Focus Keyword: NCERT Solutions Class 10 Math Chapter 11 Areas Related to Circles
Secondary Keywords & LSI: CBSE Class 10 Maths Chapter 11 solutions, Class 10 Maths Chapter 11 Exercise 11.1, area of sector and segment of circle Class 10, Areas Related to Circles class 10 CBSE marking scheme, board exam preparation Class 10 Maths
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H1 Title: NCERT Solutions for Class 10 Math Chapter 11: Areas Related to Circles (Complete Guide)
Navigating through the CBSE Class 10 Mathematics curriculum requires an in-depth understanding of planar circular geometry, arc lengths, sector partitions, and segment areas. Chapter 11 of Class 10 Mathematics, “Areas Related to Circles”, forms the foundation of modern mechanical design, civil engineering layouts, rotational dynamics, and geometric area estimation. It investigates the proportional division of a circle by radial boundaries; explores the relationship between central angles and bounded arc lengths; and details the quantitative computation of both minor and major segments formed by intersecting chords. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from foundational textbook examples to the complete 14-question set of Exercise 11.1 and an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic four-step mathematical workflows (Given Data $\rightarrow$ Formula Stated $\rightarrow$ Step-by-Step LaTeX Substitution $\rightarrow$ Final Answer with Units), and clean geometric diagrams have been highlighted to ensure students secure maximum marks in their CBSE Board Examinations.
Chapter 11: Areas Related to Circles
Master Chapter Summary & Quantitative Blueprint
In the latest rationalised NCERT textbook for Class 10 Mathematics, Chapter 11 concentrates on computing the perimeter (arc length) and area of plane regions enclosed by curves and straight lines associated with circles, specifically sectors and segments.
| Geometric Entity | Core Mathematical Definition | Standard Governing Formula | Common Board Error | CBSE Marks Weightage |
|---|---|---|---|---|
| Circumference of Circle | Total boundary length of the circular plane | $C = 2\pi r = \pi d$ | Confusing radius with diameter | 1 Mark (Prerequisite) |
| Area of Circle | Total two-dimensional space enclosed by boundary | $A = \pi r^2$ | Using diameter instead of radius squared | 1 Mark (Prerequisite) |
| Length of an Arc | Fractional part of circumference subtending angle $\theta$ | $l = \frac{\theta}{360^\circ} \times 2\pi r$ | Forgetting to divide $\theta$ by $360^\circ$ | 1 to 2 Marks |
| Area of Sector | Region bounded by two radii and their intercepted arc | $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$ | Inverting angle ratio or using $180^\circ$ | 2 to 3 Marks |
| Area of Major Sector | Unshaded remaining sector region of circle | $\text{Area} = \pi r^2 – \text{Area of Minor Sector}$ | Subtracting from semi-circle instead of full circle | 2 to 3 Marks |
| Area of Minor Segment | Region enclosed between a chord and its intercepted arc | $\text{Area} = \text{Area of Sector} – \text{Area of } \triangle$ | Mishandling triangle area when $\theta = 120^\circ$ | 3 to 5 Marks |
| Area of Major Segment | Region bounded by chord and major arc | $\text{Area} = \pi r^2 – \text{Area of Minor Segment}$ | Forgetting to subtract minor segment from total area | 3 to 5 Marks |
🧠 Examiner’s Secret: In CBSE board evaluations, if the value of $\pi$ is explicitly mentioned as $3.14$, you must use $3.14$. If it is not mentioned, always use $\frac{22}{7}$. Using $3.14$ instead of $\frac{22}{7}$ when unspecified can lead to minor fractional differences that may cost you an answer mark under strict marking keys.
Foundational Geometric Concepts and Formula Vault
Sector of a Circle
A sector of a circle is the portion of the circular region enclosed by two radii and their intercepted arc. O (Centre) θ A B Minor Arc APB Major Sector
When the central angle $\theta < 180^\circ$, the region is called the minor sector. The remaining unshaded part of the circular disc corresponding to the angle $(360^\circ – \theta)$ is designated as the major sector.
The quantitative formulas governing a sector of radius $r$ and central angle $\theta$ (in degrees) are:
$$\text{Area of Sector of Angle } \theta = \frac{\theta}{360^\circ} \times \pi r^2$$
$$\text{Length of Arc of Sector of Angle } \theta = \frac{\theta}{360^\circ} \times 2\pi r$$
Segment of a Circle
A segment of a circle is the region bounded by a chord and either of its corresponding intercepted arcs. O A B Minor Segment Major Segment
The chord $AB$ divides the circle into two parts: the minor segment (bounded by the chord and minor arc) and the major segment (bounded by the chord and major arc).
