NCERT Class 10 Maths – Exercise 11.1 Solutions

NCERT Class 10 Maths

Chapter 11 – Areas Related to Circles | Exercise 11.1

(Rationalized Syllabus 2025-26 | Formerly Ex 12.2)

💡 Important Formulas

  • Area of Sector: $\frac{\theta}{360} \times \pi r^2$
  • Length of Arc: $\frac{\theta}{360} \times 2\pi r$
  • Area of Segment: (Area of Sector) – (Area of Triangle formed by chord and radii)
Q1

Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Radius $r = 6$ cm, Angle $\theta = 60^\circ$.
$$\text{Area} = \frac{\theta}{360} \times \pi r^2$$
$$= \frac{60}{360} \times \frac{22}{7} \times 6 \times 6$$
$$= \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{22}{7} \times 6 = \frac{132}{7}$$
✔️ Area = $\frac{132}{7}$ cm² (or 18.86 cm²)
Q2

Find the area of a quadrant of a circle whose circumference is 22 cm.

Step 1: Find Radius
$$2\pi r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = \frac{7}{2} \text{ cm}$$
Step 2: Find Area of Quadrant (Quadrant is $90^\circ$ or $1/4$ circle)
$$\text{Area} = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$$
$$= \frac{1}{4} \times 22 \times \frac{1}{2} \times \frac{7}{2} = \frac{11 \times 7}{8} = \frac{77}{8}$$
✔️ Area = $\frac{77}{8}$ cm²
Q3

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Radius $r = 14$ cm.
Angle swept in 60 mins $= 360^\circ$.
Angle swept in 5 mins $\theta = \frac{360}{60} \times 5 = 30^\circ$.
$$\text{Area} = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14$$
$$= \frac{1}{12} \times 22 \times 2 \times 14 = \frac{1}{12} \times 616 = \frac{154}{3}$$
✔️ Area = $\frac{154}{3}$ cm²
Q4

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use $\pi = 3.14$)

(i) Minor Segment
$\theta = 90^\circ$, $r = 10$. Area of Segment = Area of Sector – Area of $\triangle$.
Area of Sector $= \frac{90}{360} \times 3.14 \times 100 = \frac{1}{4} \times 314 = 78.5$ cm².
Area of $\triangle$ (Right angled at center) $= \frac{1}{2} \times r \times r = \frac{1}{2} \times 100 = 50$ cm².
$$\text{Area Segment} = 78.5 – 50 = 28.5$$
✔️ Area of Minor Segment = 28.5 cm²
(ii) Major Sector
Angle for major sector $= 360 – 90 = 270^\circ$.
$$\text{Area} = \frac{270}{360} \times 3.14 \times 100 = \frac{3}{4} \times 314 = 235.5$$
✔️ Area of Major Sector = 235.5 cm²
Q5

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) length of the arc (ii) area of the sector (iii) area of the segment formed by the corresponding chord.

(i) Length of Arc
$$L = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 44 \times 3 = 22 \text{ cm}$$
(ii) Area of Sector
$$A = \frac{60}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{6} \times 22 \times 3 \times 21 = 231 \text{ cm}^2$$
(iii) Area of Segment
Area of $\triangle$ (Equilateral, since $\theta=60^\circ$) $= \frac{\sqrt{3}}{4}r^2$.
$$\text{Area } \triangle = \frac{\sqrt{3}}{4} \times 21 \times 21 = \frac{441\sqrt{3}}{4} \text{ cm}^2$$
✔️ Area of Segment = $\left(231 – \frac{441\sqrt{3}}{4}\right)$ cm²
Q6

A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Minor Segment
Area Sector $= \frac{60}{360} \times 3.14 \times 15^2 = \frac{1}{6} \times 3.14 \times 225 = 117.75$ cm².
Area $\triangle$ (Equilateral) $= \frac{1.73}{4} \times 225 = 97.3125$ cm².
Area Minor Segment $= 117.75 – 97.3125 = 20.4375$ cm².
Major Segment
Area Circle $= 3.14 \times 225 = 706.5$ cm².
Area Major = Area Circle – Area Minor.
$$706.5 – 20.4375 = 686.0625 \text{ cm}^2$$
✔️ Minor: 20.4375 cm² | Major: 686.0625 cm²
Q7

A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Area of Sector:
$$A_{sec} = \frac{120}{360} \times 3.14 \times 144 = \frac{1}{3} \times 452.16 = 150.72 \text{ cm}^2$$
Area of Triangle ($120^\circ$): Draw perpendicular from center to chord. It bisects angle (to $60^\circ$) and chord.
Area $\triangle = r^2 \sin 60^\circ \cos 60^\circ$ (or simply $1/2 \times \text{base} \times \text{height}$).
Using formula Area $= \frac{1}{2} r^2 \sin \theta = \frac{1}{2} (144) \sin 120^\circ = 72 \times \frac{\sqrt{3}}{2} = 36\sqrt{3}$.
$$36 \times 1.73 = 62.28 \text{ cm}^2$$
Area of Segment: $150.72 – 62.28 = 88.44$ cm².
✔️ Area = 88.44 cm²
Q8

A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. 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 m long instead of 5 m. (Use $\pi = 3.14$)

(i) Radius = 5m
Corner of square $= 90^\circ$, so it’s a quadrant.
$$\text{Area} = \frac{1}{4} \times 3.14 \times 5^2 = \frac{78.5}{4} = 19.625 \text{ m}^2$$
(ii) Radius = 10m
$$\text{New Area} = \frac{1}{4} \times 3.14 \times 10^2 = \frac{314}{4} = 78.5 \text{ m}^2$$
Increase $= 78.5 – 19.625 = 58.875$ m².
✔️ (i) 19.625 m² (ii) 58.875 m²
Q9

A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find: (i) total length of wire (ii) area of each sector.

(i) Length of Wire
Wire = Circumference + 5 Diameters.
$$C = \pi d = \frac{22}{7} \times 35 = 110 \text{ mm}$$
$$5 \times d = 5 \times 35 = 175 \text{ mm}$$
Total $= 110 + 175 = 285$ mm.
(ii) Area of Each Sector
Total Area divided by 10 (or $\theta = 36^\circ$).
$$\text{Area} = \frac{1}{10} \times \frac{22}{7} \times \left(\frac{35}{2}\right)^2 = \frac{1}{10} \times \frac{22}{7} \times \frac{1225}{4}$$
$$= \frac{385}{4} \text{ mm}^2$$
✔️ (i) 285 mm (ii) 96.25 mm²
Q10

An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

There are 8 sectors. Area of one sector $= \frac{1}{8}$ of circle.
$$\text{Area} = \frac{1}{8} \times \frac{22}{7} \times 45 \times 45$$
$$= \frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{22275}{28} \text{ cm}^2$$
✔️ Area = $\frac{22275}{28}$ cm²
🎉 Exercise 11.1 Completed | Chapter 11 Finished!
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