**SUBJECT –** MATHS

**CLASS –** X (10th)

**QUESTION FROM – Area Related To Circle – Chapter 12 (12.2)**

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#### Exercise 12.2 – Area Related To Circle – Chapter 12 (NCERT Solutions) Mathematics (Class 10th)

Unless stated otherwise, Use π = 22/7

**Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.**

Solution –

Given, radius of circle (r) = 6 cm

Angle of the sector (θ) = 60°

Now, area of sector = (θ/360°) * πr^{2
}= (60°/360°) * (22/7) * 6^{2
}= (1/6) * (22/7) * 6 * 6

= (22/7) * 6

= 132/7 cm^{2
}

**Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.**

Solution –

Let radius of circle be r cm.

Given, circumference of the circle = 22

=> 2πr = 22

=> 2 * (22/7) * r = 22

=> 2r/7 = 1

=> r = 7/2 cm

So, area of the quadrant (1/4)^{th} of the circle = (θ/360) * πr^{2
}= (90°/360°) * (22/7) * (7/2)^{2
}= (1/4) * (22/7) * (7/2) * (7/2)

= (1/4) * 11 * (7/2)

= 77/8 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 – Length of minute hand = radius of the circle,

=> r = 14 cm

Angle swept by the minute hand in 60 minutes = 360°

Angle swept by the minute hand in 5 minutes = (360°/60) * 5 = 6 * 5 = 30°

Now, area of the sector with r = 14 and θ = 30° is given as

= (θ/360°) * πr^{2
}= (30°/360°) * (22/7) * 14^{2
}= (1/12) * (22/7) * 14 * 14

= (11 * 14)/3

= 154/3 cm^{2
}Thus, the required area swept by the minute hand by 5 minutes = 154/3 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 π = 3.14).

Solution – Length of the radius (r) = 10 cm

Sector angle θ = 90°

Area of the sector with θ = 90° and r = 10 cm is given as

= (θ/360°) * πr^{2
}= (90°/360°) * 3.14 * 10^{2
}= (1/4) * 3.14 * 10 * 10

= 3.14 * 5 * 5

= 78.5 cm^{2}

(i) Area of the minor segment = Area of minor sector – Area of ΔAOB

= 78.5 – (1/2) * 10 * 10

= 78.5 – 50

= 28.5 cm^{2}

(ii) Area of major segment = Area of the circle] –

Area of the minor segment

= πr^{2} – 78.5

= 3.14 * 10 * 10 – 78.5

= 314 – 78.5

= 235.5 cm^{2
}

**Q5. 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 –

Given, radius = 21 cm and θ = 60°

(i) Circumference of the circle = 2πr

= 2 * (22/7) * 21

= 2 * 22 * 3 = 132 cm

Now, length of the arc APB = (60°/360°) * 132 = 132/6 = 22 cm

(ii) Area of the sector with angle 60° = (θ/360°) * πr^{2}

= (60°/360°) * (22/7) * 21^{2
}= (1/6) * (22/7) * 21 * 21

= 11 * 21

= 231 cm^{2}

(iii) Area of the segment APQ = Area of the sector AOB – Area of ΔAOB ………….1

In ΔAOB, OA = OB = 21 cm

So, ∠A = ∠B = 60° [∠O = 60°]

=> AOB is an equilateral triangle,

Hence, AB = 21 cm

Draw OM ⊥ AB such that OM/OA = sin 60°

=> OM/OA = √3/2

=> OM = (√3/2) * 21

Now, area of triangle OAB = (1/2) * AB * OM

= (1/2) * 21 * (√3/2) * 21

= 441√3/4 cm^{2} ………….2

From equation 1 and 2, we get

Area of segment = (231 – 441√3/4) cm^{2.
}

**Q6. 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 –

Here, radius (r) = 15 cm, Sector angle θ = 60°

Area of the sector with angle 60° = (θ/360°) * πr^{2
}= (60°/360°) * 3.14 * 15^{2
}= (1/6) * 3.14 * 15 * 15

= 1.57 * 5 * 15 = 117.75 cm^{2}

Since ∠O = 60° and OA = OB = 15 cm

∠AOB is an equilateral triangle.

=> AB = 15 cm and ∠A = 60°

Draw OM Ʇ AB

So, OM/OA = sin 60° = √3/2

=> OM = OA * √3/2

=> OM = 15 * √3/2

Now, AR (ΔAOB) = 1/2 * AB * OM

= 1/2 * 15 * 15 * √3/2

= 225√3/4

= (225 * 3.14)/4 = 20.4375 cm^{2}

Now area of the minor segment = (Area of minor sector) – (AR ΔAOB)

= (117.75 – 97.3125) cm^{2 }= 20.4375 cm^{2}

Area of the major segment = Area of the circle – Area of the minor segment

= πr^{2} – 20.4375

= 3.14 * 15 * 15 – 20.4375

= 706.5 – 20.4375 cm = 686.0625 cm^{2.
}

**Q7. 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 –

Given, θ = 120° and r = 12 cm

Area of the sector = (θ/360°) * πr^{2
}= (120°/360°) * 3.14 * 12^{2
}= (1/3) * 3.14 * 12 * 12