The area of the minor segment is computed by subtracting the area of the corresponding triangle $\triangle OAB$ from the area of the sector $OAPB$:
$$\text{Area of Minor Segment} = \text{Area of Sector } OAPB – \text{Area of } \triangle OAB$$
$$\text{Area of Major Segment} = \pi r^2 – \text{Area of Minor Segment}$$
Triangle Area Formulation Inside a Sector
To evaluate $\text{Area of } \triangle OAB$ where $OA = OB = r$ and central angle $\angle AOB = \theta$:
- Case I: When $\theta = 60^\circ$
$\triangle OAB$ is an equilateral triangle with side $r$:
$$\text{Area}(\triangle OAB) = \frac{\sqrt{3}}{4}r^2$$ - Case II: When $\theta = 90^\circ$
$\triangle OAB$ is a right-angled isosceles triangle:
$$\text{Area}(\triangle OAB) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times r \times r = \frac{1}{2}r^2$$ - Case III: When $\theta = 120^\circ$
Dropping a perpendicular from $O$ to chord $AB$ bisects $AB$ and angle $\theta$ ($60^\circ$ each half):
$$\text{Area}(\triangle OAB) = \frac{1}{2} r^2 \sin \theta = \frac{1}{2} r^2 \sin 120^\circ = \frac{1}{2} r^2 \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}r^2$$
💡 Did You Know?: The general formula $\text{Area}(\triangle OAB) = \frac{1}{2}r^2 \sin \theta$ holds true for any central angle $\theta$ between $0^\circ$ and $180^\circ$, making it a powerful check for any segment calculation.
[👉 Also Read: Class 10 Math Chapter 10 Circles NCERT Solutions]
Step-by-Step Solutions: NCERT Class 10 Mathematics Chapter 11 Solved Examples
Example 1 (Page 158) [CBSE 2018, 2021]
Find the area of the sector of a circle with radius $4\text{ cm}$ and of angle $30^\circ$. Also, find the area of the corresponding major sector (Use $\pi = 3.14$).
Answer:
Step 1: Identify Given Data
- Radius of the circle, $r = 4\text{ cm}$.
- Central angle of minor sector, $\theta = 30^\circ$.
- Value of $\pi = 3.14$.
Step 2: Formula and Calculation for Minor Sector Area
$$\text{Area of Minor Sector} = \frac{\theta}{360^\circ} \times \pi r^2$$
Substitute the given values:
$$\text{Area} = \frac{30^\circ}{360^\circ} \times 3.14 \times 4^2$$
$$\text{Area} = \frac{1}{12} \times 3.14 \times 16 = \frac{50.24}{12} \approx 4.19\text{ cm}^2$$
Step 3: Formula and Calculation for Major Sector Area
$$\text{Area of Major Sector} = \left(\frac{360^\circ – \theta}{360^\circ}\right) \times \pi r^2 = \pi r^2 – \text{Area of Minor Sector}$$
$$\text{Area of Complete Circle} = \pi r^2 = 3.14 \times 4^2 = 3.14 \times 16 = 50.24\text{ cm}^2$$
Subtract the minor sector area:
$$\text{Area of Major Sector} = 50.24 – 4.19 = 46.05\text{ cm}^2$$
Alternatively:
$$\text{Area} = \frac{360^\circ – 30^\circ}{360^\circ} \times \pi r^2 = \frac{330^\circ}{360^\circ} \times 3.14 \times 16 = \frac{11}{12} \times 50.24 \approx 46.05\text{ cm}^2$$
Final Answer:
The area of the minor sector is $4.19\text{ cm}^2$, and the area of the corresponding major sector is $46.05\text{ cm}^2$.
Example 2 (Page 159) [CBSE 2015, 2019, 2023 Set-1]
Find the area of the segment $AYB$ shown in the figure, if radius of the circle is $21\text{ cm}$ and $\angle AOB = 120^\circ$ (Use $\pi = \frac{22}{7}$). O A B Y M 120°
Answer:
Step 1: Identify Given Data
- Radius of the circle, $r = OA = OB = 21\text{ cm}$.
- Central angle, $\angle AOB = 120^\circ$.
- Value of $\pi = \frac{22}{7}$.
Step 2: Area of Sector $OAYB$
$$\text{Area of Sector } OAYB = \frac{\theta}{360^\circ} \times \pi r^2$$
$$\text{Area} = \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21$$
$$\text{Area} = \frac{1}{3} \times \frac{22}{7} \times 441 = \frac{1}{3} \times 22 \times 63 = 22 \times 21 = 462\text{ cm}^2$$
Step 3: Area of Triangle $\triangle OAB$
Draw $OM \perp AB$.
In isosceles $\triangle OAB$ with $OA = OB = 21\text{ cm}$, the perpendicular $OM$ bisects both the chord $AB$ and $\angle AOB$.
