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Areas related to Circles Exercise - 11.1

Hi Friends and Champs!
Before to start the next exercise 11.1 , you need to understand some basic points, which will be helpful to solve out questions in the next exercise.
* Sector of a circle: The area among the two radii and corresponding arc.
* Segment of a circle: The area between the chord and the corresponding arc.
Minor and Major sectors and segments:
In the below given figure 

Area among the radii OA, OB and arc ACB is called Monor Sector and are among the radii OA, OB and arc APB is called Major Sector. 
Area between the chord "AB" and arc ACB is called Minor Segment and area between the chord "AB" and arc APB is called Major Segment.
Area of Sector : 
area of a sector of an angle θ =  θπr²/360, After below 5 lines, you do not need to remember it.
Now how we get this formula.
As you already know the area of a circle = πr² 
and the circle have a complete revolution of 360°, that means:
area of a sector of angle 360° = πr²
⇒ area of a sector of angle 1° = πr² / 360 by unitary method
⇒ area of a sector of angle θ = πr²θ / 360 
Length of arc of Segment:
length of an arc of a sector of angle θ =2πrθ/360, After below 5 lines, you do not need to remember it.
Now as you already know the perimeter of a circle = 2πr, 
it means the perimeter of an arc of  angle 360° =  2πr
so again, by unitary method
perimeter of an arc of angle 1° degree = 2πr/360
hence, 
perimeter of an arc of angle θ = 2πrθ/360 

Exercise-11.1
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
θ = 60,
r = 6cm
area = πr²θ / 360 
⇒ area = (22/7) × 6² × 60/360
⇒ area =  ( 22/7 × 36  × 1/6 )
⇒ area =  ( 22 × 6 ) / 7
⇒ area = 132/7 cm² Ans.

2. Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
θ = 90, ( one quadrant angle = 90 degree )
circumference = 2πr = 22cm
⇒ r = 11/π
⇒ r = 11/(22/7)
⇒ r =7/2
area = πr²θ / 360 
⇒ area = π × (11/π)² × 90/360
⇒ area =  ( 121/π)  × 1/4 )
⇒ area =  ( 121 × 7) /  ( 22 × 4 )
⇒ area = 77 / 8 cm² Ans.
 
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. 
Solution:
θ = 30, ( 1 minute = 6 degree , so 30 degree area covered in 5 minute)
r = 14 cm
area = πr²θ / 360 
⇒ area = 22/7 × 14² × 30/360
⇒ area =  (22 × 14 × 14) / (7 × 12 )
⇒ area = 154 / 3 cm² Ans.

4. 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) 
Solution:
θ = 90 for minor segment
θ = 270 for major segment
r = 10 cm
(i)
area of minor segment= area of the minor sector - area of the right angle of base 10 and height 10 cm
area of minor segment =πr²θ / 360 - 1/2 × 10 × 10
⇒ area = 3.14 × 100 × 90/360 -50
⇒ area = 78.5 - 50 
⇒ area =28.5 cm² Ans.
(ii)
area of major segment = πr²θ / 360
⇒ area = 3.14 × 100 × 270/360
⇒ area =235.5 cm² Ans.

5. 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) area of the sector formed by the arc 
 (iii) area of the segment formed by the corresponding chord 
Solution:
θ = 60 
r = 21 cm
(i)
Length of the arc =2πrθ / 360 
⇒ Length = 2×22/7 × 21 × 60/360
⇒ Length =22 cm Ans.
(ii)
area of the sector = πr²θ / 360
⇒ area = 22/7 × 21 × 21 × 60/360
⇒ area =231 cm² Ans.
(iii)
Since the angle subtend is equal to the 60 degree , so the triangle formed by the segment is an equilateral triangle, hence
area of  the segment= area of the minor sector - area of the equilateral triangle of side 21 cm
area of the segment =πr²θ / 360 - √3 /4 × 21 × 21  ( area of equilateral triangle = √3/4 × side² )
⇒ area = 22/7 × 21× 21 × 60/360  - 441 √3 /4
⇒ area = 231 - 441 × √3 /4 cm² Ans.

