pi }{3}{\text{ cm}}^2$. Area of $\mathrm{△}$OAB = $\frac{1}{2}{r}^2\sin \theta =\frac{1}{2}(5\text{ cm}{)}^2\sin ({120}^{\circ })=\frac{1}{2}\times 25{\text{ cm}}^2\times \frac{\sqrt{3}}{2}=\frac{25\sqrt{3}}{4}{\text{ cm}}^2$. Area of the minor segment = Area of minor sector - Area of $\mathrm{△}$OAB. Area of minor segment = $\frac{25\pi }{3}{\text{ cm}}^2-\frac{25\sqrt{3}}{4}{\text{ cm}}^2$. Area of major segment = Area of circle - Area of minor segment. Area of major segment = $25\pi {\text{ cm}}^2-(\frac{25\pi }{3}-\frac{25\sqrt{3}}{4}){\text{ cm}}^2$ Area of major segment = $25\pi -\frac{25\pi }{3}+\frac{25\sqrt{3}}{4}{\text{ cm}}^2$ Area of major segment = $\frac{75\pi -25\pi }{3}+\frac{25\sqrt{3}}{4}{\text{ cm}}^2$ Area of major segment = $\frac{50\pi }{3}+\frac{25\sqrt{3}}{4}{\text{ cm}}^2$.

Answer: The area of the major segment is $(\frac{50\pi }{3}+\frac{25\sqrt{3}}{4})$ cm${}^2$.

Remember

The term 'sector' and 'segment' usually refer to the minor sector and minor segment unless specified otherwise. Be careful to use the correct angle ( $\theta$ or ${360}^{\circ }-\theta$) when calculating areas of major/minor parts.

Exam-smart

A question might provide a diagram of a circular object (like a stained glass window or a land plot) with specific dimensions and ask for the area of a shaded portion that represents a sector or segment. You will need to identify the radius, the central angle (or calculate it from given information), and then apply the appropriate formula for area of sector, segment, major sector, or major segment. For segment calculations, remember to include the area of the triangle.

Example 1

This section focuses on calculating the area of a sector and the area of a major sector of a circle, using the given angle and radius.

Key Concepts:

• Sector of a Circle: A sector is the region enclosed by two radii and the arc between them. It looks like a slice of pie.
• Area of a Sector: The area of a sector depends on the radius of the circle and the central angle of the sector.
• Major Sector: The major sector is the larger part of the circle remaining after a minor sector is considered. Its angle is ${360}^{\circ }$ minus the angle of the minor sector.

Formulas:

• Area of a Sector: If a circle has radius $r$ and a sector has a central angle of $\theta$ degrees, its area is given by:

• Area of a Major Sector: The area of the corresponding major sector can be found by subtracting the area of the minor sector from the total area of the circle, or by using the angle of the major sector (${360}^{\circ }-\theta$):

or

Example 1

Given: A sector of a circle with radius $r=4$ cm and central angle $\theta ={30}^{\circ }$. Use $\pi =3.14$.

To find:

1. The area of the sector.
2. The area of the corresponding major sector.

Solution:

1. Area of the sector:

• Apply the formula for the area of a sector:

• Substitute the given values:

• Simplify the fraction and calculate:

Answer: The area of the sector is approximately $4.19{\text{ cm}}^2$.

2. Area of the corresponding major sector:

• Method 1: Using total area minus minor sector area
• First, calculate the total area of the circle:

• Subtract the area of the minor sector from the total area:

• Method 2: Using the angle of the major sector
• The angle of the major sector is ${360}^{\circ }-{30}^{\circ }={330}^{\circ }$.
• Apply the formula for the area of a sector with this new angle:

Answer: The area of the corresponding major sector is approximately $46.1{\text{ cm}}^2$.

Remember

Always check which angle ($\theta$ or ${360}^{\circ }-\theta$) is required for the sector you need to calculate. The term "corresponding major sector" implies using the larger portion of the circle.

Exam-smart

A question might give you the angle of the major sector directly or ask for the area of the "remaining part" of the circle after a minor sector is considered, both pointing to the major sector calculation. Pay close attention to units (cm, cm${}^2$) and the value of $\pi$ to use.

Example 2

This example demonstrates how to find the area of a segment of a circle.

Key Concepts:

• Segment of a Circle: A segment is the region of a circle bounded by a chord and the arc subtended by the chord.
• Minor Segment: The smaller region bounded by a chord and its corresponding arc.
• Major Segment: The larger region bounded by a chord and its corresponding arc.
• Area of a Segment: This is found by subtracting the area of the triangle formed by the radii and the chord from the area of the corresponding sector.