Therefore:
$$\angle AOM = \angle BOM = \frac{120^\circ}{2} = 60^\circ$$
In right-angled triangle $\triangle OMA$:
$$\cos 60^\circ = \frac{OM}{OA} \implies \frac{1}{2} = \frac{OM}{21} \implies OM = \frac{21}{2}\text{ cm}$$
$$\sin 60^\circ = \frac{AM}{OA} \implies \frac{\sqrt{3}}{2} = \frac{AM}{21} \implies AM = \frac{21\sqrt{3}}{2}\text{ cm}$$
The total length of chord $AB$ is:
$$AB = 2 \times AM = 2 \times \frac{21\sqrt{3}}{2} = 21\sqrt{3}\text{ cm}$$
Now compute the area of $\triangle OAB$:
$$\text{Area}(\triangle OAB) = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 21\sqrt{3} \times \frac{21}{2} = \frac{441\sqrt{3}}{4}\text{ cm}^2$$
Step 4: Compute Area of Segment $AYB$
$$\text{Area of Segment } AYB = \text{Area of Sector } OAYB – \text{Area}(\triangle OAB)$$
$$\text{Area of Segment } AYB = \left(462 – \frac{441\sqrt{3}}{4}\right)\text{ cm}^2 = \frac{21}{4}(88 – 21\sqrt{3})\text{ cm}^2$$
Final Answer:
The area of the segment $AYB$ is $\left(462 – \frac{441\sqrt{3}}{4}\right)\text{ cm}^2$ (or approximately $270.97\text{ cm}^2$ if taking $\sqrt{3} \approx 1.732$).
Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 11.1
Question 1 (Page 160) [CBSE 2013, 2017, 2020 Standard]
Find the area of a sector of a circle with radius $6\text{ cm}$ if angle of the sector is $60^\circ$.
Answer:
Step 1: Identify Given Data
- Radius of circle, $r = 6\text{ cm}$.
- Central angle, $\theta = 60^\circ$.
- Unless specified, use $\pi = \frac{22}{7}$.
Step 2: Formula and Step-by-Step Substitution
$$\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2$$
$$\text{Area} = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 6^2$$
$$\text{Area} = \frac{1}{6} \times \frac{22}{7} \times 36$$
Cancel common factor $6$:
$$\text{Area} = \frac{22 \times 6}{7} = \frac{132}{7}\text{ cm}^2$$
Converting to mixed fraction / decimal:
$$\frac{132}{7}\text{ cm}^2 = 18\frac{6}{7}\text{ cm}^2 \approx 18.86\text{ cm}^2$$
Final Answer:
The area of the sector is $\frac{132}{7}\text{ cm}^2$ (or $18.86\text{ cm}^2$).
Question 2 (Page 160) [CBSE 2014, 2019, 2023]
Find the area of a quadrant of a circle whose circumference is $22\text{ cm}$.
Answer:
Step 1: Identify Given Data
- Circumference of circle, $C = 22\text{ cm}$.
- A quadrant is one-fourth of a circular disc, so its central angle is $\theta = 90^\circ$.
Step 2: Determine Radius ($r$) from Circumference
$$C = 2\pi r = 22$$
$$2 \times \frac{22}{7} \times r = 22$$
$$r = \frac{22 \times 7}{2 \times 22} = \frac{7}{2}\text{ cm} = 3.5\text{ cm}$$
Step 3: Compute Area of Quadrant
$$\text{Area of Quadrant} = \frac{1}{4}\pi r^2 = \frac{90^\circ}{360^\circ} \times \pi r^2$$
$$\text{Area} = \frac{1}{4} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2$$
$$\text{Area} = \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4}$$
$$\text{Area} = \frac{1}{4} \times 11 \times \frac{7}{2} = \frac{77}{8}\text{ cm}^2$$
Converting to decimal:
$$\frac{77}{8} = 9.625\text{ cm}^2$$
Final Answer:
The area of the quadrant is $\frac{77}{8}\text{ cm}^2$ (or $9.625\text{ cm}^2$).
Question 3 (Page 160) [CBSE 2012, 2016, 2020 Standard, 2024]
The length of the minute hand of a clock is $14\text{ cm}$. Find the area swept by the minute hand in $5\text{ minutes}$.
Answer:
Step 1: Identify Given Data
- Length of the minute hand = radius of sector, $r = 14\text{ cm}$.
- Time elapsed = $5\text{ minutes}$.
Step 2: Calculate Central Angle Subtended in 5 Minutes
A minute hand completes one full revolution ($360^\circ$) in $60\text{ minutes}$.
- Angle swept in $1\text{ minute} = \frac{360^\circ}{60} = 6^\circ$.
- Angle swept in $5\text{ minutes}$:
$$\theta = 5 \times 6^\circ = 30^\circ$$
Step 3: Compute Area Swept (Area of Sector)
$$\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$$
$$\text{Area} = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14^2$$
$$\text{Area} = \frac{1}{12} \times \frac{22}{7} \times 196$$
Cancel $196$ with $7$ ($196 / 7 = 28$):
$$\text{Area} = \frac{1}{12} \times 22 \times 28 = \frac{22 \times 7}{3} = \frac{154}{3}\text{ cm}^2$$
Converting to mixed fraction / decimal:
$$\frac{154}{3}\text{ cm}^2 = 51\frac{1}{3}\text{ cm}^2 \approx 51.33\text{ cm}^2$$
Final Answer:
The area swept by the minute hand in 5 minutes is $\frac{154}{3}\text{ cm}^2$ (or $51.33\text{ cm}^2$).
Question 4 (Page 160) [CBSE 2015, 2018, 2023 Set-2]
A chord of a circle of radius $10\text{ cm}$ subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment
(ii) major sector. (Use $\pi = 3.14$)
Answer:
Step 1: Identify Given Data
- Radius of circle, $r = 10\text{ cm}$.
