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CBSE Class 10 Chapter 11 Area 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 600

 Ans.

Given: Radius of circle OA = OB = 6 cm

Angle of the sector θ = 600

Let OAMB be a sector of a circle.

We knot that Area of sector (A) is:

A=θ3600×πr2

Putting value

A=6003600×227×(6)2cm2

A=16×227×36cm2

A=227×6=1327cm2

Q2.

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

 Ans.

Given: Circumference of circle = 22 cm

We know that Circumference of circle (C):

C=2πr

putting value

22=2πr

r=222π=11π

We know that Quadrant of a circle will subtend 36004=900 angle with centre of circle. so,

by Area of sector (A) formula:

A=θ3600×πr2

Putting value

A=9003600×π×(11π)2

A=14×π×(121π×π)

A=1214π=121×74×22=84788=778cm2

Alternative method:

We know that  the area (A) of a quadrant is one-fourth of a circle. So we can write as

14πr2

Putting value

14×π×(11π)2

A=14×π×(121π×π)

A=1214π=121×74×22=84788=778cm2

Q3.

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

 Ans.


Minute hand rotates in 1 hour = 60 minutes 3600

So in 5 minutes, minute hand will rotate=360060×5=300

So, we can say that-

Area swept by minute hand in 5 minutes = the area of a sector of 300 in a circle of 14 cm radius.

We know that Area of a sector of angle θ is

A=θ3600×πr2

Putting value

A=3003600×227×14×14

A=112×227×14×14

A=116×2×14

A=113×14=1543cm2

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)

 Ans.

As per construct figure we can visualize that
OQ=OP=radius=10cm
θ=900
So, Area of segment PMQ (Minor segment)= Area of sector OPMQ — Area of △POQ
Area of sector OPMQ :
=θ3600×πr2
=9003600×227×10×10
=14×227×100
=1114×100
=110014=78.5cm2
Area of △POQ:
=12×base×height
=12×OP×OQ
=12×10×10
=12×100=50cm2
Area of segment PMQ (Minor segment) = Area of sector OPMQ — Area of △POQ
Area of segment PMQ (Minor segment) = 78.550=28.5cm2
Area of Major segment = Area of circle — Area of miner sector OPMQ
Area of Major segment = 
=36003600×πr29003600×πr2
=(360036009003600)×πr2
=(36009003600)×πr2
=(27003600)×πr2
=34×3.14×10×10=9424=235.5cm2

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

 Ans.

Given: Radius of Circle = 21cm
Angle sbtended by arc with centre = 600
(i) Length of Arc ACB = θ3600×2πr
Putting value
=60360×2π×21
=16×2π×21
=π3×21
=7π=7×227=22cm
(ii) Area of sector OACB = θ3600×πr2
Putting value
=60360×227×21×21
=16×227×21×21
=12×227×7×21
=11×21=231cm2
(iii) In triangleOAB
OA = OB so,
A=B
A+O+B=1800
2A+600=1800
So A=600
thus we can say that OAB is an equilateral traiangle.
We know that area of equilateral traiangle (OAB)=34×(side)2
Putting value
=34×21×21=44134cm2
Thus Area of segment ACB = Area of sector OACB — Area of triangleOAB
=(23144134)cm2

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.

 Ans.

Given:
Radius of circle = OA = OB = 15 cm
Chord of circle subtends an angle of 60°
π=3.14 and 3=1.73
Area of Sector OACB :
=60360×πr2
=60360×3.14×15×15
=16×3.14×15×15
=706.56=117.75cm2
In AOB
Since OA=OB
So, A=B
A+B+O=1800
2A+600=1800
2A=1800600
2A=1200
A=600
So, AOB is a equilateral triangle.
We know that area of a equilateral triangle is: =34×(side)2
Putting value, we get
=34×15×15=225×1.734=97.3125cm2
Area of ACB = Area of sector OACB - Area of triangleAOB
=117.7597.3125=20.4375cm2
Area of Major Segment = Area of circle — Area of minor segment (ACB)
=πr220.4375
=3.14×15×1520.4375
=3.14×22520.4375
=706.520.4375
=686.0625cm2

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.

 Ans.