= 3.14 * 12 * 4

= 150.72 cm^{2 } …………..1

Now, area of ΔAOB = 1/2 * AB * OM ………..2

In ΔOAB, ∠O = 120°

=> ∠A + ∠B = 180° – 120°

=> ∠A + ∠B = 60°

Since, OB = OA = 12 cm

So, ∠A = ∠B = 30°

Again, OM/OA = sin 30° = 1/2

=> OM = OA * 1/2

=> OM = 12/2 = 6 cm

In right angle triangle AMO, AM^{2} = 12^{2} – 6^{2}

=> AM^{2} = 144 – 36

=> AM^{2} = 108

=> AM = √108

=> AM = 6√3

=> 2AM = 6√3

=> AM = 12√3

From equation 2, we get

Area of ΔAOB = 1/2 * AB * OM

= 1/2 * 12√3 * 6

= 36√3

= 36 * 1.73

= 62.28 cm^{2 } ……….3

From equation 1 and 3, we get

Area of the minor segment = Area of minor segment – Area of ΔAOB

= 150.72 – 62.28 cm = 88.44 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 **(see Fig. 12.11).** 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 –

Here, Length of the rope = 5 m

Radius of the circular region grazed by the horse = 5 m

(i) Area of the circular portion grazed = (θ/360°) * πr^{2
}= (90°/360°) * 3.14 * 5^{2 }= (1/4) * 3.14 * 5 * 5 = 78.5/4 = 19.625 m^{2 }

(ii) When length of the rope is increased to 10 m,

So, r = 10 m

Area of the circular region where θ = 90°.

= (θ/360°) * πr^{2
}= (90°/360°) * 3.14 * 10^{2 }= (1/4) * 3.14 * 10 * 10 = 314/4 = 78.5 m^{2}

So, increase in the grazing area = 78.5 – 19.625 = 58.875 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 as shown in Fig. 12.12. Find :**

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

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

Solution – Diameter of the circle = 35 mm, Radius (r) = 35/2 mm

(i) Circumference = 2πr

= 2 * (22/7) * (35/2) = 22 * 5 = 110 mm,

Length of 1 piece of wire used to make diameter to divide the circle into 10 equal sectors = 35 mm

So, length of 5 pieces = 5 * 35 = 175 mm

Hence, total length of the silver wire = 110 + 175 mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,

Hence, Sector angle θ = 360°/10 = 36°

Now, area of each sector = (θ/360°) * πr^{2}

= (36°/360°) * (22/7) * (35/2) * (35/2)

= (11 * 35)/4 = 385/4 mm^{2
}

**Q10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.**

Solution –

Here, radius (r) = 45 cm

Since circle is divided in 8 equal parts,

Now, sector angle corresponding to each part θ = 360°/8 = 45°

Now, area of a sector(part) = (θ/360°) * πr^{2
}= (45°/360°) * (22/7) * 45 * 45

= (1/8) * (22/7) * 45 * 45

= (1/4) * (11/7) * 45 * 45

= (11 * 45 * 45)/(4 * 7) = 22275/28 mm^{2
}Hence, the required area between two pins = 22275/28 mm^{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°. Find the total area cleaned at each sweep of the blades.**

Solution – Given Here, radius (r) = 25 cm

Sector angle (θ) = 115°

Area cleaned by each sweep of the blades = (θ/360°) * πr^{2 }* 2

[Since each sweep will have to and from movement]

= (115°/360°) * (22/7) * 25 * 25 * 2

= (23 * 11 * 25 * 25)/(18 * 7)

= 158125/126 mm^{2.
}

**Q12. 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 – Here, Radius (r) = 16.5 km

Sector angle (θ) = 80°

Now, Area of the sea surface over which the ships are warned

= (θ/360°) * πr2

= (80°/360°) * 3.14 * 16.5 * 16.5

= (80°/360°) * 314/100 * 165/10 * 165/10

= (157 * 11 * 11)/100

= 18997/100

= 189.97 km2

^{}

**Q13. 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 – Given, r = 28 cm

Since, the circle is divided into six equal sectors.

So, sector angle θ = 360°/6 = 60°

Area of the sector with angle 60° and radius 28 cm = (θ/360°) * πr^{2}

= (60°/360°) * (22/7) * 28 * 28

= (1/6) * 22/7 * 28 * 4 = (44 * 28)/3

= 1232/3 = 410.67 cm^{2} ……………1

Now, area of 1 design = Area of segment APB

= Area of sector – Area of ΔAOB …………..2

In ΔAOB, ∠AOB = 60°, OA = PN = 28 cm

So, ∠OAB = 60° and ∠OBA = 60°

=> ΔAOB is an equilateral triangle.

=> AB = AO = BO

=> AB = 28 cm

=> Draw OM ⊥AB

Now, in right ΔAOM, we have

OM/OA = sin 60° = √3/2

=> OM = OA * √3/2

=> OM = 28 * √3/2

=> OM = 14√3 cm

Area of ΔAOB = 1/2 * AB * OM

= 1/2 * 28 * 14√3 = 14 * 14 * 1.7

= 333.3 cm^{2} ………..3

Now, from (1), (2) and (3), we have:

Area of segment APQ = 410.67 – 333.2 = 77.47 cm^{2
}=> Area of 1 design = 77.47 cm^{2
}So, area of the 6 equal designs = 6 * 77.47 = 464.82 cm^{2
}Cost of making the design at the rate of Rs. 0.35 per cm^{2 }= Rs. 0.35 * 464.82

= Rs. 162.68

^{}

**Q14. 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 * 2 Pie R

(b) P/180 * Pie R2

(c) P/360 * 2 Pie R

(d) P/360 * 2 Pie R2

Solution – Given, radius (r) = R

Angle of sector (θ) = p

Area of the sector = (θ/360°) * πr^{2 }= (p/360°) * πR^{2
}= (2/2) * (p/360°) * πR^{2 }= (p/720°) * 2πR^{2
}Hence, option (D) is the right answer.

^{}

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