Exercise 11.1 answers: Areas Related to Circles

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1. A full turn round the centre is 360°. This sector's angle is 60°, so it takes 60/360 = 1/6 of the whole circle.
2. Area of the whole circle = π × r² = 22/7 × 6² = 22/7 × 36 cm².
3. Area of the sector = 1/6 × 22/7 × 36 = 132/7 cm².

Answer132/7 cm² (= 18⁶/₇ cm²)

Watch this explained “One worked all the way through”, 7:04 into A sector's area as its share of the full turn

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1. First find the radius from the circumference: circumference = 2 × π × r, so 22 = 2 × 22/7 × r.
2. Solving, r = 22 ÷ (44/7) = 7/2 cm.
3. A quadrant is a sector of 90°, which is 1/4 of the whole circle.
4. Area of the whole circle = π × r² = 22/7 × (7/2)² = 77/2 cm².
5. Area of the quadrant = 1/4 × 77/2 = 77/8 cm².

Answer77/8 cm² (= 9⁵/₈ cm²)

Watch this explained “Running the rule backwards”, 10:02 into The arc's length by the same share argument

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1. The minute hand's length is the radius of the circle it sweeps: r = 14 cm.
2. In 60 minutes the hand sweeps a full 360°. In 5 minutes it sweeps 5/60 × 360° = 30°.
3. Area swept = 30/360 × π × r² = 1/12 × 22/7 × 196 cm².
4. This works out to 154/3 cm².

Answer154/3 cm² (= 51¹/₃ cm²)

Watch this explained “Getting to the angle first”, 8:50 into A sector's area as its share of the full turn

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(i) minor segment

1. The chord's sector has angle 90° (a right angle), so its area is 90/360 × π × r² = 1/4 × 3.14 × 100 = 78.5 cm².
2. Since the angle at the centre is 90°, the triangle formed by the two radii and the chord is a right-angled triangle with both legs equal to the radius, 10 cm.
3. Area of this triangle = 1/2 × 10 × 10 = 50 cm².
4. Minor segment = sector − triangle = 78.5 − 50 = 28.5 cm².

Answer57/2 cm² (= 28.5 cm²)

(ii) major sector

1. Area of the whole circle = π × r² = 3.14 × 100 = 314 cm².
2. The major sector is everything except the minor sector (angle 90°): major sector = whole circle − minor sector = 314 − 78.5 = 235.5 cm².

Answer471/2 cm² (= 235.5 cm²)

Watch this explained “Two more that need no ratio table”, 11:38 into A segment as what is left when the triangle is taken away

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(i) the length of the arc

1. Length of arc = 60/360 × 2 × π × r = 1/6 × 2 × 22/7 × 21.
2. This works out to 22 cm.

Answer22 cm

(ii) area of the sector formed by the arc

1. Area of the sector = 60/360 × π × r² = 1/6 × 22/7 × 441.
2. This works out to 231 cm².

Answer231 cm²

(iii) area of the segment formed by the corresponding chord

1. The two radii (21 cm each) and the chord form a triangle. Since the angle between the two equal radii is 60°, this triangle is equilateral with side 21 cm.
2. Area of the equilateral triangle = (√3/4) × 21² = 441√3/4 cm².
3. Segment area = sector − triangle = 231 − 441√3/4 cm².

Answer(231 − 441√3/4) cm²

Watch this explained “One circle, both answers, one fraction”, 5:27 into The arc's length by the same share argument

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1. Area of the sector = 60/360 × π × r² = 1/6 × 3.14 × 225 = 117.75 cm².
2. Since the angle is 60°, the triangle on the two radii and the chord is equilateral with side 15 cm.
3. Area of this triangle = (√3/4) × 15² = (1.73/4) × 225 = 97.3125 cm².
4. Minor segment = sector − triangle = 117.75 − 97.3125 = 20.4375 cm².
5. Area of the whole circle = π × r² = 3.14 × 225 = 706.5 cm².
6. Major segment = whole circle − minor segment = 706.5 − 20.4375 = 686.0625 cm².

AnswerMinor segment = 327/16 cm² (= 20.4375 cm²); major segment = 10977/16 cm² (= 686.0625 cm²)

Watch this explained “The other piece the chord makes”, 10:06 into A segment as what is left when the triangle is taken away

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1. Area of the sector = 120/360 × π × r² = 1/3 × 3.14 × 144 = 150.72 cm².
2. Let O be the centre and AB the chord. Draw OM perpendicular to AB. Triangles OMA and OMB are congruent (RHS: OA = OB, OM is common), so M is the midpoint of AB and ∠AOM = ∠BOM = 120°/2 = 60°.
3. In right triangle OMA: OM = OA × cos 60° = 12 × 1/2 = 6 cm, and AM = OA × sin 60° = 12 × √3/2 = 6√3 cm. So AB = 2 × 6√3 = 12√3 cm.
4. Area of triangle OAB = 1/2 × AB × OM = 1/2 × 12√3 × 6 = 36√3 = 36 × 1.73 = 62.28 cm².
5. Area of the segment = sector − triangle = 150.72 − 62.28 = 88.44 cm².