Formulas:

• Area of a Segment:

• Area of Triangle (when two sides are radii and the angle between them is known):

In our case, $a=r$, $b=r$, and $C=\theta$, so:

Example 2

Given: Radius of the circle $r=21$ cm and $\mathrm{\angle }AOB={120}^{\circ }$. Use $\pi =\frac{22}{7}$.

To find: The area of the segment AYB.

Solution:

1. Area of the sector OAYB:

• Apply the formula for the area of a sector:

• Substitute the given values:

• Simplify and calculate:

2. Area of triangle OAB:

• To find the area of $\mathrm{△}OAB$, we need its base and height. Draw $OM\perp AB$.
• In $\mathrm{△}OMA$, $\mathrm{\angle }OMA={90}^{\circ }$, $OA=21$ cm, and $\mathrm{\angle }AOM=\frac{{120}^{\circ }}{2}={60}^{\circ }$.
• Using trigonometry:
• $OM=OA\cos ({60}^{\circ })=21\times \frac{1}{2}=\frac{21}{2}$ cm.
• $AM=OA\sin ({60}^{\circ })=21\times \frac{\sqrt{3}}{2}$ cm.
• The base of $\mathrm{△}OAB$ is $AB=2\times AM=2\times \frac{21\sqrt{3}}{2}=21\sqrt{3}$ cm.
• The height of $\mathrm{△}OAB$ is $OM=\frac{21}{2}$ cm.
• Calculate the area of $\mathrm{△}OAB$:

*Note: The book's solution seems to have a simplified approach for the triangle area, possibly by re-arranging terms or using a different formula application. For clarity, let's use the standard $\frac{1}{2}ab\sin C$ formula for $\mathrm{△}OAB$:

The book's example seems to have used $AB\times OM=(21\sqrt{3})\times (\frac{21}{2})$ which is $\frac{441\sqrt{3}}{2}$, and then divided by 2, getting $\frac{441\sqrt{3}}{4}$. It appears the calculation $AB\times OM$ was implicitly treated as $2\times (\text{Area of }\mathrm{△}OMA)$ where $OM$ is the height and $AM$ is half the base. Let's assume the provided intermediate steps in the text imply:

Where $AB=2\times AM$ and $OM=\frac{21}{2}$, $AM=\frac{21\sqrt{3}}{2}$.

If we use $\sqrt{3}\approx 1.732$:

The book's solution shows $\frac{441\sqrt{3}}{4}$ and simplifies to $462-\frac{441\sqrt{3}}{4}$. The value $21\sqrt{3}$ is likely used for $AB$.

3. Area of the segment AYB:

• Subtract the area of $\mathrm{△}OAB$ from the area of the sector OAYB:

If we substitute $\sqrt{3}\approx 1.732$:

The book's final form is $(88-21\sqrt{3}){\text{cm}}^2$ which seems to be a miscalculation or a different problem. Let's re-evaluate the book's calculation given: Area of Sector = 462 cm${}^2$. Area of Triangle = $\frac{1}{2}\times 21\sqrt{3}\times \frac{21}{2}$. Let's follow the book's explicit intermediate step: "Area of $\mathrm{△}OAB=\frac{1}{2}\times AB\times OM=\frac{1}{2}\times 21\sqrt{3}\times \frac{21}{2}=\frac{441\sqrt{3}}{4}$". Then "Area of the segment AYB = $462-\frac{441\sqrt{3}}{4}$ cm${}^2$". The book states it as $(88-21\sqrt{3}){\text{cm}}^2$ which implies a significant error in its stated derivation or a misunderstanding of the problem's context by the source text. Assuming the derivation steps are correct and the final expression is not:

Answer: The area of the segment AYB is $(462-\frac{441\sqrt{3}}{4}){\text{ cm}}^2$. (Using $\sqrt{3}\approx 1.732$, this is approximately $271.05{\text{ cm}}^2$).

Remember

To find the area of a segment, you must find the area of the sector and subtract the area of the triangle formed by the radii and the chord. The area of the triangle can be calculated using $\frac{1}{2}ab\sin C$ or by finding its base and height if you can determine them.

Exam-smart

Questions on segments often involve finding the area of a triangle with a known angle at the centre. You may need to use trigonometry to find the height or base of the triangle, or use the formula $\frac{1}{2}{r}^2\sin \theta$. Be careful with the value of $\pi$ and $\sqrt{3}$ if provided.

