Exercise 11.1 Practice
Sector and Segment Areas
Q1: Sector Area
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Find the area of a sector of a circle with radius 7 cm if the angle of the sector is $45^\circ$.
Formula: Area = $\frac{\theta}{360} \times \pi r^2$
Substitution: Area = $\frac{45}{360} \times \frac{22}{7} \times 7 \times 7$
Calculation: $\frac{1}{8} \times 22 \times 7 = \frac{154}{8} = 19.25$
Area is 19.25 cm².
Q2: Quadrant Area
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Find the area of a quadrant of a circle whose circumference is 44 cm.
Find Radius: $2\pi r = 44 \Rightarrow 2 \times \frac{22}{7} \times r = 44 \Rightarrow r = 7$ cm.
Find Area: Quadrant angle is $90^\circ$. Area = $\frac{1}{4} \pi r^2$.
Area = $\frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = \frac{154}{4} = 38.5$ cm².
Area is 38.5 cm².
Q3: Clock Hand
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The length of the minute hand of a clock is 21 cm. Find the area swept by the minute hand in 10 minutes.
Find Angle: Minute hand covers $360^\circ$ in 60 mins. In 10 mins: $\frac{360}{60} \times 10 = 60^\circ$.
Find Area: Area = $\frac{60}{360} \times \frac{22}{7} \times 21 \times 21$.
Area = $\frac{1}{6} \times 22 \times 3 \times 21 = 231$ cm².
Area swept is 231 cm².
Q4: Chord Segments
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A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment
(ii) major sector
(Use $\pi = 22/7$)
(i) minor segment
(ii) major sector
(Use $\pi = 22/7$)
(i) Minor Segment: Area Sector - Area Triangle.
Sector (90°): $\frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154$.
Triangle (Right $\Delta$): $\frac{1}{2} \times 14 \times 14 = 98$.
Segment Area = $154 - 98 = 56$ cm².
Sector (90°): $\frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154$.
Triangle (Right $\Delta$): $\frac{1}{2} \times 14 \times 14 = 98$.
Segment Area = $154 - 98 = 56$ cm².
(ii) Major Sector: Total Area - Minor Sector Area.
Total Area = $\pi r^2 = 616$.
Major Sector = $616 - 154 = 462$ cm².
Total Area = $\pi r^2 = 616$.
Major Sector = $616 - 154 = 462$ cm².
Minor Segment: 56 cm²; Major Sector: 462 cm².
Q5: Arc and Sector
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In a circle of radius 35 cm, an arc subtends an angle of $60^\circ$ at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord
(Use $\pi = 22/7, \sqrt{3} = 1.73$)
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord
(Use $\pi = 22/7, \sqrt{3} = 1.73$)
(i) Length of Arc: $\frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 35 = \frac{1}{6} \times 220 = 36.67$ cm.
(ii) Area of Sector: $\frac{60}{360} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 35 \times 35 = \frac{1}{6} \times 3850 = 641.67$ cm².
(iii) Area of Segment: Area Sector - Area Triangle.
Area $\Delta$ (Equilateral, 60°): $\frac{\sqrt{3}}{4} a^2 = \frac{1.73}{4} \times 35^2 = 529.81$ cm².
Segment = $641.67 - 529.81 = 111.86$ cm².
Area $\Delta$ (Equilateral, 60°): $\frac{\sqrt{3}}{4} a^2 = \frac{1.73}{4} \times 35^2 = 529.81$ cm².
Segment = $641.67 - 529.81 = 111.86$ cm².
Q6: Minor/Major Segments
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A chord of a circle of radius 10 cm subtends an angle of $60^\circ$ at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
Area of Sector: $\frac{60}{360} \times 3.14 \times 100 = 52.33$ cm².
Area of Triangle: $\frac{\sqrt{3}}{4} \times 100 = 43.25$ cm².
Minor Segment: $52.33 - 43.25 = 9.08$ cm².
Major Segment: $\pi r^2 - \text{Minor} = 314 - 9.08 = 304.92$ cm².
Q7: 120 Degree Chord
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A chord of a circle of radius 10 cm subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle.
(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
Area of Sector: $\frac{120}{360} \times 3.14 \times 100 = 104.67$ cm².
Area of Triangle: Use formula $r^2 \sin(\theta/2) \cos(\theta/2)$ or construction.
Area = $100 \sin 60 \cos 60 = 100 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = 25\sqrt{3} = 43.25$ cm².
Area = $100 \sin 60 \cos 60 = 100 \times \frac{\sqrt{3}}{2} \times \frac{1}{2} = 25\sqrt{3} = 43.25$ cm².
Area of Segment: $104.67 - 43.25 = 61.42$ cm².
