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**NCERT Class 10 Mathematics Chapter 11 Areas Related to Circles**

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**Solutions****NCERT Class 10 Mathematics Chapter 11 Areas Related to Circles****Areas Related to Circles**

**Areas Related to Circles****Chapter – 11**

Exercise 11.1 |

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

Ans:

Let OACB be a sector of the circle making 60° angle at centre O of the circle

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

Ans:

Let the radius of the circle be r

Circumference = 22

2Пr = 22

r = 22/ 2П = 11/ П

= 1/ 4П = (11/22 )2

Quadrant of will subtend 90^{ o} angles at the circle.

Area of such quadrant of the circle = 90o/360^{o} x X r^{2}

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

Ans: We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.

In 5 minutes, minute hand will rotate = 360^{o}/60 x 5 = 30

Therefore, the area swept by the minute hand in minute in will be the area of a sector of sectors of 30^{o i }n a circle of 14 cm radius

Area of sector of 30^{o }= θ/360^{o} ✖ Пr^{2}

Area of sector of 30^{o }= 30o /360ox 22/7 x 14 x 14

= 22/12 x 2 x 14

= 11x 14 / 3

= 154/3 cm^{2}

Therefore, the areas swept by the hand in 5 minutes is 154/3 cm^{2}

**4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:**

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.**

Ans: Length of the arc 60/360 x 2 x 22/7 x 21 cm = ⅙ x 2 x 22x 3cm^{2}=22cm

**(ii) Area of the sector formed by the arc.**

Ans: The area of the sectors formed by the arc corresponding to sector angle 60^{o}

= 60/360x πr2=⅙ x 22/7 x (21)2 cm2 = ⅙ 22/7 x 21 x 21 cm = 231cm2

**(iii) Area of the segment forced by the corresponding chord.**

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]**

Ans:

Radius (r) of circle = 15 cm

Area of sector OPRQ = 60/360^{o }x πr^{2 }

= 117.75cm^{2}

In ∆OPQ,

∠OPQ = ∠OQP (As OP = OQ)

∠OPQ + ∠OQP + ∠POQ = 180°

∠OPQ = 120°

∠OPQ = 60°

∆OPQ is an equilateral triangle.

Area of major segment PSQ = Area of circle − Area of segment PRQ

= 117.75 – 97,3125

= 20,4375cm^{2}

= 3.14 x 225 – 20,4375

= 706.5 – 20,4375

= 686.0625 cm^{2}

**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]**

Ans:

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.

SV = VT

In ΔOVS,

OV/OS = cos60o

OV/12 = ½

OV = 6cm

SV/SO = sin 60o = √3/2

SV/12 = √3/2

SV = 6√3cm

ST = 2SV = 2 X 6√3 cm

Area of ∆OST = ½ x ST X OV

= ½ X 12√3 X6

= 36 √3 = 36 X 1.73 = 62.28 cm^{2}

Area of segment sut = area of sector sout – area of ∆OST

= 150.72 – 32,28

= 88.44cm^{2}

**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 the given figure). Find [Use π = 3.14]**

(i) The area of that part of the field in which the horse can graze.

(ii) The increase in the grazing area of the rope were 10 m long instead of 5 m.

Ans: From the figure, it can be observed that the horse can graze a sector of 90° in a circle of 5 m radius.

(i) Area that can be grazed by horse = Area of sector OACB

= 90°/360°πr^{2 }

= ¼ x 3.14 x (5)^{2}

= 19.625m^{2}

Area that can be grazed by the horse = Area of sector OABC

= 90°/360° πr^{2 }

¼ X 3.14 X (5)^{ 2}

= 19.625m^{2}

Area that can be grazed by the horse when length of rope is 10 m long

= 90^{o }/ 360^{o}x π x (10)^{2 }

= ¼ x 3.14 x 100

= 78.875m^{2}

(ii) Increase in grazing area = (78.5 − 19.625) m^{2}

= 58.875 m^{2}

**9. A brooch is made with silver wire in the form of a circle with a 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.

Ans: (i) Diameter of the circle = 35mm

35/2 Then radius r = mm

The total length of wire required = The length of wire required for making 5 diameters + the length of wire required for making the perimeter of the circle.

5x35mm + 2π×35/2mm^{2}

= (175 + 22 /7 x 35)]mm -x 35 mm = 285mm

(ii) Here r = 35/2mm

The sector angle of each brooch = 360/10 degrees . i.e θ = 36^{o}

Therefore , the area of each brooch

Here, r = 45cm

θ = sector angle between two consecutive ribs.

= 360^{o}/8 = 450 [There are 8 sector of same size]

Therefore, the area between two consecutive ribs of the umbrella = The area of one sector

= 45/360 xπr^{2 }= ⅛ x 22/7 x (45)^{2}cm^{2}

= 11/28 x 45 x cm^{2} = 22275/28 cm^{2}

**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.**

**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.**

**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 warned. [Use π = 3.14]**

Ans: Here, r = 16.5km =-km 33/2

θ = 80°

The area of the sea over which the ship are warned

= 820/360 x πr2 2/9 x 3 .14 x (93/2)^{2} km^{2}

= 1.57 x 121 km^{2 }= 189.97

**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 Rs.0.35 per cm2. [𝑈𝑠𝑒 √3 = 1.7]**

Ans: Here r = 285cm

**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) 9/180 x 2π R

(b) P/180 X πR^{2}

(c) p/360 x 2π R

(d) P/720 x 2πR

Ans: (d) P/720 x 2πR^{2}