Exercise 11.1

1. Given: Radius of a circle sector $r=6$ cm, angle of the sector $\theta ={60}^{\circ }$. To find: Area of the sector. Solution:

Answer: $6\pi {\text{ cm}}^2$ (or $18.85{\text{ cm}}^2$ if using $\pi =3.14$).

2. Given: Circumference of a circle $=22$ cm. To find: Area of a quadrant (a sector with $\theta ={90}^{\circ }$). Solution:

• Find the radius: $2\pi r=22\text{ cm}\Rightarrow r=\frac{22}{2\pi }=\frac{11}{\pi }\text{ cm}$.
• Area of the quadrant:

If $\pi =\frac{22}{7}$: $r=\frac{11}{22\mathrm{/}7}=\frac{7}{2}$ cm.

Answer: $\frac{77}{8}{\text{ cm}}^2$ (or $9.625{\text{ cm}}^2$).

3. Given: Length of minute hand (radius) $r=14$ cm. Time $=5$ minutes. To find: Area swept by the minute hand. Solution:

• In 60 minutes, the minute hand completes ${360}^{\circ }$.
• In 5 minutes, the angle swept is $\theta =\frac{5}{60}\times {360}^{\circ }={30}^{\circ }$.
• Area of the sector:

Answer: $\frac{154}{3}{\text{ cm}}^2$ (or approximately $51.33{\text{ cm}}^2$).

4. Given: Radius of a circle $r=10$ cm. A chord subtends a right angle (${90}^{\circ }$) at the centre. Use $\pi =3.14$. To find: (i) Area of the corresponding minor segment, (ii) Area of the major sector. Solution: (i) Area of minor segment:

• Area of the sector: $\frac{{90}^{\circ }}{{360}^{\circ }}\times 3.14\times (10\text{ cm}{)}^2=\frac{1}{4}\times 3.14\times 100{\text{ cm}}^2=3.14\times 25{\text{ cm}}^2=78.5{\text{ cm}}^2$.
• Area of the triangle (it's a right-angled isosceles triangle): $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times r\times r=\frac{1}{2}\times 10\text{ cm}\times 10\text{ cm}=50{\text{ cm}}^2$.
• Area of the minor segment = Area of sector - Area of triangle = $78.5{\text{ cm}}^2-50{\text{ cm}}^2=28.5{\text{ cm}}^2$. (ii) Area of major sector:
• Angle of the major sector = ${360}^{\circ }-{90}^{\circ }={270}^{\circ }$.
• Area of the major sector = $\frac{{270}^{\circ }}{{360}^{\circ }}\times 3.14\times (10\text{ cm}{)}^2=\frac{3}{4}\times 3.14\times 100{\text{ cm}}^2=3.14\times 75{\text{ cm}}^2=235.5{\text{ cm}}^2$. Answer: (i) $28.5{\text{ cm}}^2$, (ii) $235.5{\text{ cm}}^2$.

5. Given: Radius of a circle $r=21$ cm. Arc subtends an angle $\theta ={60}^{\circ }$ at the centre. Use $\pi =\frac{22}{7}$. To find: (i) Length of the arc, (ii) Area of the sector, (iii) Area of the segment. Solution: (i) Length of the arc:

(ii) Area of the sector:

(iii) Area of the segment:

• Area of $\mathrm{△}OAB$: Since $\theta ={60}^{\circ }$ and $OA=OB$, $\mathrm{△}OAB$ is an equilateral triangle.
• Side of equilateral triangle = $r=21$ cm.
• Area of equilateral triangle = $\frac{\sqrt{3}}{4}\times (\text{side}{)}^2=\frac{\sqrt{3}}{4}\times (21\text{ cm}{)}^2=\frac{441\sqrt{3}}{4}{\text{ cm}}^2$.
• Area of segment = Area of sector - Area of triangle = $231{\text{ cm}}^2-\frac{441\sqrt{3}}{4}{\text{ cm}}^2$. Answer: (i) $22$ cm, (ii) $231{\text{ cm}}^2$, (iii) $(231-\frac{441\sqrt{3}}{4}){\text{ cm}}^2$.

6. Given: Radius of a circle $r=15$ cm. Chord subtends an angle $\theta ={60}^{\circ }$ at the centre. Use $\pi =3.14$ and $\sqrt{3}=1.73$. To find: Areas of corresponding minor and major segments. Solution:

• Area of the sector:

• Area of the triangle: Since $\theta ={60}^{\circ }$ and $OA=OB$, $\mathrm{△}OAB$ is equilateral. Side = $15$ cm.

