Solve the followings Questions.

Unless stated otherwise, use π =22/7.

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Answer:

Area of the sector making angle θ
= (θ/360°)×π r²
Area of the sector making angle 60°
= (60°/360°)×π r² cm²
= (1/6)×6²π  
= 36/6 π cm²
= 6 × 22/7 cm²
= 132/7 cm²

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer:

Quadrant of a circle means sector is making angle 90°.
Circumference of the circle = 2πr
= 22 cm
Radius of the circle = r
= 22/2π cm
= 7/2 cm
Area of the sector making angle 90°
= (90°/360°)×π r² cm²
= (1/4)×(7/2)²π  
= (49/16) π cm²
= (49/16) × (22/7) cm²
= 77/8 cm²

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer:

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°
In 5 minutes, minute hand will rotate = 360°/ 60 x 5 = 30°
Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area of sector of angle θ = θ/360° . π. r²
Area of sector of 30° = 30°/360° . 22/7 . 14 .14
→ = 22/12 . 2 .14

= ((11).(14))/3
= 154/3 cm²
= 51.33 cm²

Therefore, the area swept by the minute hand in 5 minutes is 51.33 cm²

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)

Answer:

Radius of the circle = 10 cm


Major sector is making 360° − 90° = 270°


Area of the sector making angle 270°


=(270°/360°) × πr² cm²


=(3/4) × 10²π = 75πcm²


=75 × 3.14cm² = 235.5cm²

∴ Area of the major sector = 235.5cm²


Height of ΔAOB = OA = 10 cm


Base of ΔAOB = OB = 10 cm


Area of ΔAOB = 1/2 × OA × OB


= 1/2 × 10 × 10 = 50cm²


Minor sector is making 90°


Area of the sector making angle 90°


= (90°/360°) × πr² cm²


= (1/4) × 10²π = 25πcm²


= 25 × 3.14 cm²

= 78.5 cm²


Area of the minor segment = Area of the sector making angle 90° − Area of ΔAOB


= 78.5 cm² − 50cm² = 28.5cm²

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

Answer:

In the mentioned figure,
O is the centre of circle,
AB is a chord
AXB is a major arc,
OA=OB= radius =21 cm
Arc AXB subtends an angle 60° at O.

i) Length of an arc AXB = 60/360 × 2π × r

= 1/6 × 2 × 22/7 × 21

= 22cm

ii) Area of sector AOB = 60/360 × π × r²

= 1/6 × 22/7 × (21)²

= 231cm²

(iii) Area of equilateral ΔAOB = √3/4 × (OA)² = √3/4 × 212 = (441√3)/4 cm²

Area of the segment formed by the corresponding chord
= Area of the sector formed by the arc – Area of equilateral ΔAOB
= 231 cm² – (441√3)/4 cm²

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)

Answer:

Radius of the circle = 15 cm
ΔAOB is isosceles as two sides are equal.
∴ ∠A = ∠B
Sum of all angles of triangle = 180°
∠A + ∠B + ∠C = 180°
⇒ 2 ∠A = 180° – 60°
⇒ ∠A = 120°/2
⇒ ∠A = 60°
Triangle is equilateral as ∠A = ∠B = ∠C = 60°
∴ OA = OB = AB = 15 cm
Area of equilateral ΔAOB
= √3/4 × (OA)² = √3/4 × 15²
= (225√3)/4 cm² = 97.3 cm²

Angle subtend at the centre by minor segment = 60°
Area of Minor sector making angle 60°
= (60°/360°) × π r² cm²
= (1/6) × 15² π  cm² =  225/6 π  cm²
=  (225/6) × 3.14 cm² = 117.75  cm²

Area of the minor segment = Area of Minor sector – Area of equilateral ΔAOB
= 117.75  cm² – 97.3 cm² = 20.4 cm²

Angle made by Major sector = 360° – 60° = 300°
Area of the sector making angle 300°
= (300°/360°) × π r² cm²
= (5/6) × 15² π  cm² =  1125/6 π  cm²
=  (1125/6) × 3.14 cm² = 588.75  cm²

Area of major segment = Area of Minor sector + Area of equilateral ΔAOB
= 588.75  cm² + 97.3 cm² = 686.05 cm²

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)

Answer:

We have,
In a circle,
Radius
r = 12cm.
θ = 120°
Area of segment APB = Area of sector OAPB – Area of ΔOAB
Area of sector OAPB
= θ/360° × πr²
= 120°/360 × 3.14 × (12)²
= 1/3 × 3.14 × 144
= 3.14 × 4 ×12
= 3.14 × 48
= 150.72 cm²
Finding area of triangle ΔAOB
Area of ΔAOB = 12 × Base × Height
We draw
OM⊥AB
∴ ∠OMB = ∠OMA = 90°
In
ΔOMA and ΔOMB
∠OMA = ∠OMB (Both 90°)
OM = OM( Both radius)
OA = OB (Common line)
∴ ΔOMA ≅ ΔOMB (S.A.S. Congurency)
⇒ ∠AOM = ∠BOM
Then,
BM = AM = 1/2 AB ……. (1)
Now,
∠AOM = ∠BOM = 1/2 ∠BOA
= 1/2 × 1200
= 60°

