Exercise 11.1 — Areas Related to Circles

Area of sector and segment, arc length — clocks, wipers, brooches, table covers, and one MCQ

14 questions✓ Free · step-by-step

12 marksNCERT Cl-10 Maths, Ex 11.1, Q1
Find the area of a sector of a circle with radius 6 cm if the angle of the sector is 60°.
Hint. The sector is a fraction of the whole circle — the fraction is the angle out of 360°.
Step 1 — Write the sector-area formula. Area = (θ/360°) × πr²

Step 2 — Substitute θ = 60°, r = 6. Area = (60/360) × (22/7) × 6²

Step 3 — Simplify the fraction first, since 60/360 reduces cleanly and keeps the numbers smaller. 60/360 = 1/6

Step 4 — Compute. Area = (1/6) × (22/7) × 36 = (22 × 36)/(7 × 6) = 22 × 6/7 = 132/7

✦ Answer: 132/7 cm² ≈ 18.86 cm²
Where students slip. Forgetting to reduce 60/360 to 1/6 before multiplying, which makes the arithmetic with 22/7 harder than it needs to be.
Another way. 60° is one-sixth of a full turn, so the sector is exactly one-sixth of the whole circle's area: (1/6) × (22/7) × 36, the same computation reached by recognising the fraction directly.
23 marksNCERT Cl-10 Maths, Ex 11.1, Q2
Find the area of a quadrant of a circle whose circumference is 22 cm.
Hint. A quadrant is a quarter of the circle — angle 90°. First recover the radius from the circumference.
Step 1 — Find the radius from the circumference, since the sector formula needs r, not the circumference. 2πr = 22 2 × (22/7) × r = 22 r = 22 × 7 / (2 × 22) = 7/2 = 3.5

Step 2 — A quadrant is a 90° sector. Area = (90/360) × πr² = (1/4) × (22/7) × (3.5)²

Step 3 — Compute (3.5)² = 12.25. Area = (1/4) × (22/7) × 12.25

Step 4 — Simplify. 12.25/7 = 1.75, so Area = (1/4) × 22 × 1.75 = (1/4) × 38.5 = 9.625

✦ Answer: 9.625 cm² (equivalently 77/8 cm²)
Where students slip. Using the circumference (22 cm) directly as if it were the radius or diameter. The circumference has to be converted to a radius through 2πr = C before anything else can be computed.
Another way. Keep everything as fractions: r = 7/2, so r² = 49/4, and the quadrant area is (1/4)(22/7)(49/4) = (22 × 49)/(7 × 16) = (22 × 7)/16 = 154/16 = 77/8.
33 marksNCERT Cl-10 Maths, Ex 11.1, Q3
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Hint. The minute hand sweeps the full 360° in 60 minutes. Work out what fraction of that 5 minutes represents.
Step 1 — Convert the time to an angle. The minute hand turns 360° in 60 minutes, so in 5 minutes it turns (5/60) × 360° = 30°

Step 2 — The hand traces out a sector of radius 14 cm and angle 30°. Area = (30/360) × (22/7) × 14²

Step 3 — Simplify 30/360 = 1/12. Area = (1/12) × (22/7) × 196

Step 4 — 196/7 = 28, so Area = (1/12) × 22 × 28 = (22 × 28)/12 = 616/12

Step 5 — Reduce. 616/12 = 154/3

✦ Answer: 154/3 cm² ≈ 51.33 cm²
Where students slip. Using 5° or 5/360 directly, forgetting that the question describes *minutes*, not degrees. The clock face has to be converted from a time interval to an angle first.
Another way. 5 minutes is 1/12 of an hour, and the minute hand's tip sweeps the whole circle in an hour — so the area swept is exactly 1/12 of the full circle's area, (1/12)(22/7)(196), reaching the same number without separately computing the angle in degrees.
44 marksNCERT Cl-10 Maths, Ex 11.1, 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 π = 3.14)
Hint. A right angle at the centre makes the triangle a simple one — half of a square, in effect. Find the sector, then the triangle, then subtract; the major sector is just what's left of the full circle.
Step 1 — Area of the sector (angle 90°, radius 10). sector = (90/360) × 3.14 × 10² = (1/4) × 314 = 78.5 cm²

