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NCERT Solutions · Class 10 Maths · Chapter 11

Chapter 11: Areas Related to Circles (Mensuration)

Step-by-step NCERT solutions for Class 10 Maths Chapter 11, Areas Related to Circles (2026-27 reprint): Exercise 11.1 on the area of a sector and a segment, arc length, grazing area, brooch, umbrella, wipers, lighthouse and table-cover problems. All 14 questions are answered, with the key answer highlighted.

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-10-maths/chapter-11-areas-related-to-circles

For a sector of angle $θ$ (in degrees) in a circle of radius $r$ : arc length $= \frac{θ}{360} \times 2 π r$ , area of sector $= \frac{θ}{360} \times π r^{2}$ . Area of a segment $=$ area of the sector $-$ area of the triangle formed by the chord and the two radii. Use $π = \frac{22}{7}$ unless stated otherwise.

Exercise 11.1

Find the area of a sector of radius 6 cm and angle 60°.

Solution

$\frac{60}{360} \times \frac{22}{7} \times 36 = \frac{132}{7}$

132/7 cm² (≈ 18.86 cm²)

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

Solution

$2 π r = 22 \Rightarrow r = \frac{7}{2}$ cm. Quadrant $= \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{77}{8}$ .

77/8 cm² (= 9.625 cm²)

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

Solution

In 60 minutes it turns 360°, so in 5 minutes it turns 30°.

$\frac{30}{360} \times \frac{22}{7} \times 196 = \frac{154}{3}$

154/3 cm² (≈ 51.33 cm²)

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find (i) the minor segment (ii) the major sector (π = 3.14).

Solution

Minor sector $= \frac{1}{4} \times 3.14 \times 100 = 78.5$ cm²
Right triangle $= \frac{1}{2} \times 10 \times 10 = 50$ cm²

Minor segment $= 78.5 - 50 = 28.5$ cm²
Major sector $= 314 - 78.5 = 235.5$ cm²

(i) 28.5 cm² (ii) 235.5 cm²

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

Solution

$\frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22$ cm
$\frac{60}{360} \times \frac{22}{7} \times 441 = 231$ cm²
The triangle with two radii at 60° is equilateral, of area $\frac{3}{4} \times 2 1^{2} = \frac{441 3}{4}$ cm², so the segment $= (231 - \frac{441 3}{4})$ cm².

(i) 22 cm (ii) 231 cm² (iii) (231 – 441√3/4) cm² (≈ 40.05 cm²)

A chord of a circle of radius 15 cm subtends 60° at the centre. Find the minor and major segments (π = 3.14, √3 = 1.73).

Solution

Sector $= \frac{1}{6} \times 3.14 \times 225 = 117.75$ cm²
Equilateral triangle $= \frac{1.73}{4} \times 225 = 97.3125$ cm²
Minor segment $= 117.75 - 97.3125 = 20.4375$ cm²
Major segment $= 3.14 \times 225 - 20.4375 = 706.5 - 20.4375 = 686.0625$ cm²

Minor 20.4375 cm², major 686.0625 cm²

A chord of a circle of radius 12 cm subtends 120° at the centre. Find the area of the segment (π = 3.14, √3 = 1.73).

Solution

Sector $= \frac{120}{360} \times 3.14 \times 144 = 150.72$ cm²
Triangle $= \frac{1}{2} \times 12 \times 12 \times sin 120° = 72 \times \frac{3}{2} = 363 = 62.28$ cm²

(Or: the perpendicular from O to the chord is $12 cos 60° = 6$ and half the chord is $63$ .)

Segment $= 150.72 - 62.28 = 88.44$ cm²

88.44 cm²

A horse is tied to a corner of a square field of side 15 m by a 5 m rope. Find (i) the area it can graze (ii) the increase if the rope were 10 m long (π = 3.14).

Solution

The horse grazes a quadrant (the corner angle is 90°).

$\frac{1}{4} \times 3.14 \times 25 = 19.625$ m²
With 10 m: $\frac{1}{4} \times 3.14 \times 100 = 78.5$ m², an increase of $78.5 - 19.625 = 58.875$ m².

(i) 19.625 m² (ii) 58.875 m²

A brooch is a circle of diameter 35 mm made of silver wire, with 5 diameters dividing it into 10 equal sectors. Find (i) the total length of wire (ii) the area of each sector.

Solution

Circumference $+ 5$ diameters $= \frac{22}{7} \times 35 + 5 \times 35 = 110 + 175 = 285$ mm.
Each sector is $\frac{1}{10}$ of the circle: $\frac{1}{10} \times \frac{22}{7} \times (\frac{35}{2})^{2} = \frac{385}{4}$ mm².

(i) 285 mm (ii) 385/4 mm² (= 96.25 mm²)

An umbrella with 8 equally spaced ribs is a flat circle of radius 45 cm. Find the area between two consecutive ribs.

Solution

$\frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{22275}{28}$

22275/28 cm² (≈ 795.5 cm²)

A car has two non-overlapping wipers, each 25 cm long, sweeping 115°. Find the total area cleaned per sweep.

Solution

$2 \times \frac{115}{360} \times \frac{22}{7} \times 625 = \frac{158125}{126}$

158125/126 cm² (≈ 1254.96 cm²)

A lighthouse spreads red light over a sector of 80° to a distance of 16.5 km. Find the area of sea warned (π = 3.14).

Solution

$\frac{80}{360} \times 3.14 \times 16. 5^{2} = \frac{2}{9} \times 3.14 \times 272.25 = 189.97$

≈ 189.97 km²

A round table cover of radius 28 cm has six equal designs (the segments between the circle and an inscribed regular hexagon, Fig. 11.11). Find the cost of the designs at ₹0.35 per cm² (√3 = 1.7).

Solution

The six designs are the circle minus the regular hexagon, which is made of 6 equilateral triangles of side 28 cm.