Question Bankc Mathematics

8/12/2019 Question Bankc Mathematics

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AREAS RELATED TO CIRCLES

A. SUMMATIVE ASSESSMENT

12.1 PERIMETER AND AREA OF A CIRCLE A REVIEW

12

1. Perimeter (circumference) of a circle

with diameterd(d= 2r, whereris the radius) is

given by C = d= 2r.

2. Perimeter of a semicircle with radius r

= 2r+ r= r( + 2).

3. Area of a circle with radius ris given by

A = r2.

4. Area of a semicircle of radius r=2

2

r.

5. Area of a ring whose outer and inner

radii are R and rrespectively

= (R2 r2) = (R + r) (R r)

6. If two circles touch internally, then the

distance between their centres is equal to the

difference of their radii.

7. If two circles touch externally, then the

distance between their centres is equal the sum of

their radii.

8. The distance moved by a rotating wheel

in one revolution is equal to the circumference of

the wheel.

9. The number of revolutions completed by

a rotating wheel in one minute

= Distance moved in one minute

.Circumference of the wheel

(Unless stated otherwise, use22

7 = )

Q.1. The radii of two circles are 19 cm and

9 cm respectively. Find the radius of the circle

which has circumference equal to the sum of the

circumferences of the two circles. [Imp.]

Sol. Radius of 1st circle = r1

= 19 cm

Radius of 2nd circle = r2

= 9 cm

Circumference of 1st circle = 2r1 = 2(19) cm

Circumference of 2nd circle = 2r2

= 2(9) cm

Let R be the radius of the circle which has

circumference equal to the sum of the circumferences

of 1st and 2nd circle.

So, 2R = 2 (19) + 2(9) R = 19 + 9

R = 28

Hence, the required radius is 28 cm.

TEXTBOOK'S EXERCISE 12.1

Q.2. The radii of two circles are 8 cm and

6 cm respectively. Find the radius of the circle

having area equal to the sum of the areas of the

two circles. [Imp.]

Sol. Radius of 1st circle = r1

= 8 cm

Radius of 2nd circle = r2

= 6 cm

Let R be the radius of the required circle.

We know that area of a circle = 2

r

Area of 1st circle = r1

2 = 2 (8) cm2

Area of 2nd circle = 2

2r = 2

(6) cm2

Area of the required circle

= Area of 1st circle + Area of 2nd circle

R2 = (8)2 + (6)2

R2 = (8)2 + (6)2

R2 = 64 + 36 R2 = 100 R = 10

Hence, the required radius is 10 cm.

Question Bank In Mathematics Class X (TermII)


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Q.3. Figure depicts an archery target

marked with its five scoring areas from thecentre outwards as Gold, Red, Blue, Black and

White. The diameter of the region representing

Gold score is 21cm and each of the other bands

is10.5cm wide. Find the area of each of the five

scoring regions.

Sol. r1

=21

2cm = 10.5 cm,

r2

= (10.5 + 10.5) cm = 21 cm

r3

= (21 + 10.5) cm = 31.5 cm

r4

= (31.5 + 10.5) cm = 42 cm and

r5

= (42 + 10.5) cm = 52.5 cm

Gold

Red

Blue

Black

White

r5

r2

r1

r3

Let the areas of the gold, red, blue, black and

white regions be A1, A

2, A

3, A

4and A

5respectively.

A1

= r1

2 =22

7 (10.5)2 cm2 = 346.5 cm2

A2

= 2 2

2 1( ) r r

=22

7{(21)2 (10.5)2} cm2

= 222 [441 110.25] cm

7

=22

7 330.75 cm2 = 1039.5 cm2

A3

= 2 2 2 2 23 222

( ) [(31.5) (21) ] cm

7

r r =

= 222

[992.25 441] cm7

=22

7 551.25 cm2 = 1732.5 cm2

A4

= 2 24 3( ) r r

=22

7[(42)2 (31.5)2] cm2

=22

7 [1764 992.25] cm2

=22

7 771.75 cm2 = 2425.5 cm2

A5

= 2 2

5 4( )r r =22

7 [(52.5)2 (42)2] cm2

=22

7 [2756.25 1764] cm2

= 3118.5 cm2

Area of Gold region = 346.5 cm2

Area of Red region = 1039.5 cm2

Area of Blue region = 1732.5 cm2

Area of Black region = 2425.5 cm2

Area of White region = 3118.5 cm2.

Q.4. The wheels of a car are of diameter

80cm each. How many complete revolutions

does each wheel make in 10 minutes when the

car is travelling at a speed of 66 km per hour?