Step 2 — Area of the triangle formed by the two radii and the chord. The angle between the two radii is 90°, so this triangle is right-angled with both legs equal to the radius: triangle = ½ × 10 × 10 = 50 cm²

Step 3 — Minor segment = sector − triangle. minor segment = 78.5 − 50 = 28.5 cm²

Step 4 — Major sector = full circle − minor sector. full circle = 3.14 × 10² = 314 cm² major sector = 314 − 78.5 = 235.5 cm²

✦ Answer: minor segment = 28.5 cm²; major sector = 235.5 cm²
Where students slip. Computing the triangle's area with ½r²sin θ as 50sin90° and then, in a moment of carelessness, using sin 90° = 0 (confusing it with cos 90°). Since sin 90° = 1, the triangle area is exactly 50, not 0.
Another way. Because the angle is exactly 90°, the triangle area can be seen directly as half of a 10×10 square — no sine needed at all, which is why this particular question is often used as the first example of the segment method.
55 marksNCERT Cl-10 Maths, Ex 11.1, Q5
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find (i) the length of the arc, (ii) the area of the sector formed by the arc, (iii) the area of the segment formed by the corresponding chord.
Hint. A 60° angle with two radii makes an equilateral triangle — both radii equal, and the included angle 60° forces the third side to equal the radius too.
(i) Arc length. arc = (60/360) × 2 × (22/7) × 21 = (1/6) × 2 × 22 × 3 = (1/6) × 132 = 22

✦ Arc length = 22 cm

(ii) Sector area. sector = (60/360) × (22/7) × 21² = (1/6) × (22/7) × 441

441/7 = 63, so sector = (1/6) × 22 × 63 = (22 × 63)/6 = 1386/6 = 231

✦ Sector area = 231 cm²

(iii) Segment area. With the included angle 60° and both sides equal to the radius (21 cm), the triangle formed is equilateral with side 21 — because an isosceles triangle with a 60° apex angle must have its base angles equal to (180° − 60°)/2 = 60° too, making all three angles, and hence all three sides, equal.

triangle area = (√3/4) × 21² = (√3/4) × 441 ≈ 0.4330 × 441 ≈ 190.98

segment = sector − triangle = 231 − 190.98 ≈ 40.02

✦ Segment area ≈ 40.04 cm² (taking √3 ≈ 1.732; the exact value is 231 − (441√3)/4 cm²)
Where students slip. Using ½r²sin θ for the triangle but forgetting that at 60° this reduces to the equilateral-triangle formula — both are the same number, but recognising the equilateral shape is faster and less error-prone than computing sin 60° from scratch.
Another way. ½r²sin 60° = ½ × 441 × (√3/2) = 441√3/4 — exactly the equilateral-triangle formula (√3/4)s² with s = r = 21, confirming the two routes agree.
65 marksNCERT Cl-10 Maths, Ex 11.1, 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 π = 3.14 and √3 = 1.73)
Hint. Same structure as the previous question — 60° gives an equilateral triangle — but this time both segments are wanted, and the given constants must be used exactly as stated.
Step 1 — Sector area (angle 60°, radius 15, π = 3.14). sector = (60/360) × 3.14 × 15² = (1/6) × 3.14 × 225

3.14 × 225 = 706.5, so sector = 706.5/6 = 117.75 cm²

Step 2 — Triangle area. The 60° angle between two equal radii makes an equilateral triangle of side 15, so triangle = (√3/4) × 15² = (1.73/4) × 225 = 1.73 × 56.25 = 97.3125 cm²

Step 3 — Minor segment = sector − triangle. minor segment = 117.75 − 97.3125 = 20.4375 cm²

Step 4 — Major segment = full circle − minor segment. full circle = 3.14 × 225 = 706.5 cm² major segment = 706.5 − 20.4375 = 686.0625 cm²