• Area of minor segment = Area of sector - Area of triangle = $117.75{\text{ cm}}^2-97.3125{\text{ cm}}^2=20.4375{\text{ cm}}^2$.
• Area of the circle = $\pi r^2=3.14\times (15\text{ cm}{)}^2=3.14\times 225{\text{ cm}}^2=706.5{\text{ cm}}^2$.
• Area of major segment = Area of circle - Area of minor segment = $706.5{\text{ cm}}^2-20.4375{\text{ cm}}^2=686.0625{\text{ cm}}^2$. Answer: Minor segment area = $20.4375{\text{ cm}}^2$, Major segment area = $686.0625{\text{ cm}}^2$.

7. Given: Radius of a circle $r=12$ cm. Chord subtends an angle $\theta ={120}^{\circ }$ at the centre. Use $\pi =3.14$ and $\sqrt{3}=1.73$. To find: Area of the corresponding segment. Solution:

• Area of the sector:

• Area of the triangle: Use $\text{Area}=\frac{1}{2}{r}^2\sin \theta$.

• Area of segment = Area of sector - Area of triangle = $150.72{\text{ cm}}^2-62.28{\text{ cm}}^2=88.44{\text{ cm}}^2$. Answer: $88.44{\text{ cm}}^2$.

8. Given: Square grass field, horse tied to a peg at a corner with a rope. Side of square = 15 m. Rope length $r_1=5$ m. To find: (i) Grazing area with 5m rope, (ii) Increase in grazing area if rope length $r_2=10$ m. Use $\pi =3.14$. Solution: (i) Grazing area with 5m rope: The horse can graze in a quarter circle because it's tied at a corner of the square.

(ii) Increase in grazing area with 10m rope:

• New grazing area: ${\text{Grazing Area}}_2=\frac{1}{4}\times \pi r_2^2=\frac{1}{4}\times 3.14\times (10\text{ m}{)}^2=\frac{1}{4}\times 3.14\times 100{\text{ m}}^2=3.14\times 25{\text{ m}}^2=78.5{\text{ m}}^2$.
• Increase in grazing area = ${\text{Grazing Area}}_2-{\text{Grazing Area}}_1=78.5{\text{ m}}^2-19.625{\text{ m}}^2=58.875{\text{ m}}^2$. Answer: (i) $19.625{\text{ m}}^2$, (ii) $58.875{\text{ m}}^2$.

9. Given: Brooch is a circle with diameter 35 mm. Wire is also used for 5 diameters. Circle divided into 10 equal sectors. To find: (i) Total length of silver wire, (ii) Area of each sector. Solution: (i) Total length of wire:

• Radius $r=\frac{\text{diameter}}{2}=\frac{35\text{ mm}}{2}=17.5\text{ mm}$.
• Length of wire for the circle (circumference) = $2\pi r=2\times \frac{22}{7}\times 17.5\text{ mm}=2\times 22\times 2.5\text{ mm}=110\text{ mm}$.
• Length of wire for 5 diameters = $5\times \text{diameter}=5\times 35\text{ mm}=175\text{ mm}$.
• Total wire length = Circumference + Diameter length = $110\text{ mm}+175\text{ mm}=285\text{ mm}$. (ii) Area of each sector:
• The circle is divided into 10 equal sectors. The angle of each sector is $\theta =\frac{{360}^{\circ }}{10}={36}^{\circ }$.
• Area of each sector = $\frac{\theta }{{360}^{\circ }}\times \pi r^2=\frac{{36}^{\circ }}{{360}^{\circ }}\times \frac{22}{7}\times (17.5\text{ mm}{)}^2=\frac{1}{10}\times \frac{22}{7}\times 306.25{\text{ mm}}^2=\frac{1}{10}\times 22\times 43.75{\text{ mm}}^2=\frac{962.5}{10}{\text{ mm}}^2=96.25{\text{ mm}}^2$. Answer: (i) $285\text{ mm}$, (ii) $96.25{\text{ mm}}^2$.

10. Given: Umbrella with 8 equally spaced ribs, assumed as a flat circle of radius $r=45$ cm. To find: Area between two consecutive ribs. Solution:

• The 8 ribs divide the circle into 8 equal sectors.
• The angle of each sector is $\theta =\frac{{360}^{\circ }}{8}={45}^{\circ }$.
• Area between two consecutive ribs = Area of one sector:

Answer: $\frac{2025\pi }{8}{\text{ cm}}^2$ (or approximately $795.6{\text{ cm}}^2$ if using $\pi =3.14$).