Given:
Radius of circle = OA = OB = 12 cm
Chord of circle subtends an angle of 120°
π=3.14 and 3=1.73
1. Area of the Sector :
Area of sector=θ360×πr2
putting given values:
Area of sector=120360×3.14×(12)2
Area of sector=13×3.14×144
Area of sector=150.72cm2
2. Area of the Triangle:
The triangle is an equilateral triangle because the central angle is 120 degrees.
Area of triangle=34×a2
In this case, the side length is the radius r=12 cm:
Area of triangle=34×(12)2
Area of triangle=34×144
Area of triangle=363cm2
3. Area of the Segment:
Area of segment=Area of sectorArea of triangle
Area of segment=150.72363cm2
Using the given approximation 31.73:
Area of segment150.7236×1.73cm2
Area of segment150.7262.28cm2
Area of segment88.44cm2
So, the area of the corresponding segment of the circle is approximately 88.44cm2.

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. 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)

 Ans.

Given:
Radius of circle = length of rope = OA = OB = 5 m
As per construction θ=1200
π=3.14 \)
(i) Part of the area where horse can graze = Area of sector OACB
Area of sector OACB=θ360×πr2
putting given values:
Area of sector OACB=90360×3.14×(5)2
Area of sector OACB=14×3.14×25
Area of sector OACB=19.625m2
(ii) Part of the area where horse can graze if rope length is 10 m long = Area of sector OACB when r = 10 m
Area of sector OACB=θ360×πr2
putting given values:
Area of sector OACB=90360×3.14×(10)2
Area of sector OACB=14×3.14×100
Area of sector OACB=78.5m2
Change in graze are when rope length changed 5 m to 10 m = 78.519.625=58.875m2

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. 11.9. Find :

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

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

 Ans.

Given:
Diameter of circle = 35 mm
So, Radius of circle =352mm
We know that circumference of circle =2πr
putting value we get
=2×227×(352)
=2××11×5=110mm
Length of wire required=5×Diameter+circumference
Putting value
Length of wire required=5×35+110
Length of wire required=175+110=285 mm
Each of 10 sector of circle subtending 36010=360 at center of circle
So Area of each sector of the brooch = 3603600×πr2
Putting value
=3603600×227×(352)×(352)=3854mm2 

Q10.

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.

 Ans.


Given:
Ribs in Umbrella = 8
So, Radius of circle = 45 cm
We have to find Area between two consecutive ribs.
Angle from the centre (θ)=36008=450
So, the area of two consecutive ribs =θ360×πr2
Putting value
=45360×227×45×45
=18×227×45×45
=1128×2025
=2227528=795.53cm2

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.

 Ans.


Given:
Length of wiper blade = radius  = 25 cm
Angle from the centre (θ)=1150
Area cleaned at each sweep of the blades = Area of sector
So, The area cleaned at each sweep of the blade =θ360×πr2
Putting value
=115360×227×25×25
=2372×227×25×25
=158125252cm2
Total area swept by both blade=2×158125252=158125126=1254.96cm2

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)

 Ans.


Given:
Distance of light spread = radius  = 16.5 km
Sector of Angle (θ)=800
Area of sector = Area of the sea over which the ships are warned
So, The area of sector =θ360×πr2
Putting value
=80360×3.14×16.5×16.5
=29×3.14×16.5×16.5
=189.97km2

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 Rs. 0.35 per cm2.

 Ans.

Given:
radius  = 28 cm
As per construction we can say table design in hexagon shape and design are segment of a circle.
Lets take a segment PBC. Chord BC is a side of hexagon.
Each chord will substitute 3606=600 at the centre of circle.
In OBC
Since OB=OC
So, B=C
B+C+O=1800
2B+600=1800
2B=1800600
2B=1200
B=600
So, OBC is a equilateral triangle.
We know that area of a equilateral triangle is: =34×(side)2
Putting value, we get
=34×28×28

=1963=333.2cm2

Area of sector OBPC =θ360×πr2

Putting value

=6003600×227×28×28

=16×227×28×28

=113×4×28

=11×4×283=12323=410.6667cm2

Area of segment BPC = Area of sector OBPC - Area of OBC

Area of segment BPC=410.6667 - 333.2 = 77.4667cm2

Total area of design=6×77.4667=464.8cm2

Cost for making 1cm2 designs = Rs. 0.36

Cost for making 464.8cm2 designs =464.8×0.35=Rs.162.68

Hence, The cost of making such type designs is 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)=p180×2πR   (B)=p180×πR2

(C)=p360×2πR    (D)=p720×2πR2

 Ans.

Given: Radius =R
Angle with Center θ=p
We know that area of sector = θ3600×πR2
Putting value
 =p03600×πR2
By multiplying numerator and denominator by 2, we get
=p720×2πR2
Hence, right answer is option (D).

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