✦ Answer: minor segment ≈ 20.44 cm²; major segment ≈ 686.06 cm²
Where students slip. Computing the major segment as (full circle − sector) rather than (full circle − segment). The major *segment* is the complement of the minor *segment*, not of the minor sector — those are different quantities and only differ by the small triangle.
Another way. Since minor segment = sector − triangle, you can also write major segment = full circle − sector + triangle = major sector + triangle — a useful cross-check: 588.75 + 97.3125 = 686.0625 ✓ (where major sector = 706.5 − 117.75 = 588.75).
74 marksNCERT Cl-10 Maths, Ex 11.1, 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 π = 3.14 and √3 = 1.73)
Hint. 120° is obtuse, so the triangle is no longer equilateral. Use ½r²sin θ directly, remembering sin 120° = sin 60°.
Step 1 — Sector area (angle 120°, radius 12, π = 3.14). sector = (120/360) × 3.14 × 12² = (1/3) × 3.14 × 144

3.14 × 144 = 452.16, so sector = 452.16/3 = 150.72 cm²

Step 2 — Triangle area, using ½r²sin θ. sin 120° = sin(180° − 120°) = sin 60° = √3/2

triangle = ½ × 12² × sin 120° = ½ × 144 × (√3/2) = 36√3 = 36 × 1.73 = 62.28 cm²

Step 3 — Segment = sector − triangle. segment = 150.72 − 62.28 = 88.44 cm²

✦ Answer: 88.44 cm²
Where students slip. Treating 120° like 60° and reaching for the equilateral-triangle formula. The triangle here is isosceles but *not* equilateral — the two radii are equal, but the angle between them is 120°, so the base is longer than the two equal sides. ½r²sin θ is the general tool that works regardless.
Another way. sin 120° = sin 60° by the supplementary-angle identity, so the numeric value of the triangle's area at 120° is exactly the same as it would be at 60° for the same radius — only the sector fraction (120/360 rather than 60/360) is what changes between the two cases.
85 marksNCERT Cl-10 Maths, Ex 11.1, Q8 (Fig. 11.8)
A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by a 5 m long rope. Find (i) the area the horse can graze, (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
Hint. At a corner, the interior angle of the square is 90°, so the horse can graze exactly a quarter-circle — as long as the rope is shorter than the sides, which it is in both cases here.
(i) Grazing area with a 5 m rope.

Step 1 — The horse is tied at a corner of the square, where the interior angle is 90°. Since the rope (5 m) is shorter than the sides (15 m), the horse's grazing region is a quarter-circle of radius 5 m, entirely within the field.

Step 2 — Compute. area = (90/360) × 3.14 × 5² = (1/4) × 3.14 × 25 = 78.5/4 = 19.625 m²

✦ (i) = 19.625 m²

(ii) Increase when the rope becomes 10 m.

Step 1 — With a 10 m rope, the same reasoning applies — 10 m is still less than the 15 m side, so the region is still a clean quarter-circle, this time of radius 10 m. new area = (1/4) × 3.14 × 10² = 314/4 = 78.5 m²

Step 2 — The increase is the new area minus the old. increase = 78.5 − 19.625 = 58.875 m²

✦ (ii) = 58.875 m²
Where students slip. Assuming the grazing region is a full circle rather than a quarter-circle. The peg is at a *corner* of the square, so the field itself blocks three-quarters of the circle the horse could otherwise reach — only the 90° wedge between the two sides is available.
Another way. Since both areas share the same 90°/360° = 1/4 factor and the same π, the increase can be found by factoring: increase = (1/4)(3.14)(10² − 5²) = (1/4)(3.14)(75) = 58.875, without separately computing both areas first.
95 marksNCERT Cl-10 Maths, Ex 11.1, Q9 (Fig. 11.9)
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) the total length of the silver wire required, (ii) the area of each sector of the brooch.
Hint. The wire forms the circle's boundary plus five full diameters — not five radii. Ten equal sectors means each one has angle 36°.
(i) Total wire length.

Step 1 — The circle's circumference, with diameter 35 mm (so radius 17.5 mm). circumference = πd = (22/7) × 35 = 110 mm

Step 2 — Five diameters, each a full 35 mm line across the circle. wire in diameters = 5 × 35 = 175 mm

Step 3 — Add the two parts. total wire = 110 + 175 = 285 mm

✦ (i) = 285 mm

(ii) Area of each sector.