6. 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) 
Solution:
r = 15 cm
θ = 60 degree for minor segment and,
θ = 300 degree for major segment
Since the angle subtend is equal to the 60 degree , so the triangle formed by the minor segment is an equilateral triangle, hence
area of  the minor segment= area of the minor sector - area of the equilateral triangle of side 15 cm
area of the minor segment =πr²θ / 360 - √3 /4 × 15 × 15  ( area of equilateral triangle = √3/4 × side² )
⇒ area = 3.14 × 15× 15 × 60/360  - 225 × 1.73/4
⇒ area = 20.4375 cm² Ans.
Area of major segment = Area of circle - Area of minor segment
= πr² - area of minor segment
= 3.14 × 15 × 15 - 20.4375
=686.0625 cm² Ans.
7. 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) 
Solution:
r = 12 cm
θ = 120 degree
area of  the sector= πr²θ / 360
area of the sector =3.14 × 12 × 12 × 120/360
⇒ area of the sector= 150.72

since ΔOAB is an isosceles triangle , let OC is the perpendicular to AB ,therefore ∠AOC = 60°.
so by Pythagoras theorem
sin 60°= CA/OA
 √3 / 2 =CA/12
⇒CA = 6√3 
OC² = OA² - CA²
OC² = 12² - (6√3)² 
OC² =144 - 108
OC² =36
OC = 6
area of ΔAOC = 1/2 × CA × OC
area of ΔAOC =1/2  × 6√3 × 6
area of ΔAOC =18√3 
hence, area of ΔOAB = 2 × 18√3 = 36√3 = 36 × 1.73 = 62.28
Area of segment = Area of sector - Area of ΔOAB
= 150.72 - 62.28
= 88.44 cm² Ans.

8. 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 (see Fig. 11.8). 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 π = 3.14) 
Solution:
r = 5 m     ( length of the rope )
θ = 90 degree    ( since horse tied to a peg at one corner of a square shaped field)

(i)
area of  the grazed part= πr²θ / 360
area of the grazed part =3.14 × 5 × 5 × 90/360
⇒ area of the sector= 19.625 m² Ans.
(ii)
new radius of the grazed sector = 10 m
hence the area is = 3.14 × 10 × 10 × 90/360
                            = 78.5 m²

Increase in the grazing area = 78.5 - 19.625 = 58.875 m² Ans

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 as shown in Fig. 11.9. Find : (i) the total length of the silver wire required. (ii) the area of each sector of the brooch. 
Solution:
r = 35/2 mm    ( since diameter is 35 mm )
θ = 36             ( since diameter divide in 10 equal part , so one part angle is 360/10 )

(i)
total length "L"  of the silver wire required 
L = perimeter of the circle made + 5 × length of one diameter
   = 2πr + 5 × length of one diameter
   = 2×22/7 ×35/2 + 5×35
   = 285 mm Ans.
(ii)        
area of  the sector =πr²θ / 360
⇒ area = 22/7 × 35/2× 35/2 × 36/360
⇒ area = 96.25 mm² Ans.
10. An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Solution:
r = 45 cm
θ = 45        ( since umbrella has 8 ribs which are equally spaced , one sector angle made = 360/8 )
area between two consecutive ribs  =πr²θ / 360
⇒ area = 22/7 × 45× 45 × 45/360
⇒ area = 795.5357 cm² Ans.

11. 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. 
Solution:
r = 25 cm        (wiper blade length )
θ = 115        
wiped area by one wiper =πr²θ / 360
Total wiped area by two wiper =2πr²θ / 360
⇒ Total area = 2 × 22/7 × 25× 25 × 115/360
⇒ area = 1254.96 cm² Ans.

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured 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) 
Solution:
r = 16.5 km       (distance covered by light )
θ = 80 degree        
Area of the sea covered by the light =πr²θ / 360
Area =3.14×16.5×16.5× 80/360
Area = 189.97 km² Ans.

13. 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 √3 = 1.7)
Solution:
r = 28 cm       (radius of the cover )
θ = 60 degree  ( 6 equal parts so angle at centre made by one part 360/6 )       
Area of one design segment = Area of one sector - Area of the triangle made by one sector
Area = πr²θ / 360 - √3/4 × side²     ( since the triangle made is an equilateral triangle )  
Area =22/7× 28×28× 60/360 - 1.7/4 × 28×28
Area = 77.46667 cm²
Total Area of design = 6 × area of one segment
                                 = 6 × 77.46667
                                 = 464.8
Total cost of design = 0.35 * total area
                                = 0.35 * 464.8
                                = 162.68 Ans.
14. Tick the correct answer in the following : Area of a sector of angle p (in degrees) of a circle with radius R is
(A) 2πRp / 180  (B) πR²p / 180 (C) 2πRp / 360 (D) 2πR²p / 720
Solution:
r = R
θ = p degree         
Area of the sector = πr²θ / 360
Area =πR²p / 360 
or 
Area = 2πR²p / 720        ( multiply and divide by 2 to match the answer given in options)
so option (D) is correct.


I hope all the questions are properly readable and understandable, in case of some confusion, kindly let me know in comment section 🙏🙏🙏

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