- Central angle, $\theta = 90^\circ$.
- Value of $\pi = 3.14$.
Step 2: Compute Area of Minor Sector
$$\text{Area of Minor Sector} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 10^2 = \frac{1}{4} \times 314 = 78.5\text{ cm}^2$$
Step 3: Compute Area of Triangle $\triangle AOB$
Since $\angle AOB = 90^\circ$, base and height are both equal to radius $r = 10\text{ cm}$:
$$\text{Area}(\triangle AOB) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 10 = 50\text{ cm}^2$$
Step 4: Compute Area of Minor Segment
$$\text{Area of Minor Segment} = \text{Area of Minor Sector} – \text{Area}(\triangle AOB)$$
$$\text{Area of Minor Segment} = 78.5 – 50 = 28.5\text{ cm}^2$$
Step 5: Compute Area of Major Sector
The central angle for the major sector is $360^\circ – 90^\circ = 270^\circ$:
$$\text{Area of Major Sector} = \frac{270^\circ}{360^\circ} \times \pi r^2 = \frac{3}{4} \times 3.14 \times 100 = \frac{3}{4} \times 314 = 3 \times 78.5 = 235.5\text{ cm}^2$$
Alternatively:
$$\text{Area of Major Sector} = \pi r^2 – \text{Area of Minor Sector} = 314 – 78.5 = 235.5\text{ cm}^2$$
Final Answer:
(i) The area of the minor segment is $28.5\text{ cm}^2$.
(ii) The area of the major sector is $235.5\text{ cm}^2$.
Question 5 (Page 160) [CBSE 2014, 2017, 2020 Standard]
In a circle of radius $21\text{ cm}$, an arc subtends an angle of $60^\circ$ at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord.
Answer:
Step 1: Identify Given Data
- Radius of circle, $r = 21\text{ cm}$.
- Central angle, $\theta = 60^\circ$.
- Value of $\pi = \frac{22}{7}$.
Step 2: (i) Length of the Arc
$$l = \frac{\theta}{360^\circ} \times 2\pi r$$
$$l = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21$$
$$l = \frac{1}{6} \times 2 \times 22 \times 3 = \frac{1}{6} \times 132 = 22\text{ cm}$$
Step 3: (ii) Area of the Sector Formed by the Arc
$$\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2$$
$$\text{Area} = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21$$
$$\text{Area} = \frac{1}{6} \times 22 \times 3 \times 21 = \frac{1}{6} \times 1386 = 231\text{ cm}^2$$
Alternatively, using arc length $l$:
$$\text{Area} = \frac{1}{2} \times l \times r = \frac{1}{2} \times 22 \times 21 = 231\text{ cm}^2$$
Step 4: (iii) Area of the Segment Formed by the Chord
In $\triangle AOB$, $OA = OB = 21\text{ cm}$ and $\angle AOB = 60^\circ$.
Because $OA = OB$, $\angle OAB = \angle OBA = \frac{180^\circ – 60^\circ}{2} = 60^\circ$.
Thus, $\triangle AOB$ is an equilateral triangle with side $a = 21\text{ cm}$.
$$\text{Area}(\triangle AOB) = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 21^2 = \frac{441\sqrt{3}}{4}\text{ cm}^2$$
Now compute the segment area:
$$\text{Area of Segment} = \text{Area of Sector} – \text{Area}(\triangle AOB)$$
$$\text{Area of Segment} = \left(231 – \frac{441\sqrt{3}}{4}\right)\text{ cm}^2$$
Evaluating with $\sqrt{3} \approx 1.732$:
$$\text{Area} = 231 – \frac{441 \times 1.732}{4} = 231 – 190.95 = 40.05\text{ cm}^2$$
Final Answer:
(i) Length of the arc = $22\text{ cm}$.
(ii) Area of the sector = $231\text{ cm}^2$.
(iii) Area of the segment = $\left(231 – \frac{441\sqrt{3}}{4}\right)\text{ cm}^2$ (or $40.05\text{ cm}^2$).
Question 6 (Page 160) [CBSE 2011, 2016, 2019 Set-3]
A chord of a circle of radius $15\text{ cm}$ 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 $\sqrt{3} = 1.73$)
Answer:
Step 1: Identify Given Data
- Radius of circle, $r = 15\text{ cm}$.
- Central angle, $\theta = 60^\circ$.
- Values: $\pi = 3.14$, $\sqrt{3} = 1.73$.