Answer2211/25 cm² (= 88.44 cm²)

Watch this explained “Two ratios at the half angle”, 7:16 into A segment as what is left when the triangle is taken away

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(i) the area of … the field in which the horse can graze

1. The peg is at a corner of the square field, and a square's corner angle is 90°, so the horse can only graze inside a 90° sector.
2. With rope length 5 m as the radius, grazing area = 90/360 × π × r² = 1/4 × 3.14 × 25 = 19.625 m².

Answer157/8 m² (= 19.625 m²)

(ii) increase in the grazing area if the rope were 10 m long

1. With rope length 10 m (which still fits, since 10 m is less than the 15 m side), grazing area = 1/4 × 3.14 × 100 = 78.5 m².
2. Increase in grazing area = 78.5 − 19.625 = 58.875 m².

Answer471/8 m² (= 58.875 m²)

Watch this explained “Getting to the angle first”, 8:50 into A sector's area as its share of the full turn

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(i) the total length of the silver wire required

1. Diameter = 35 mm, so radius = 17.5 mm.
2. Length of the circular rim = π × diameter = 22/7 × 35 = 110 mm.
3. The 5 diameters add 5 × 35 = 175 mm of wire.
4. Total wire = 110 + 175 = 285 mm.

Answer285 mm

(ii) the area of each sector of the brooch

1. 5 diameters through the centre cut the circle into 10 equal sectors.
2. Each sector's angle = 360°/10 = 36°.
3. Area of the whole circle = π × r² = 22/7 × 17.5² = 962.5 mm².
4. Area of each sector = 962.5/10 = 96.25 mm².

Answer385/4 mm² (= 96.25 mm²)

Watch this explained “One curved length and five straight ones”, 11:08 into The arc's length by the same share argument

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1. 8 equally spaced ribs divide the flat circle into 8 equal sectors.
2. Each sector's angle = 360°/8 = 45°.
3. Area between two consecutive ribs = 45/360 × π × r² = 1/8 × 22/7 × 45².
4. This works out to 22275/28 cm².

Answer22275/28 cm² (≈ 795.54 cm²)

Watch this explained “Getting to the angle first”, 8:50 into A sector's area as its share of the full turn

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1. Each blade sweeps a sector of radius 25 cm and angle 115°.
2. Area cleaned by one blade = 115/360 × π × r² = 115/360 × 22/7 × 625 = 158125/252 cm².
3. The two wipers do not overlap, so they clean two separate regions, and the total area is twice the area cleaned by one blade.
4. Total area cleaned = 2 × 158125/252 = 158125/126 cm².

Answer158125/126 cm² (≈ 1254.96 cm²)

Watch this explained “Two blades, or one longer sweep?”, 9:54 into A sector's area as its share of the full turn

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1. The lit-up region is a sector with radius 16.5 km (the distance the light reaches) and angle 80°.
2. Area = 80/360 × π × r² = 80/360 × 3.14 × 16.5².
3. This works out to 18997/100 km².

Answer18997/100 km² (= 189.97 km²)

Watch this explained “One worked all the way through”, 7:04 into A sector's area as its share of the full turn

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1. Six equal designs round the edge mean six equal chords, so each design's angle at the centre = 360°/6 = 60°.
2. Area of the sector for one design = 60/360 × π × r² = 1/6 × 22/7 × 784 = 1232/3 cm².
3. Since the angle is 60°, the triangle on the two radii and the chord is equilateral with side 28 cm.
4. Area of this triangle = (√3/4) × 28² = (1.7/4) × 784 = 333.2 cm².
5. Area of one design (the segment) = sector − triangle = 1232/3 − 333.2 = 1162/15 cm².
6. Area of all 6 designs = 6 × 1162/15 = 2324/5 cm² = 464.8 cm².
7. Cost = 464.8 × ₹0.35 = ₹162.68.

Answer₹4067/25 (= ₹162.68)

Watch this explained “Six of them, and then the bill”, 12:22 into A segment as what is left when the triangle is taken away

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1. The sector-area rule is: area = p/360 × π × R².
2. Check each option's powers of R first: options (A) and (C) have R to the power 1, which is a length, not an area — so they are ruled out straight away.
3. Between the two area options, (B) p/180 × πR² is p/360 × πR² doubled, which is twice too big.
4. (D) p/720 × 2πR² simplifies to p/360 × πR², which matches the rule exactly.

Answer(D) p/720 × 2πR²

Watch this explained “Which of these could even be an area?”, 10:55 into A sector's area as its share of the full turn