Q8: Grazing Horse
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A horse is tied to a peg at one corner of a square shaped grass field of side 20 m by means of a 7 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 14 m long instead of 7 m.
(Use $\pi = 22/7$)
(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 14 m long instead of 7 m.
(Use $\pi = 22/7$)
Corner of square = 90°. Grazing area = Quadrant of circle.
(i) Area with 7m rope: $\frac{90}{360} \pi (7)^2 = \frac{1}{4} \times \frac{22}{7} \times 49 = 38.5$ m².
(ii) Increase: New Area (14m) - Old Area.
New Area = $\frac{1}{4} \times \frac{22}{7} \times 14^2 = 154$ m².
Increase = $154 - 38.5 = 115.5$ m².
New Area = $\frac{1}{4} \times \frac{22}{7} \times 14^2 = 154$ m².
Increase = $154 - 38.5 = 115.5$ m².
Q9: Silver Brooch
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A brooch is made with silver wire in the form of a circle with diameter 42 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.
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
(i) Total Wire: Circumference + 5 Diameters.
Circumference = $\pi d = \frac{22}{7} \times 42 = 132$ mm.
5 Diameters = $5 \times 42 = 210$ mm.
Total = $132 + 210 = 342$ mm.
Circumference = $\pi d = \frac{22}{7} \times 42 = 132$ mm.
5 Diameters = $5 \times 42 = 210$ mm.
Total = $132 + 210 = 342$ mm.
(ii) Sector Area: Total Area / 10.
Radius = 21 mm. Area = $\pi r^2 = \frac{22}{7} \times 21 \times 21 = 1386$ mm².
Each sector = $1386 / 10 = 138.6$ mm².
Radius = 21 mm. Area = $\pi r^2 = \frac{22}{7} \times 21 \times 21 = 1386$ mm².
Each sector = $1386 / 10 = 138.6$ mm².
Q10: Umbrella
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An umbrella has 10 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 50 cm, find the area between the two consecutive ribs of the umbrella.
Radius = 50 cm. Number of sectors = 10.
Area of one sector = $\frac{1}{10} \times \pi r^2$.
Area = $\frac{1}{10} \times \frac{22}{7} \times 50 \times 50 = \frac{55000}{70} = 785.71$ cm².
Q11: Car Wipers
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A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle of $120^\circ$. Find the total area cleaned at each sweep of the blades.
Radius = 21 cm, Angle = 120°. Two wipers.
Total Area = $2 \times \frac{120}{360} \times \frac{22}{7} \times 21 \times 21$.
$= 2 \times \frac{1}{3} \times 22 \times 3 \times 21 = 924$ cm².
Q12: Lighthouse
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To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $72^\circ$ to a distance of 15 km. Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$)
Radius = 15 km, Angle = 72°.
Area = $\frac{72}{360} \times 3.14 \times 15 \times 15$.
$= \frac{1}{5} \times 3.14 \times 225 = 3.14 \times 45 = 141.3$ km².
Q13: Table Cover Designs
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A round table cover has six equal designs as shown in figure (segments). If the radius of the cover is 30 cm, find the cost of making the designs at the rate of $0.50$ per cm². (Use $\sqrt{3} = 1.73$)
Area of design = Area of circle - Area of hexagon. Or 6 segments.
Angle of each sector = 60°.
Area of 1 segment = Sector - Equilateral Triangle.
Sector: $\frac{60}{360} \times 3.14 \times 900 = 471$ cm².
Triangle: $\frac{1.73}{4} \times 900 = 389.25$ cm².
Sector: $\frac{60}{360} \times 3.14 \times 900 = 471$ cm².
Triangle: $\frac{1.73}{4} \times 900 = 389.25$ cm².
Segment Area = $471 - 389.25 = 81.75$ cm².
Total Area = $6 \times 81.75 = 490.5$ cm². Cost = $490.5 \times 0.50 = 245.25$.
Cost is 245.25.
Q14: MCQ
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Tick the correct answer: Area of a sector of angle p (in degrees) of a circle with radius R is:
(A) $\frac{p}{180} \times 2\pi R$
(B) $\frac{p}{180} \times \pi R^2$
(C) $\frac{p}{360} \times 2\pi R$
(D) $\frac{p}{720} \times 2\pi R^2$
(A) $\frac{p}{180} \times 2\pi R$
(B) $\frac{p}{180} \times \pi R^2$
(C) $\frac{p}{360} \times 2\pi R$
(D) $\frac{p}{720} \times 2\pi R^2$
Formula is $\frac{p}{360} \times \pi R^2$.
Check Option D: $\frac{p}{720} \times 2\pi R^2 = \frac{p}{360} \times \pi R^2$.
Option (D) is correct.