In right angle triangle OMA
sin O= AM/AO
sin 60° = AM/12
√3/2 = AM/12
AM = 6√3
Again, In right angle triangle OMA
cos O = OM/AO
cos 60° = OM/AO
1/2 = OM/12
OM = 6cm.
Now, from equation (1) and we get,
AM = 1/2 AB
2AM = AB
AB = 2AM
AB = 2 × 6√3      ∴ AM = 6√3
AB = 12√3 cm.
Now,
Area of ΔAOB = 1/2 × Base × Height
= 1/2 × AB × OM
= 1/2 × 12√3 × 6
= 36√3
= 36 × 1.73
62.28 cm²
Hence, the Area of segment of
APB = Area of sector OAPB − Area of ΔOAB
= 150.72 − 62.28
= 88.44 cm²
Hence, this is the answer.

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 figure). Find:

Answer:

Side of square field = 15 m
Length of rope is the radius of the circle, r = 5 m
Since, the horse is tied at one end of square field; it will graze only quarter of the field with radius 5 m.
(i) Area of circle = π r² = 3.14 × 5² = 78.5 m²
Area of that part of the field in which the horse can graze = 1/4 of area of the circle

= 78.54 = 19.625 m²


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 figure. Find:

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Answer:

Diameter =35mm


∴ Circumference of circle
= 2 × π × r = 2 × π × 17.5 = 110 mm

i) The total length of the silver wire required


= Circumference of circle + (5 × 35)


= 110 + 175 = 285mm


(ii) The area of each sector of the brooch


= Area of circle/10


= (π ×17.5 × 17.5)/10 
 = 385/4 mm²
 

10. An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer:

Here, area between two consecutive ribs = 1/8 × Area of umbrella
Radius = r = 45cm
Area of umbrella = πr²
= 22/7 × 45 × 45 cm²
= 22/7 × 2025 cm²

Required area = 1/8 × Area of umbrella
= 1/8 × 22/7 × 2025 cm²
= 22275/28 cm²
= 795.54 cm²
Hence, area between two consecutive ribs is 795.54 cm²
 

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.

Answer:

Angle of the sector of circle made by wiper = 115°
Radius of wiper = 25 cm
Area of the sector made by wiper
= (115°/360°) × π r² cm²
= 23/72 × 22/7 × 25²
=  23/72 × 22/7 × 625 cm²
= 158125/252 cm²

Total area cleaned at each sweep of the blades
= 2 ×158125/252 cm²
=  158125/126 = 1254.96 cm²

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)

Answer:

Let O bet the position of Lighthouse.
Distance over which light spread i.e. radius, r = 16.5 km
Angle made by the sector = 80°
Area of the sea over which the ships are warned = Area made by the sector.
Area of sector = (80°/360°) × π r² km²
= 2/9 × 3.14 × (16.5)² km²
=  189.97 km²

13. A round table cover has six equal designs as shown in figure. 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)

Answer:

r = 28 cm, θ = 360°/6 = 60°

Area OAMB = θ/360° × πr² = 60°/360° × 22/7 × 28 × 28

= 1232/3 cm² = 410.67 cm²

Now, in ΔONA and ΔONB
i) ∠ONA = ∠ONB [each 90°]
ii) OA = OB [Radii of common circle]
iii) ON = ON [common]

∴ ΔONA ≅ ΔONB [By RHS congruency]
Hence, AN = NB = 1/2 AB

& ∠AON = ∠BON = 1/2 ∠AOB = 60°/2 = 30°

Now in ΔONA, cos 30° = ON/OA ⇒ √3/2 = ON/28 ⇒ ON = 14√3cm

& sin 30° = AN/OA ⇒ 1/2 = AN/28 ⇒ AN = 14cm

& 2AN = 14 × 2 = 28cm = AB

∴  arΔAOB = 1/2 × AB × ON = 1/2 × 28 × 14√3 = 196√3 = 196 × 1.7 = 333.2 cm²

∴ Area of one design = 410.67 − 333.2 = 77.47 cm²

∴ Area of six design = 6 × 77.47 = 464.82 cm²

Therefore, Cost of making the designs = Rs. (464.82 × 0.35)

= Rs.162.68

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) p/180 × 2πR
(B) p/180 × π R²
(C) p/360 × 2πR
(D) p/720 × 2πR²

Answer:
Area of a sector of angle p
= p/360 × π R²
= p/360 × 2/2 × π R²
=  2p/720 × 2πR²
Hence, Option (D) is correct.

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2