Step 1 — Five diameters through the centre create 10 equal sectors, so each has angle 360°/10 = 36°

Step 2 — Apply the sector formula with radius 17.5 mm. area = (36/360) × (22/7) × 17.5² = (1/10) × (22/7) × 306.25

Step 3 — 306.25/7 = 43.75, so area = (1/10) × 22 × 43.75 = 962.5/10 = 96.25

✦ (ii) = 96.25 mm² per sector
Where students slip. Using 5 × (radius) = 5 × 17.5 = 87.5 mm for the diameters' wire, mistaking radii for diameters. Each of the five lines drawn is a *full diameter* (35 mm), not a radius — they run all the way across the circle through the centre.
Another way. Check part (ii) against the whole circle: 10 sectors of 96.25 mm² should total the full circle's area. Circle area = (22/7)(17.5)² = 962.5 mm², and 10 × 96.25 = 962.5 ✓
103 marksNCERT Cl-10 Maths, Ex 11.1, Q10 (Fig. 11.10)
An umbrella has 8 ribs which are equally spaced. Assuming the umbrella is a flat circle of radius 45 cm, find the area between two consecutive ribs.
Hint. Eight equally spaced ribs divide the circle into eight equal sectors — you don't need to know the angle explicitly if you think in terms of a fraction of the whole area.
Step 1 — Eight ribs radiating from the centre create 8 equal sectors, so the area between two consecutive ribs is 1/8 of the full circle.

Step 2 — Compute the full circle's area. full area = (22/7) × 45² = (22/7) × 2025

Step 3 — Divide by 8. area between ribs = (1/8) × (22/7) × 2025 = (22 × 2025)/(7 × 8) = 44550/56

Step 4 — Simplify. 44550/56 = 795.5357...

✦ Answer: ≈ 795.54 cm² (exactly 44550/56 = 22275/28 cm²)
Where students slip. Computing the angle 360°/8 = 45° and then re-deriving the sector formula from scratch. That is correct but slower — since the ribs are equally spaced, 'one of eight equal slices' is already the whole story.
Another way. As a check, 8 × 795.54 ≈ 6364.3 cm², which should match the full circle's area (22/7)(2025) ≈ 6364.3 cm² ✓
113 marksNCERT Cl-10 Maths, Ex 11.1, Q11
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Hint. Each wiper cleans a sector of radius 25 cm and angle 115°. Since the two wipers don't overlap, the total is simply twice one sector's area.
Step 1 — Area cleaned by one wiper (radius 25, angle 115°). one wiper = (115/360) × (22/7) × 25²

Step 2 — Compute the pieces. 25² = 625, and 115 × 625 = 71875. one wiper = (71875 × 22) / (360 × 7) = 1581250/2520

Step 3 — Simplify. 1581250/2520 ≈ 627.48

Step 4 — Since the wipers do not overlap, the total cleaned area is twice this. total = 2 × 627.48 ≈ 1254.96

✦ Answer: ≈ 1254.96 cm² (one wiper ≈ 627.48 cm², two wipers together)
Where students slip. Halving the angle or the radius by mistake, thinking the two wipers 'share' the sweep somehow. Because the question states they do not overlap, the two areas are simply added — neither wiper's area is affected by the other's existence.
Another way. Compute the doubled quantity directly to avoid rounding twice: total = 2 × (115/360)(22/7)(625) = (115 × 22 × 625 × 2)/(360 × 7) = 3162500/2520 ≈ 1254.96, matching the two-step method.
123 marksNCERT Cl-10 Maths, Ex 11.1, Q12
To warn ships of underwater rocks, a lighthouse spreads a red light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Hint. The 16.5 km is the radius of the lit sector — the lighthouse is the centre.
Step 1 — Identify the sector's radius and angle. radius = 16.5 km, angle = 80°

Step 2 — Apply the sector formula. area = (80/360) × 3.14 × 16.5²

Step 3 — Compute 16.5² = 272.25, and simplify 80/360 = 2/9. area = (2/9) × 3.14 × 272.25

