Chapter 11: Areas Related to Circles
Exercise 11.1
Q1
Find the area of a sector of a circle with radius 6 cm if angle of the sector is \( 60^{\circ} \).
Solution:
Given: Radius \( r = 6 \) cm, Angle \( \theta = 60^{\circ} \).
Area of sector \( = \frac{\theta}{360^{\circ}} \times \pi r^2 \)
\[ = \frac{60}{360} \times \frac{22}{7} \times 6 \times 6 \]
\[ = \frac{1}{6} \times \frac{22}{7} \times 36 \]
\[ = \frac{132}{7} \text{ cm}^2 \]
Answer: \( \frac{132}{7} \text{ cm}^2 \)
Q2
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Given Circumference \( = 22 \) cm.
\[ 2\pi r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \]
\[ r = \frac{7}{2} = 3.5 \text{ cm} \]
A quadrant is a sector with angle \( 90^{\circ} \). Area \( = \frac{1}{4}\pi r^2 \).
\[ \text{Area} = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \]
\[ = \frac{1}{8} \times 22 \times \frac{7}{2} \times \frac{2}{7} \times \frac{7}{2} \text{ (simplifying)} \]
\[ = \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{77}{8} \text{ cm}^2 \]
Answer: \( \frac{77}{8} \text{ cm}^2 \)
Q3
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Radius \( r = 14 \) cm (length of minute hand).
Angle swept in 60 minutes = \( 360^{\circ} \).
Angle swept in 5 minutes \( \theta = \frac{360}{60} \times 5 = 30^{\circ} \).
Area swept = Area of sector with \( \theta = 30^{\circ} \).
\[ \text{Area} = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14 \]
\[ = \frac{1}{12} \times 22 \times 2 \times 14 \]
\[ = \frac{154}{3} \text{ cm}^2 \]
Answer: \( \frac{154}{3} \text{ cm}^2 \)
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 \( \pi = 3.14 \))
Solution:
Given \( r = 10 \) cm, \( \theta = 90^{\circ} \).
(i) Area of Minor Segment
Area of minor sector = \( \frac{90}{360} \times 3.14 \times 10^2 = \frac{1}{4} \times 314 = 78.5 \text{ cm}^2 \).
Area of \( \Delta AOB \) (Right angled at O) = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2 \).
Area of Segment = Area of Sector - Area of Triangle
\[ = 78.5 - 50 = 28.5 \text{ cm}^2 \]
(ii) Area of Major Sector
Angle for major sector = \( 360^{\circ} - 90^{\circ} = 270^{\circ} \).
Area = \( \frac{270}{360} \times 3.14 \times 100 \)
\[ = \frac{3}{4} \times 314 = 235.5 \text{ cm}^2 \]
(Alternatively: Area of Circle - Area of Minor Sector = \( 314 - 78.5 = 235.5 \))
Q5
In a circle of radius 21 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
(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
Solution:
Given \( r = 21 \) cm, \( \theta = 60^{\circ} \).
(i) Length of the Arc
\[ l = \frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 \]
\[ = \frac{1}{6} \times 44 \times 3 = 22 \text{ cm} \]
(ii) Area of the Sector
\[ A = \frac{\theta}{360} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 21 \times 21 \]
\[ = \frac{1}{6} \times 22 \times 3 \times 21 = 231 \text{ cm}^2 \]
(iii) Area of the Segment
Since \( \theta = 60^{\circ} \), the triangle formed by the chord and radii is equilateral.
Area of equilateral \( \Delta = \frac{\sqrt{3}}{4}r^2 = \frac{\sqrt{3}}{4} \times 21 \times 21 = \frac{441\sqrt{3}}{4} \text{ cm}^2 \).
Area of Segment = Area of Sector - Area of Triangle
\[ = \left( 231 - \frac{441\sqrt{3}}{4} \right) \text{ cm}^2 \]
Q6
A chord of a circle of radius 15 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 \( \sqrt{3} = 1.73 \))
Solution:
Given \( r = 15 \) cm, \( \theta = 60^{\circ} \).
1. Minor Segment:
Area of Sector = \( \frac{60}{360} \times 3.14 \times 15 \times 15 = \frac{1}{6} \times 706.5 = 117.75 \text{ cm}^2 \).
Area of \( \Delta \) (Equilateral) = \( \frac{\sqrt{3}}{4}r^2 = \frac{1.73}{4} \times 225 = 97.3125 \text{ cm}^2 \).
Area of Minor Segment = \( 117.75 - 97.3125 = 20.4375 \text{ cm}^2 \).
2. Major Segment:
Area of Circle = \( \pi r^2 = 3.14 \times 225 = 706.5 \text{ cm}^2 \).
Area of Major Segment = Area of Circle - Area of Minor Segment
\[ = 706.5 - 20.4375 = 686.0625 \text{ cm}^2 \]
Q7
A chord of a circle of radius 12 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 \( \sqrt{3} = 1.73 \))
Solution:
Given \( r = 12 \) cm, \( \theta = 120^{\circ} \).
Area of Sector = \( \frac{120}{360} \times 3.14 \times 12 \times 12 = \frac{1}{3} \times 452.16 = 150.72 \text{ cm}^2 \).