Step 2: Area of Sector of $60^\circ$
$$\text{Area of Sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times 3.14 \times 15^2 = \frac{1}{6} \times 3.14 \times 225 = \frac{706.5}{6} = 117.75\text{ cm}^2$$
Step 3: Area of Equilateral Triangle $\triangle AOB$
Since $\angle AOB = 60^\circ$ and $OA = OB$, $\triangle AOB$ is equilateral with side $15\text{ cm}$:
$$\text{Area}(\triangle AOB) = \frac{\sqrt{3}}{4} \times 15^2 = \frac{1.73}{4} \times 225 = \frac{389.25}{4} = 97.3125\text{ cm}^2$$
Step 4: Compute Area of Minor Segment
$$\text{Area of Minor Segment} = \text{Area of Sector} – \text{Area}(\triangle AOB)$$
$$\text{Area of Minor Segment} = 117.75 – 97.3125 = 20.4375\text{ cm}^2$$
Step 5: Compute Area of Major Segment
$$\text{Total Area of Circle} = \pi r^2 = 3.14 \times 15^2 = 3.14 \times 225 = 706.5\text{ cm}^2$$
$$\text{Area of Major Segment} = \text{Total Area} – \text{Area of Minor Segment}$$
$$\text{Area of Major Segment} = 706.5 – 20.4375 = 686.0625\text{ cm}^2$$
Final Answer:
The area of the minor segment is $20.4375\text{ cm}^2$, and the area of the major segment is $686.0625\text{ cm}^2$.
Question 7 (Page 160) [CBSE 2012, 2017, 2020 Standard, 2024]
A chord of a circle of radius $12\text{ cm}$ subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
Answer:
Step 1: Identify Given Data
- Radius of circle, $r = 12\text{ cm}$.
- Central angle, $\theta = 120^\circ$.
- Values: $\pi = 3.14$, $\sqrt{3} = 1.73$.
Step 2: Area of Sector of $120^\circ$
$$\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2$$
$$\text{Area} = \frac{120^\circ}{360^\circ} \times 3.14 \times 12^2 = \frac{1}{3} \times 3.14 \times 144 = 48 \times 3.14 = 150.72\text{ cm}^2$$
Step 3: Area of Triangle $\triangle AOB$ with $\theta = 120^\circ$
Draw perpendicular $OM \perp AB$.
In right triangle $\triangle OMA$:
$$\angle AOM = 60^\circ$$
$$OM = OA \cos 60^\circ = 12 \times \frac{1}{2} = 6\text{ cm}$$
$$AM = OA \sin 60^\circ = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}\text{ cm}$$
$$AB = 2 \times AM = 12\sqrt{3}\text{ cm}$$
Now calculate the area of $\triangle AOB$:
$$\text{Area}(\triangle AOB) = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 12\sqrt{3} \times 6 = 36\sqrt{3}\text{ cm}^2$$
Substitute $\sqrt{3} = 1.73$:
$$\text{Area}(\triangle AOB) = 36 \times 1.73 = 62.28\text{ cm}^2$$
Step 4: Compute Area of Corresponding Segment
$$\text{Area of Segment} = \text{Area of Sector} – \text{Area}(\triangle AOB)$$
$$\text{Area of Segment} = 150.72 – 62.28 = 88.44\text{ cm}^2$$
Final Answer:
The area of the corresponding segment is $88.44\text{ cm}^2$.
Question 8 (Page 161) [CBSE 2015, 2018, 2023 Set-3]
A horse is tied to a peg at one corner of a square shaped grass field of side $15\text{ m}$ by means of a $5\text{ m}$ long rope (see Fig. 11.8). Find:
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were $10\text{ m}$ long instead of $5\text{ m}$. (Use $\pi = 3.14$) Peg Square Field (15m x 15m) r = 5m Corner Angle = 90°
Answer:
Step 1: Identify Given Data
- Side of square grass field = $15\text{ m}$.
- Because the peg is fixed at the corner of a square, the angle grazed is a quadrant: $\theta = 90^\circ$.
- Initial rope length, $r_1 = 5\text{ m}$.
- Extended rope length, $r_2 = 10\text{ m}$.
- Value of $\pi = 3.14$.
Step 2: (i) Area Horse Can Graze with $5\text{ m}$ Rope
$$\text{Grazing Area}_1 = \frac{90^\circ}{360^\circ} \times \pi r_1^2 = \frac{1}{4} \times 3.14 \times 5^2$$
$$\text{Grazing Area}_1 = \frac{1}{4} \times 3.14 \times 25 = \frac{78.5}{4} = 19.625\text{ m}^2$$
Step 3: (ii) Area Horse Can Graze with $10\text{ m}$ Rope
$$\text{Grazing Area}_2 = \frac{90^\circ}{360^\circ} \times \pi r_2^2 = \frac{1}{4} \times 3.14 \times 10^2$$
$$\text{Grazing Area}_2 = \frac{1}{4} \times 3.14 \times 100 = \frac{314}{4} = 78.5\text{ m}^2$$
Step 4: Compute the Increase in Grazing Area
$$\text{Increase in Area} = \text{Grazing Area}_2 – \text{Grazing Area}_1$$
$$\text{Increase in Area} = 78.5 – 19.625 = 58.875\text{ m}^2$$
Alternatively:
$$\text{Increase} = \frac{1}{4}\pi (r_2^2 – r_1^2) = \frac{1}{4} \times 3.14 \times (100 – 25) = \frac{3.14 \times 75}{4} = 58.875\text{ m}^2$$
Final Answer:
(i) The area of the field the horse can graze is $19.625\text{ m}^2$.
(ii) The increase in the grazing area is $58.875\text{ m}^2$.