[2011 (T-II)]

Sol. Radius of a wheel =Diameter 80

cm2 2

=

= 40 cm

Distance travelled in one revolution= Circumference of wheel

= = 22 1760

2 2 40 cm = cm7 7

r

Distance travelled in 10 minutes =66

60 10 km

= 11 km = 1100000 cm

Number of complete revolutions made by the


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wheel in 10 minutes

= Distance covered by wheel in 10 minutesCircumference of the wheel

= = =

1100000 1100000 74375.

1760 1760

7

Q.5.Tick the correct answer in the following

and justify your choice :

If the perimeter and the area of a circle are

numerically equal, then the radius of the circle is

(a) 2 units (b) units(c) 4 units (c) 7 unitsSol. Let the radius of the circle be r units.

Its circumference = 2 r units

And its area = 2r sq. units

As per condition : = 22 r r r = 2 unitsHence, the correct answer is (a).

OTHER IMPORTANT QUESTIONS

Q.1. If the diameter of the wheel of a cycle

is 7 cm, then its area is :

(a) 77 cm2 (b) 22 cm2

(c) 277

cm2

(d) 277 cm

Sol.(c) Here, diameter = 7 cmradius = 7

cm.2

Area = 2r = 22 7 7

7 2 2cm2 =

277 cm2

.

Q.2. If a copper wire of length 88 cm is bent

in the form of a circle, then the radius of the

circle is :

(a) 7 cm (b) 14 cm

(c) 21 cm (d) 22 cm

Sol. (b) Total length of copper wire = 88 cm

2 r = 88 cm r = 88 7

2 22= 14 cm.

Q.3. If the diameter of a protractor is 8 cm,

then its perimeter is :

(a) (4+ 2) cm (b) 4( + 2) cm

(c) ( +8) cm (d) 4( + 4) cm

Sol. (b) The perimeter of the protractor

= r( + 2) = 4( + 2) cm.

Q.4.If the diameter of a wheel is 84 cm, then

the distance travelled in 10 complete revolutions

is approximately :

(a) 20 m (b) 26 m

(c) 30 m (d) 32 m

Sol. (b) Diameter of the wheel = 84 cm

r = 42 cm

Circumference = 2r = 22

2 427

cm

= 44 6 cm = 264 cm.

Distance travelled in 10 complete revolutions

(264 10) cm = 2640 cm = 26 m (approx)

Q.5. The sum of the circumferences of the

circles of radii 3 cm and 4 cm is equal to thecircumference of a circle of radius :

(a) 7 cm (b) 14 cm

(c) 21 cm (d) none of these

Sol.(a) Sum of the circumferences of two circles

of radii 3 cm and 4 cm = (2.3 + 2.4) cm

= (6 + 8) cm = 14 cm = 2.7 cm

Hence, the required radius of larger circle

= 7 cm.

Q.6.If the radii of two circles are in the ratio

of4 : 3, then their areas are in the ratio of:(a) 4 : 3 (b) 8 : 3

(c) 16 : 9 (c) 9 : 16

Sol. (c) Let 4x and 3x be the radii of two circles,

then

= =

21

22

A (4 ) 16

A 9(3 )

x

x

Q.7. If a steel wire in the form of a rhombus

of side11cm is rebent in the circular form, then


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the radius of the circle is :

(a) 14 cm (b) 7 cm(c) 6 cm (d) 21 cm

Sol. (b) Perimeter of the rhombus

= 4 11 cm = 44 cm

Circumference = 2 r

=2 44r r =

=

44 77cm.

2 22

Q.8. The radius of the wheel of a cycle is

25 cm. The number of revolutions it will take to

cover a distance of 22 m is :

(a) 7 (b) 14(c) 28 (d) 21

Sol. (b) Circumference of the wheel

= 2 25 cm = 50 cm = 22

507

cm

No. of revolutions for covering 22 m

=

=

22 100 714

50 22.