Area of Triangle:
For \( \theta = 120^{\circ} \), drop a perpendicular from the centre to the chord. This bisects the angle into \( 60^{\circ} \) and \( 60^{\circ} \).
Base of triangle = \( 2r \sin(60^{\circ}) \). Height = \( r \cos(60^{\circ}) \).
Area = \( \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times (2 \times 12 \times \frac{\sqrt{3}}{2}) \times (12 \times \frac{1}{2}) \)
\[ = 36\sqrt{3} = 36 \times 1.73 = 62.28 \text{ cm}^2 \]
Area of Segment = Sector - Triangle
\[ = 150.72 - 62.28 = 88.44 \text{ cm}^2 \]
Q8
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 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 10 m long instead of 5 m. (Use \( \pi = 3.14 \))
(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 10 m long instead of 5 m. (Use \( \pi = 3.14 \))
Solution:
The horse grazes a quadrant of a circle (since the corner of a square is \( 90^{\circ} \)).
(i) Area with 5 m rope:
\[ r = 5 \text{ m} \]
\[ \text{Area} = \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 25 \]
\[ = 19.625 \text{ m}^2 \]
(ii) Increase in area with 10 m rope:
New radius \( R = 10 \) m.
\[ \text{New Area} = \frac{1}{4} \times 3.14 \times 100 = 78.5 \text{ m}^2 \]
\[ \text{Increase} = 78.5 - 19.625 = 58.875 \text{ m}^2 \]
Q9
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.
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Solution:
Diameter \( d = 35 \) mm, Radius \( r = \frac{35}{2} \) mm.
(i) Total length of wire:
Wire is used for the circumference and 5 diameters.
\[ \text{Length} = 2\pi r + 5d \]
\[ = \left( \frac{22}{7} \times 35 \right) + (5 \times 35) \]
\[ = 110 + 175 = 285 \text{ mm} \]
(ii) Area of each sector:
The circle is divided into 10 equal sectors.
\[ \text{Area} = \frac{1}{10} \times \pi r^2 \]
\[ = \frac{1}{10} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} \]
\[ = \frac{1}{10} \times 11 \times 5 \times \frac{35}{2} \]
\[ = \frac{385}{4} \text{ mm}^2 \text{ (or } 96.25 \text{ mm}^2) \]
Q10
An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Solution:
There are 8 equal sectors. Radius \( r = 45 \) cm.
Area between two consecutive ribs = Area of one sector.
\[ \text{Area} = \frac{1}{8} \times \pi r^2 \]
\[ = \frac{1}{8} \times \frac{22}{7} \times 45 \times 45 \]
\[ = \frac{22275}{28} \text{ cm}^2 \text{ (approx } 795.53 \text{ cm}^2) \]
Q11
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of \( 115^{\circ} \). Find the total area cleaned at each sweep of the blades.
Solution:
Radius \( r = 25 \) cm, Angle \( \theta = 115^{\circ} \).
Total area = 2 \( \times \) Area of one sector.
\[ \text{Total Area} = 2 \times \frac{115}{360} \times \frac{22}{7} \times 25 \times 25 \]
\[ = 2 \times \frac{23}{72} \times \frac{22}{7} \times 625 \]
\[ = \frac{158125}{126} \text{ cm}^2 \text{ (approx } 1254.96 \text{ cm}^2) \]
Q12
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle \( 80^{\circ} \) to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use \( \pi = 3.14 \))
Solution:
Radius \( r = 16.5 \) km, Angle \( \theta = 80^{\circ} \).
\[ \text{Area} = \frac{80}{360} \times 3.14 \times 16.5 \times 16.5 \]
\[ = \frac{2}{9} \times 3.14 \times 272.25 \]
\[ = 189.97 \text{ km}^2 \]
Q13
A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use \( \sqrt{3} = 1.7 \))
Solution:
There are 6 equal segments. Angle of each sector \( \theta = \frac{360}{6} = 60^{\circ} \).
Radius \( r = 28 \) cm.
Area of one segment = Area of Sector - Area of Equilateral Triangle.
\[ \text{Sector Area} = \frac{60}{360} \times \frac{22}{7} \times 28^2 = \frac{1}{6} \times 22 \times 4 \times 28 = \frac{1232}{3} = 410.67 \text{ cm}^2 \]
\[ \text{Triangle Area} = \frac{\sqrt{3}}{4} r^2 = \frac{1.7}{4} \times 28 \times 28 = 1.7 \times 7 \times 28 = 333.2 \text{ cm}^2 \]
Area of 1 design = \( 410.67 - 333.2 = 77.47 \text{ cm}^2 \).
Total Area of 6 designs = \( 6 \times 77.47 = 464.82 \text{ cm}^2 \).
Cost = \( 464.82 \times 0.35 = 162.68 \) (approx ₹ 162.69).
Q14
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) \( \frac{p}{720} \times 2\pi R^2 \)
Solution:
Formula for area of sector = \( \frac{\theta}{360} \times \pi R^2 \).
Here \( \theta = p \). So, Area = \( \frac{p}{360} \pi R^2 \).
Checking option (D): \( \frac{p}{720} \times 2\pi R^2 = \frac{p}{360} \times \pi R^2 \).
Answer: (D)