NCERT Mathematics · Class 10 · English · 2026-27

Areas Related to Circles — Exercise 11.1

Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.

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Question 1

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

Step-by-step solution

Use π=22/7 unless a different value is specified.

A sector occupies θ/360 of the circle. Area=(60/360)π×6²=6π.

Answer: 132/7 cm² (approximately 18.86 cm²).

Resource ID: CB-10-M-CH11-EX11.1-Q1

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Question 2

Find a quadrant’s area when the circle circumference is 22 cm.

Step-by-step solution

Use π=22/7 unless a different value is specified.

2πr=22 gives r=22/[2(22/7)]=3.5 cm. A quadrant is one quarter of a circle, so area=π(3.5)²/4.

Answer: 9.625 cm².

Resource ID: CB-10-M-CH11-EX11.1-Q2

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Question 3

A 14 cm minute hand moves for five minutes. Find the swept area.

Step-by-step solution

Use π=22/7 unless a different value is specified.

Five minutes is 5/60 of one revolution, so θ=30°. Swept area=(30/360)π×14²=(1/12)(22/7)196.

Answer: 154/3 cm² (approximately 51.33 cm²).

Resource ID: CB-10-M-CH11-EX11.1-Q3

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Question 4

A radius-10 cm circle has a chord subtending 90°. Find the minor segment and major sector, using π=3.14.

Step-by-step solution

Use π=22/7 unless a different value is specified.

The 90° sector area is 314/4=78.5 cm². Its triangle has two perpendicular radius legs, giving area 10×10/2=50 cm². Minor segment=78.5−50. Major sector has angle 270°, so area=3×314/4.

Answer: (i) 28.5 cm²; (ii) 235.5 cm².

Resource ID: CB-10-M-CH11-EX11.1-Q4

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Question 5

A 60° arc belongs to a circle of radius 21 cm. Find arc length, sector area and minor segment area.

Step-by-step solution

Use π=22/7 unless a different value is specified.

Arc length=(60/360)2π×21=22 cm. Sector area=(60/360)π×21²=231 cm². The radii and chord make an equilateral triangle of side 21; its area=(√3/4)21². Subtract it from the sector.

Answer: (i) 22 cm; (ii) 231 cm²; (iii) 231−441√3/4 cm².

Resource ID: CB-10-M-CH11-EX11.1-Q5

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Question 6

A chord subtends 60° in a radius-15 cm circle. Find both segment areas, using π=3.14 and √3=1.73.

Step-by-step solution

Use π=22/7 unless a different value is specified.

Sector area=(60/360)3.14×225=117.75. The central triangle is equilateral, area=(1.73/4)225=97.3125. Minor segment=117.75−97.3125=20.4375. Whole circle=3.14×225=706.5; major segment is the remainder.

Answer: Minor 20.4375 cm²; major 686.0625 cm².

Resource ID: CB-10-M-CH11-EX11.1-Q6

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Question 7

Find the minor segment for radius 12 cm and central angle 120°, using π=3.14 and √3=1.73.

Step-by-step solution

Use π=22/7 unless a different value is specified.

Sector area=(120/360)3.14×144=150.72. Triangle area=(1/2)12²sin120°=36√3=62.28. Subtract triangle from sector.

Answer: 88.44 cm².

Resource ID: CB-10-M-CH11-EX11.1-Q7

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Question 8

A horse is tied at a corner of a 15 m square field. Compare grazing with 5 m and 10 m ropes, using π=3.14.

Step-by-step solution

Use π=22/7 unless a different value is specified.

Both ropes are shorter than the field side, so each reachable region is a quadrant. With 5 m: area=3.14×25/4=19.625 m². With 10 m: area=3.14×100/4=78.5 m². Increase=78.5−19.625.

Answer: (i) 19.625 m²; (ii) increase 58.875 m².

Resource ID: CB-10-M-CH11-EX11.1-Q8

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Question 9

A diameter-35 mm wire brooch contains its circumference and five diameters, making ten sectors. Find wire length and sector area.

Step-by-step solution

Use π=22/7 unless a different value is specified.

Circumference=πd=(22/7)35=110 mm. Five diameters add 5×35=175 mm, giving 285 mm total. Radius=17.5 mm; each sector has one tenth of the circle area: (1/10)(22/7)(17.5)².

Answer: (i) 285 mm; (ii) 96.25 mm².

Resource ID: CB-10-M-CH11-EX11.1-Q9

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Question 10

A flat circular umbrella has radius 45 cm and eight equally spaced ribs. Find one inter-rib area.

Step-by-step solution

Use π=22/7 unless a different value is specified.

Eight ribs divide the circle into eight equal sectors, each of angle 45°. Area=π×45²/8=(22/7)2025/8.

Answer: 22275/28 cm², approximately 795.54 cm².

Resource ID: CB-10-M-CH11-EX11.1-Q10

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Question 11

Two non-overlapping 25 cm wipers each sweep 115°. Find total cleaned area.

Step-by-step solution

Use π=22/7 unless a different value is specified.

Each sweep is modelled as a sector of radius 25 cm. With no overlap, add both areas: 2(115/360)(22/7)25².

Answer: 158125/126 cm², approximately 1254.96 cm².

Resource ID: CB-10-M-CH11-EX11.1-Q11

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Question 12

A lighthouse warns ships over an 80° sector reaching 16.5 km. Find sea area using π=3.14.

Step-by-step solution

Use π=22/7 unless a different value is specified.

Use sector area=(80/360)πr². Substitution gives (2/9)×3.14×16.5²=(2/9)×3.14×272.25.

Answer: 189.97 km².

Resource ID: CB-10-M-CH11-EX11.1-Q12

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Question 13

A radius-28 cm circular table cover has six segment designs outside an inscribed regular hexagon. Find design cost at ₹0.35/cm², using √3=1.7.

Step-by-step solution

Use π=22/7 unless a different value is specified.

A regular hexagon consists of six equilateral triangles with side equal to radius 28. Circle area=(22/7)28²=2464. Hexagon area=6(1.7/4)28²=1999.2. Design area=2464−1999.2=464.8 cm². Multiply by ₹0.35.

Answer: ₹162.68.

Resource ID: CB-10-M-CH11-EX11.1-Q13

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Question 14

Select the correct sector-area formula for angle p degrees and radius R.

Step-by-step solution

Use π=22/7 unless a different value is specified.

The usual formula is (p/360)πR². Option D, (p/720)×2πR², simplifies to this same expression; the other options do not.

Answer: Option D.

Resource ID: CB-10-M-CH11-EX11.1-Q14

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