Q.9. If the diameter of a semicircular

protractor is 14 cm, then the perimeter of the

protractor is :(a) 26cm (b) 14cm

(c) 28cm (d) 36cm [2011 (T-II)]

Sol. (d) Radius of the protractor =14

2cm = 7 cm

Perimeter of the semicircular protractor

= 2 r + r =22

14 7 36 cm7

+ =

Q.10. The number of rounds that a wheel of

diameter 7

11m will make in going 4 km is :

(a) 1500 (b) 1700

(c) 2000 (d) 2500 [2011 (T-II)]

Sol. (c) Circumference of the wheel

=d =22 7

2 m7 11

=

No. of rounds made to cover 4 km

=4000

20002

=

Q.11. If the area and circumference of a

circle are numerically equal, then the diameterof the circle is :

(a) 3 units (b) 5 units

(c) 4 units (d) 2 units[2011 (T-II)]

Sol. (c) r2 = 2r r2 2r = 0

r(r 2) = 0 r 2 units

d = 2r d= 2 2 units = 4 units

Q.12. If the radius of a circle is doubled,

then its area will become :

(a) double (b) triple

(c) four times (d) same

Sol. (c) Let radius = x, then new radius = 2x

Area = 2 ,x New area = = 2 2(2 ) 4x x .

Q.13. The length of a wire in the form of an

equilateral triangle is 44 cm. If it is rebent into

the form of a circle, then area of the circle is :

(a) 484cm2 (b) 176cm2

(c) 154cm2 (d) 44cm2

Sol. (c) Perimeter of the triangle = circumference

of the circle

44 = 2 r r = 44 7

7 cm2 22

=

Area of the circle = 2

r =22

7 77

cm2

= 154 cm2.

Q.14. If the diameter of a semi-circular

protractor is 14 cm, then find its perimeter.

Sol. Diameter of the protractor = 14 cm

Radius of protractor = 14

cm 7 cm2

=

Perimeter of the semi-circular protractor

= 2r + r = +

2214 7

7 cm = 36 cm.

Q.15.Find the circumference and area of the

circle whose diameter measures 14 cm.

Sol. Diameter (d) = 14 cm radius = 7 cm

Circumference of the circle = d

= =22

14 cm 44 cm7


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Area of the circle =r2

= 2 222 7 7 cm 154 cm .7

Q.16. Find the radius of the circle which has

circumference of 100 cm.

Sol. C = 2r r=

C 100 700

222 442

7

= 15.9 cm.

Q.17.A wheel has diameter84 cm. Find how

many complete revolutions must it make to cover

792 m. [2011 (T-II)]Sol. Diameter of the wheel = 84 cm = 0.84 m.

Circumference of the wheel = d

= 0.84 metres

No. of revolutions made to cover 792 metres

=792 792 7

0.84 0.84 22=

= 300.

Q.18. The difference between circumference

and diameter of a circle is 105 cm. Find the

radius of the circle.

Sol. Circumference = 2 RadiusDiameter = 2 Radius

According to the question,

2r 2r = 105

2r( 1) = 105

22

2 1 1057

r =

15 105 7

2 1057 2 15

r r

= =

r = 24.5 cm.

Hence, the radius of the circle is 24.5 cm.Q.19. A race track is in the form of a ring

whose inner circumference is 352 m and outercircumference is 396 m. Find the width of the

track.22

Use =7

[2011 (T-II)]

Sol. Let the outer and inner radii of the ring be R

metres and rmetres respectively. Then,

2R = 396 and 2r = 352

22 22

2 R = 396 and 2 3527 7

=r

7 1 7 1

R 396 and = 352

22 2 22 2

r=

R = 63 m and r = 56 m

Hence, width of the track = (R r) metres

= (63 56) metres = 7 metres.

Q.20. The difference between circumference

and diameter of a circle is 135 cm. Find the

radius of the circle.22

Use =7

[2011 (T-II)]

Sol. Circumference = 2 Radius

Diameter = 2 Radius


2r 2r = 135

2r( 1) = 135

2r22

17

= 135

15 135 72 135

7 15 2r r

= =

r =63

31.5 cm2

=

Hence, the radius of the circle is 31.5 cm.

Q.21. How many times will the wheel of a

car rotate in a journey of 2002 m, if the radius

of the wheel is 49 cm. [2011 (T-II)]

Sol. Radius of the wheel = 49 cm.

Diameter of the wheel = 49 2 cm

= 98 cm = 0.98 m

Circumference of the wheel =d= 0.98 metres

No. of revolutions made to cover 2002 m


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2002 2002 7650

0.98 0.98 22

= = =

.

Q.22. Two circles touch externally. The sum

of their areas is 130 sq. cm and the distance

between their centres is 14 cm. Find the radii of

the circles. [Imp.]

Sol. Let the radii of the two circles be R cm and

rcm respectively. Let C1

and C2

be the centres of the

given circles, then,

C1C

2 = R + r

R + r = 14 (i)

[C1C2 = 14 cm (given)]

Given that the sum of the areas of two circles

is equal to 130 sq. cm.