Step 4 — Multiply through, since both factors are now plain decimals. 3.14 × 272.25 = 854.865 area = (2/9) × 854.865 = 1709.73/9 ≈ 189.97

✦ Answer: ≈ 189.97 km²
Where students slip. Treating 16.5 km as a diameter and halving it before use. The question describes light spreading 'to a distance of 16.5 km' from the lighthouse, which is exactly the radius of the illuminated sector, with no halving needed.
Another way. Keep the fraction 80/360 = 2/9 unreduced through the whole calculation and multiply at the end: (2 × 3.14 × 272.25)/9 = 1709.73/9, which avoids an extra rounding step in the middle.
135 marksNCERT Cl-10 Maths, Ex 11.1, Q13 (Fig. 11.11)
A round table cover has six equal designs, as shown, cut from a circle of radius 28 cm. Find the cost of making the designs at ₹0.35 per cm². (Use √3 = 1.7)
Hint. Six equal designs means each occupies a 60° sector, and the six straight cuts form a regular hexagon inscribed in the circle. Each design is a segment.
Step 1 — Six equal designs split the circle into six 60° sectors. Because a regular hexagon inscribed in a circle has each side equal to the radius, the triangle cut off by each chord (side of the hexagon) is equilateral with side 28 cm — this is the same 60°-gives-equilateral fact used in Q5 and Q6.

Step 2 — Area of one sector (radius 28, angle 60°). sector = (60/360) × (22/7) × 28² = (1/6) × (22/7) × 784

784/7 = 112, so sector = (1/6) × 22 × 112 = 2464/6 = 410.6667 cm²

Step 3 — Area of the equilateral triangle (side 28). triangle = (√3/4) × 28² = (1.7/4) × 784 = 1.7 × 196 = 333.2 cm²

Step 4 — Each design is a segment: sector − triangle. one design = 410.6667 − 333.2 = 77.4667 cm²

Step 5 — All six designs. total design area = 6 × 77.4667 = 464.8 cm²

Step 6 — Cost at ₹0.35 per cm². cost = 464.8 × 0.35 = 162.68

✦ Answer: the cost of making the designs is ₹162.68
Where students slip. Computing the area of the whole hexagon and calling that the design area. The designs are the six *segments* — the curved slivers between the hexagon's sides and the circle's arc — not the hexagon itself.
Another way. Total design area also equals (full circle) − (hexagon area): circle = (22/7)(784) = 2464 cm², hexagon = 6 × 333.2 = 1999.2 cm², difference = 2464 − 1999.2 = 464.8 cm² ✓ — the same total reached from the other direction.
141 markNCERT Cl-10 Maths, Ex 11.1, Q14
Tick the correct answer: the area of a sector of angle p (in degrees) of a circle with radius R is: (A) (p/180) × 2πR (B) (p/180) × πR² (C) (p/360) × 2πR (D) (p/720) × 2πR².
Hint. Start from the formula you already know, (p/360)πR², and check which option matches it exactly.
Step 1 — Recall the standard formula. area of sector = (p/360) × πR²

Step 2 — Compare each option to this. (A) has 2πR, which has units of length, not area — dimensionally wrong for an area. (C) also has 2πR, and additionally uses p/360 rather than converting it — same dimensional problem. (B) is (p/180)πR², which is exactly twice the correct formula. (D) is (p/720) × 2πR². Simplify: (p/720) × 2 = 2p/720 = p/360, so this becomes (p/360)πR² — exactly the standard formula.

✦ Answer: (D)
Where students slip. Picking (B) because p/180 'looks similar' to p/360 without checking the factor of 2 carefully. (p/180)πR² is double the correct area, since 180 is half of 360.
Another way. Test with a known case: a semicircle has p = 180°. The correct area is ½πR². Option (D) gives (180/720)(2πR²) = (1/4)(2πR²) = πR²/2 ✓, while option (B) gives (180/180)πR² = πR², which is the area of a *full* circle, not a semicircle — confirming (B) is wrong and (D) is right.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (this chapter now has a single exercise, 11.1, with 14 questions; areas of combinations of plane figures, with the old second exercise, are no longer part of this chapter). Questions are referenced from the NCERT textbook for identification.

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