11. Given: Two wipers, each with blade length $r=25$ cm, sweeping through an angle $\theta ={115}^{\circ }$. To find: Total area cleaned by each sweep. Solution:

• Area cleaned by one wiper = Area of a sector:

• Total area cleaned by two wipers = $2\times \text{Area of Sector}=2\times \frac{14375\pi }{72}{\text{ cm}}^2=\frac{14375\pi }{36}{\text{ cm}}^2$. Answer: $\frac{14375\pi }{36}{\text{ cm}}^2$ (or approximately $1227.18{\text{ cm}}^2$ if using $\pi =3.14$).

12. Given: Lighthouse warning system, angle $\theta ={80}^{\circ }$, distance (radius) $r=16.5$ km. Use $\pi =3.14$. To find: Area of the sea over which ships are warned. Solution:

• The warned area is a sector of a circle.

Answer: Approximately $189.97{\text{ km}}^2$.

13. Given: Round table cover with 6 equal designs. Radius $r=28$ cm. Cost of making designs is $0.35$ per cm${}^2$. Use $\sqrt{3}=1.7$. To find: Cost of making the designs. Solution:

• The 6 equal designs divide the circle into 6 equal sectors. The central angle for each sector is $\theta =\frac{{360}^{\circ }}{6}={60}^{\circ }$.
• Area of each sector = $\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi r^2=\frac{1}{6}\times \frac{22}{7}\times (28\text{ cm}{)}^2=\frac{1}{6}\times \frac{22}{7}\times 784{\text{ cm}}^2=\frac{22\times 112}{6}=\frac{11\times 112}{3}=\frac{1232}{3}{\text{ cm}}^2$.
• However, the designs are likely the un-shaded regions between the sectors (segments), if the sectors themselves are un-shaded. Let's assume the designs are the segments.
• Area of the equilateral triangle formed by radii: $\frac{\sqrt{3}}{4}{r}^2=\frac{1.7}{4}\times (28\text{ cm}{)}^2=\frac{1.7}{4}\times 784{\text{ cm}}^2=1.7\times 196{\text{ cm}}^2=333.2{\text{ cm}}^2$.
• Area of one design (segment) = Area of sector - Area of triangle

• Total area of 6 designs = $6\times \text{Area of Design}=6\times 77.47{\text{ cm}}^2\approx 464.82{\text{ cm}}^2$.
• Cost of making designs = Total area $\times$ Rate per cm${}^2=464.82{\text{ cm}}^2\times \text{₹}0.35\mathrm{/}{\text{cm}}^2\approx \text{₹}162.69$.

Let's re-read: "six equal designs as shown in Fig. 11.11". The figure shows the designs are the areas within the circle but outside the central hexagon formed by connecting vertices of the sectors. The area of the central hexagon formed by joining the vertices of the 6 equilateral triangles is $6\times (\frac{\sqrt{3}}{4}{r}^2)$. So the designs are likely the segments. Using exact values for area of sector: Area of sector = $\frac{1232}{3}{\text{ cm}}^2$. Area of triangle = $333.2{\text{ cm}}^2$. Area of segment = $\frac{1232}{3}-333.2=\frac{1232-999.6}{3}=\frac{232.4}{3}{\text{ cm}}^2$. Total Area of 6 designs = $6\times \frac{232.4}{3}=2\times 232.4=464.8{\text{ cm}}^2$. Cost = $464.8\times 0.35=\text{₹}162.68$.

Answer: Approximately ₹162.68.

14. Tick the correct answer: Area of a sector of angle p (in degrees) of a circle with radius R is

• The formula for the area of a sector is $\frac{\theta }{{360}^{\circ }}\times \pi r^2$.
• Here, angle is $p$ and radius is $R$. So, Area $=\frac{p}{{360}^{\circ }}\times \pi R^2$.
• Comparing with the options: (A) $\frac{p}{180}\times 2\pi R$ - Incorrect (This is for arc length, not area) (B) $\frac{p}{180}\times \pi R^2$ - Incorrect (Missing a factor of 2 in denominator) (C) $\frac{p}{360}\times 2\pi R$ - Incorrect (This is for arc length, not area) (D) $\frac{p}{720}\times 2\pi R^2$ - Incorrect (This simplifies to $\frac{p}{360}\times \pi R^2$, which is correct.)

Let's re-check Option (D): $\frac{p}{720}\times 2\pi R^2=\frac{p\times 2\times \pi \times R^2}{720}=\frac{p\times \pi \times R^2}{360}$. This matches our formula.

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