R2 +r2 = 130

R2 + r2 = 130 (ii)

(R + r)2 2Rr = 130

(14)2 2Rr = 130

2Rr = 196 130 = 66

Rr = 33

=33

Rr

(iii)

Substituting =33

Rr

in (ii), we get

233

r

+ r2

= 130

2

2

1089130r

r

+ =

4

2

1089 r

r

+= 130

1089 + r4 = 130r2

r4 130r2 + 1089 = 0

r4 121r2 9r2 + 1089 = 0

r2(r2 121) 9 (r2 121) = 0

(r2 121)(r2 9) = 0

r2 = 121 or r2 = 9

r = 11 or 3

Substituting the values of r in (i), we get

R = 3 or 11.

Radii of two circles are 11 cm and 3 cm.

Q.23. In Radhikas house there is a flower

pot. The sum of radii of circular top and bottom

of the flower pot is 140 cm and the difference of

their circumferences i s 88 c m. F in d t he

diameters of the circular top and bottom.

[Imp.]

Sol. Sum of radii of circular top and bottom

= 140 cm.

Let radius of the top = r cm.

Therefore, the radius of the bottom

= (140 r) cm.

The circumference of top = 2r cm.

The circumference of bottom = 2(140 r) cm.

Difference of circumferences

= [2r 2 (140 r)] cm.

By the given condition,2r 2(140 r) = 88 (Given)

2(r 140 + r) = 88

88 88

2 140 14222

27

r = = =

2r = 140 + 14 = 154

r =154

2 cm = 77 cm

Radius of the top = 77 cm.

Diameter of the top = 2 77 cm = 154 cm.

Radius of bottom = 140 r= (140 77) cm

= 63 cm.

Diameter of bottom = 2 63 cm = 126 cm.

Q.24. A field is in the form of a circle. The

cost of ploughing the field at Rs 1.50 per m2 is

Rs5775. Find the cost of fencing the field at Rs

8.50 per metre.


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Sol. Area of the field =2

Total cost of ploughing

Rate per m

=5775

1.5

m2

= 2

57753

m2 = 3850 m2

Let the radius of the field be r metres.

Then, area of circle = r2

2 222

3850 38507

r r = =

73850 175 7 35 35

22r= = =

= 35.

Circumference of the field = 2r units

222 35 m = 220 m.

7

=

Hence, cost of fencing the field

= Rs17

2202

= Rs 1870.

Q.25. A blacksmith Rajesh bent a steel wire,in the form of a square, encloses an area of 121

sq cm. The same wire he bent in the form of a

circle. Find the area of the circle. [HOTS]

Sol. Area of square = 121 cm2

Side of the square = 121 cm = 11 cm.

Perimeter of the square = (4 11) cm = 44 cm.

Length of the wire = 44 cm.

Circumference of the circle

= length of the wire = 44 cm.

Let the radius of the circle = r cm.

Then, 22

2 44 2 44 7.7

r r r = = =

Hence, area of the circle

2 22 7 77

r = =

cm2 = 154 cm2.

Q.26. The diameter of a circular pond is

17.5m. It is surrounded by a path of width3.5 m. Find the area of the path. [Imp.]

Sol. Diameter of circular pond = 17.5 m

Width of the path = 3.5 m

Diameter of the pond including path

= (17.5 + 7) m = 24.5 m.

Area of the circular pond = r2

222 17.5

7 2

= m2

22 17.5 17.5

7 2 2= m2

11 17.5 2.5

2

= m2 = 240.625 m2

Area of the pond + path = R2

22 2 4.5 24.5

7 2 2

= m2

11 3.5 24.5

2

= m2 = 471.625 m2

Area of the path = (471.625 240.625)

= 231 m2.

Q.27. Two circles touch internally. The sumof their areas is 116 square cm and distance

between their centres is 6 cm. Find the radii of the circles. [HOTS]

Sol. Let the radius of circle having centre O be

R and the radius of circle having centre O be r.


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The sum of the areas = 116 cm2 [Given]

+ = 2 2

R 116r

+ =2 2R 116r (i)Distance between the centres = 6 cm [Given]

OO = 6 cm R r = 6 cm (ii)

= + =2 2 2(R ) 36 R 2R 36r r r = = =116 2R 36 2R 116 36 80r r

+ = + + = + =2 2 2(R ) R 2R 116 80 196r r rR + r = 14 (iii)

Solving (ii) and (iii), we get R = 10 and r = 4.