Question 9 (Page 161) [CBSE 2014, 2019 Set-1, 2024]
A brooch is made with silver wire in the form of a circle with diameter $35\text{ mm}$. The wire is also used in making $5\text{ diameters}$ which divide the circle into $10\text{ equal sectors}$ as shown in Fig. 11.9. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Answer:
Step 1: Identify Given Data
- Diameter of brooch, $d = 35\text{ mm}$.
- Radius of brooch, $r = \frac{35}{2}\text{ mm}$.
- Number of diameters made of silver wire = $5$.
- Total equal sectors = $10$.
- Value of $\pi = \frac{22}{7}$.
Step 2: (i) Total Length of Silver Wire Required
The silver wire is used along the circular circumference and for the 5 internal diameters:
$$\text{Total Wire Length} = \text{Circumference} + (5 \times \text{Diameter})$$
$$\text{Circumference} = \pi d = \frac{22}{7} \times 35 = 22 \times 5 = 110\text{ mm}$$
$$\text{Length of 5 Diameters} = 5 \times 35 = 175\text{ mm}$$
$$\text{Total Wire Length} = 110 + 175 = 285\text{ mm}$$
Step 3: (ii) Area of Each Sector of the Brooch
Because the circle is partitioned into $10$ equal sectors:
$$\text{Central Angle of each sector, } \theta = \frac{360^\circ}{10} = 36^\circ$$
$$\text{Area of Each Sector} = \frac{1}{10} \times \text{Total Circular Area} = \frac{1}{10} \times \pi r^2$$
$$\text{Area} = \frac{1}{10} \times \frac{22}{7} \times \left(\frac{35}{2}\right) \times \left(\frac{35}{2}\right)$$
Cancel $35$ with $7$ ($35 / 7 = 5$):
$$\text{Area} = \frac{1}{10} \times \frac{22 \times 5 \times 35}{4} = \frac{1}{10} \times \frac{110 \times 35}{4} = \frac{11 \times 35}{4} = \frac{385}{4}\text{ mm}^2$$
Converting to decimal:
$$\frac{385}{4} = 96.25\text{ mm}^2$$
Final Answer:
(i) The total length of the silver wire required is $285\text{ mm}$.
(ii) The area of each sector of the brooch is $\frac{385}{4}\text{ mm}^2$ (or $96.25\text{ mm}^2$).
Question 10 (Page 161) [CBSE 2013, 2017, 2020 Standard]
An umbrella has $8\text{ ribs}$ which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius $45\text{ cm}$, find the area between the two consecutive ribs of the umbrella.
Answer:
Step 1: Identify Given Data
- Radius of flat circular umbrella, $r = 45\text{ cm}$.
- Number of equally spaced ribs = $8$.
- Value of $\pi = \frac{22}{7}$.
Step 2: Determine Central Angle Between Consecutive Ribs
$$\theta = \frac{360^\circ}{8} = 45^\circ$$
Step 3: Compute Area Between Two Consecutive Ribs (One Sector)
$$\text{Area} = \frac{1}{8} \times \pi r^2 = \frac{45^\circ}{360^\circ} \times \frac{22}{7} \times 45^2$$
$$\text{Area} = \frac{1}{8} \times \frac{22}{7} \times 2025$$
Divide $22$ and $8$ by common factor $2$:
$$\text{Area} = \frac{11 \times 2025}{4 \times 7} = \frac{22275}{28}\text{ cm}^2$$
Converting to decimal:
$$\frac{22275}{28} \approx 795.54\text{ cm}^2$$
Final Answer:
The area between two consecutive ribs is $\frac{22275}{28}\text{ cm}^2$ (or approximately $795.54\text{ cm}^2$).
Question 11 (Page 161) [CBSE 2016, 2020 Standard, 2023]
A car has two wipers which do not overlap. Each wiper has a blade of length $25\text{ cm}$ sweeping through an angle of $115^\circ$. Find the total area cleaned at each sweep of the blades.
Answer:
Step 1: Identify Given Data
- Length of each blade = radius, $r = 25\text{ cm}$.
- Sweeping angle of each wiper, $\theta = 115^\circ$.
- Total number of non-overlapping wipers = $2$.
- Value of $\pi = \frac{22}{7}$.
Step 2: Formula for Total Area Cleaned by Both Blades
$$\text{Total Area} = 2 \times \left(\frac{\theta}{360^\circ} \times \pi r^2\right)$$
Step 3: Step-by-Step Substitution
$$\text{Total Area} = 2 \times \left(\frac{115^\circ}{360^\circ} \times \frac{22}{7} \times 25 \times 25\right)$$
Simplify the fractions:
$$\text{Total Area} = 2 \times \left(\frac{23}{72} \times \frac{22}{7} \times 625\right)$$
Cancel $2$ with $72$ ($72 / 2 = 36$):
$$\text{Total Area} = \frac{23}{36} \times \frac{22}{7} \times 625$$
Divide $22$ and $36$ by $2$:
$$\text{Total Area} = \frac{23 \times 11 \times 625}{18 \times 7} = \frac{253 \times 625}{126} = \frac{158125}{126}\text{ cm}^2$$
Converting to decimal:
$$\frac{158125}{126} \approx 1254.96\text{ cm}^2$$
Final Answer:
The total area cleaned at each sweep of the blades is $\frac{158125}{126}\text{ cm}^2$ (or approximately $1254.96\text{ cm}^2$).
Question 12 (Page 161) [CBSE 2012, 2018, 2024 Set-2]
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^\circ$ to a distance of $16.5\text{ km}$. Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$)
Answer:
Step 1: Identify Given Data
- Radius of the circular light beam sector, $r = 16.5\text{ km}$.
- Sector angle, $\theta = 80^\circ$.
- Value of $\pi = 3.14$.
Step 2: Formula and Calculation
$$\text{Area of Sea Warned} = \frac{\theta}{360^\circ} \times \pi r^2$$
$$\text{Area} = \frac{80^\circ}{360^\circ} \times 3.14 \times (16.5)^2$$
Simplify the angle ratio:
$$\frac{80}{360} = \frac{2}{9}$$
Compute $(16.5)^2$:
$$(16.5)^2 = 272.25$$
Substitute back:
$$\text{Area} = \frac{2}{9} \times 3.14 \times 272.25$$
Divide $272.25$ by $9$:
$$\frac{272.25}{9} = 30.25$$
$$\text{Area} = 2 \times 3.14 \times 30.25 = 6.28 \times 30.25 = 189.97\text{ km}^2$$
Final Answer:
The area of the sea over which the ships are warned is $189.97\text{ km}^2$.
Question 13 (Page 161) [CBSE 2015, 2017, 2019, 2023 Standard]
A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is $28\text{ cm}$, find the cost of making the designs at the rate of ₹$0.35\text{ per cm}^2$. (Use $\sqrt{3} = 1.7$) O 6 Equal Minor Segments
Answer:
Step 1: Identify Given Data
- Radius of round table cover, $r = 28\text{ cm}$.
- Number of equal designs = $6$.
- Rate of making designs = ₹$0.35\text{ per cm}^2$.
- Values: $\pi = \frac{22}{7}$, $\sqrt{3} = 1.7$.
Step 2: Characterise the Geometric Shape of the Designs
The $6$ equal designs correspond to the $6$ equal minor segments formed by the sides of a regular inscribed hexagon.
$$\text{Central angle subtended by each segment, } \theta = \frac{360^\circ}{6} = 60^\circ$$
Step 3: Compute Area of One Sector ($60^\circ$)
$$\text{Area of Sector} = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 28 \times 28$$
Cancel $28$ with $7$ ($28 / 7 = 4$):
$$\text{Area of Sector} = \frac{1}{6} \times 22 \times 4 \times 28 = \frac{2464}{6} = \frac{1232}{3}\text{ cm}^2$$
Step 4: Compute Area of Corresponding Equilateral Triangle
Since central angle is $60^\circ$ and two sides are radii ($28\text{ cm}$), the triangle is equilateral:
$$\text{Area}(\triangle) = \frac{\sqrt{3}}{4} \times r^2 = \frac{1.7}{4} \times 28 \times 28 = 1.7 \times 7 \times 28 = 1.7 \times 196 = 333.2\text{ cm}^2$$
Step 5: Compute Area of One Minor Segment (One Design)
$$\text{Area of One Design} = \frac{1232}{3} – 333.2\text{ cm}^2$$
Step 6: Compute Total Area of All 6 Designs
$$\text{Total Area} = 6 \times \left(\frac{1232}{3} – 333.2\right) = \left(6 \times \frac{1232}{3}\right) – (6 \times 333.2)$$
$$\text{Total Area} = (2 \times 1232) – 1999.2 = 2464 – 1999.2 = 464.8\text{ cm}^2$$
Step 7: Calculate Total Cost
$$\text{Cost} = \text{Total Area} \times \text{Rate} = 464.8 \times 0.35 = ₹162.68$$
Final Answer:
The cost of making the designs is ₹$162.68$.
Question 14 (Page 162) [CBSE 2018, 2020 Standard Objective]
Tick the correct answer in the following:
Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is:
(A) $\frac{p}{180} \times 2\pi R$
(B) $\frac{p}{180} \times \pi R^2$
(C) $\frac{p}{360} \times 2\pi R$
(D) $\frac{p}{720} \times 2\pi R^2$
Answer:
The correct option is (D) $\frac{p}{720} \times 2\pi R^2$.
Step-by-Step Mathematical Justification:
- The standard formula for the area of a sector of angle $p$ and radius $R$ is:
$$\text{Area of Sector} = \frac{p}{360} \times \pi R^2$$ - Multiplying both numerator and denominator by $2$:
$$\text{Area of Sector} = \frac{p}{360 \times 2} \times 2\pi R^2 = \frac{p}{720} \times 2\pi R^2$$ - Evaluating other options:
- (A) represents an arc length multiplied by an incorrect denominator.
- (B) has denominator $180$ instead of $360$ for area.
- (C) represents the arc length formula, not area.
- Hence, option (D) is mathematically identical to the sector area formula.
Final Answer:
Option (D).
[👉 Also Read: Class 10 Math Chapter 12 Surface Areas and Volumes NCERT Solutions]
Master High-Yield Board FAQs (Rank Math FAQ Schema Ready)
What is the exact formula for the area of a sector of a circle?
The area of a sector of a circle having radius $r$ and central angle $\theta$ (in degrees) is given by $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2$. If the angle $\theta$ is provided in radians, the formula simplifies to $\text{Area} = \frac{1}{2} r^2 \theta$.
How do you find the area of a minor segment without advanced trigonometry?
To find the area of a minor segment, compute the area of the entire sector using $\frac{\theta}{360^\circ} \times \pi r^2$, and subtract the area of the isosceles triangle formed by the two radii and the chord. For a $60^\circ$ central angle, use the equilateral triangle formula $\frac{\sqrt{3}}{4}r^2$. For a $90^\circ$ central angle, use the right-triangle formula $\frac{1}{2}r^2$.
How do you calculate the area of a triangle with a 120-degree central angle?
For a triangle with two equal sides of length $r$ and an included angle of $120^\circ$, draw an altitude from the centre to the chord. This altitude bisects the angle into two $60^\circ$ angles and divides the chord into two equal halves. The altitude equals $r \cos 60^\circ = \frac{r}{2}$ and the base equals $2r \sin 60^\circ = r\sqrt{3}$. Thus, the area is $\frac{1}{2} \times r\sqrt{3} \times \frac{r}{2} = \frac{\sqrt{3}}{4}r^2$.
What is the difference between a sector and a segment of a circle?
A sector is a pie-shaped region bounded by two distinct radii and their intercepted arc, with its vertex located at the centre of the circle. A segment is a region bounded strictly by a chord and its intercepted arc, lying entirely along the perimeter without connecting to the centre.
How are arc length and sector area mathematically related?
Arc length $l$ and sector area $A$ are directly proportional. Since $l = \frac{\theta}{360^\circ} \times 2\pi r$ and $A = \frac{\theta}{360^\circ} \times \pi r^2$, dividing the area by the arc length yields $\frac{A}{l} = \frac{r}{2}$. Rearranging this provides the direct relation $A = \frac{1}{2}lr$.
When should students substitute 3.14 for pi instead of 22/7?
Students should use $\pi = 3.14$ only when the examination question paper explicitly specifies it. Under CBSE marking guidelines, if no value is indicated in the problem statement, students must use $\pi = \frac{22}{7}$.
What angle is swept by the minute hand of a clock in one minute?
The minute hand of a clock completes a full revolution of $360^\circ$ in $60\text{ minutes}$. Therefore, the angle swept by the minute hand in one single minute is $\frac{360^\circ}{60} = 6^\circ$. In $t$ minutes, the angle swept is $(6t)^\circ$.
What angle does the hour hand of a clock sweep in one minute?
The hour hand covers $360^\circ$ in $12\text{ hours}$ ($720\text{ minutes}$). Therefore, the angular speed of the hour hand is $\frac{360^\circ}{720} = 0.5^\circ$ per minute. In 60 minutes (one hour), it sweeps $30^\circ$.
How do you find the area of a major segment?
The area of a major segment is calculated by subtracting the area of the corresponding minor segment from the total area of the circle: $\text{Area of Major Segment} = \pi r^2 – \text{Area of Minor Segment}$.
What is a quadrant of a circle and what is its area formula?
A quadrant is a sector representing one-fourth of a complete circular disc, defined by a right-angle central angle of $\theta = 90^\circ$. Its area formula is $\text{Area} = \frac{90^\circ}{360^\circ}\pi r^2 = \frac{1}{4}\pi r^2$.
What is the perimeter of a sector of a circle?
The perimeter of a sector includes the curved arc length plus the two straight bounding radii: $\text{Perimeter} = l + 2r = \left(\frac{\theta}{360^\circ} \times 2\pi r\right) + 2r$. Students often forget to add $2r$ and mistakenly calculate only the arc length.
Can the area of a sector be larger than the area of a semicircle?
Yes. If the central angle $\theta$ is greater than $180^\circ$, the sector is classified as a major sector, and its area will exceed half the total area of the circle ($\frac{1}{2}\pi r^2$).
What units must be written for arc length and sector area?
Arc length is a one-dimensional linear measure and must be expressed in units such as centimetres ($\text{cm}$), metres ($\text{m}$), or millimetres ($\text{mm}$). Sector area is a two-dimensional surface measure and must always be expressed in squared units, such as $\text{cm}^2$, $\text{m}^2$, or $\text{mm}^2$.
How should students tackle multi-blade wiper or rib problems?
In wiper or umbrella problems, find the area of a single sector using the given blade length (radius) and sweep angle. Then multiply that individual sector area by the total number of non-overlapping blades or ribs to determine the total area.
What common mistake should students avoid in round table cover design questions?
In table cover design questions like Exercise 11.1 Question 13, a common mistake is multiplying decimal approximations too early. Always express the area of one design algebraically as $(\text{Area of Sector} – \text{Area of Triangle})$, multiply by $6$ first to clear denominators, and then substitute decimal values like $\sqrt{3} = 1.7$ at the final stage to avoid compounding rounding errors.