Hence, the radii of the given circles are 10 cm

and 4 cm respectively.Q.28. Ram Prakash walks around a circular

park of area88704 m2. How long will he take to

walk10 rounds at the speed of 4.5 km hr1?

Sol. Let rbe the radius of the circular park,

Area =r2 = 88704 m2

= =2 7

88704 2822422

r

r = 168 mCircumference of the circular field

222 2 168 m 1056 m.

7

r= = =

Distance covered in 10 rounds = 10 circum-

ference of the circular park

= 10 1056 m = 10560 m.

Time taken to cover 4.5 km = 1 hour

Time taken to cover =10560

km1000

=10560

4.5 1000 hours

176hours

75=

= 2 hours 20 minutes 48 seconds.

Q.29. In the given figure, OPQR is a

rhombus whose three vertices P, Q, R lie on acircle of radius8 cm. Find the area of the shaded

region. [2011 (T-II)]

Sol. Clearly, OP = OQ = OR = 8 cm

Let OQ and PR intersect at S

Since the diagonals of a rhombus bisect each

other at right angles, we have, OS = 4 cm and

OSP = 90

Now, PR = 2PS

and RS = 2 2 2 2OR OS cm = 8 4 cm

(64 16) cm = 48 cm 4 3 cm= =

PR = 2RS PR = 2 4 3cm = 8 3cm

The area of the rhombus (shaded area) OPQR

= 2 21 1

OQ PR = 8 8 3 cm 32 3 cm2 2

= .

Q.30. Find the difference between the areaof a regular hexagonal plot each of whose side is

72m and the area of the circular swimming tank

inscribed in it.22

.7

Take =

Sol. Side of hexagonal plot = 72 m.

Area of equilateral triangle OAB

23(Side)

4=

2 23(72) m

4

=

21296 3 m=

Area of hexagonal plot= 6 area of triangle OAB

= 6 1296 3 m2

= 7776 1.732 m2 = 13468.032 m2

=[ 3 1.732]

= =

22 2 2 2 72

OC OA AC (72)2

2

725184 5184 12962

= = = 3888

2 3888r = Area of inscribed circular swimming tank

2 222(3888) m

7r= =

212219.429 m=

Required difference= 13468.032 m2 12219.429 m2

= 1248.603 m2.


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PRACTICE EXERCISE 12.1A

Choose the correct option (Q. 1 4) :

1. If the diameter of a circle is 3.5 cm, then

its circumference will be :

(a) 11 cm (b) 22 cm

(c) 33 cm (d) 44 cm

2. If the radius of a circle is doubled, then itscircumference will be :

(a) half (b) twice

(c) thrice (d) one-fourth

3. If the area of a circle is 4 cm2

, then itsradius is :

(a) 1 cm (b) 2 cm

(c) 3 cm (d) 4 cm

4. What will be the cost of polishing a

circular table-top of radius 2 m at the rate of Rs

7 per m2?

(a) Rs 44 (b) Rs 88

(c) Rs 49 (d) Rs 84

5. Find the circumference and area of a

circle of radius 8.4 cm.

6. Fi nd t he area of a ci rcle whosecircumference is 22 cm.

7. The radius of a wheel is 84 cm. How

many revolutions will it make to go 52.8 km?

8. The difference between the area of a

circle and the square of its radius is 16.8 m2. Findthe radius of the circle.

9. The diameter of a cart wheel is 1.4 m.Find the distance to which the cart moves when

wheel makes 1000 revolutions.

10. Sum of radii of two circles is 140 cm andthe difference of their circumferences is

88 cm. Find the diameters of the circles.

11. A circular flower bed lies inside a

rectangular field of size 25 m 18 m. The areaof the field excluding the flower bed is 296 m 2.

Find the diameter of the flower bed.

12. A larger wheel of diameter 50 cm is

a tta ch ed to a s ma ller w he el o f d ia me te r

30 cm. Find the number of revolutions made by

the smaller wheel, when the larger one makes 15

revolutions.

13. From an equilateral triangle of side 24

cm, a circle of radius 7 cm is cut off. Find the

area of remaining portion of the triangle.

14. The short and long hand of a clock are

4 cm and 6 cm respectively. Find the sum of

distances travelled by their tips in one day.

12.2 AREAS OF SECTOR AND SEGMENT

OF A CIRCLE

1. Length of an arc which subtends an angle

of at the centre =2

360 180

r r =

.

2. Sector of a circle is a region enclosed byan arc of a circle and its two bounding radii.

(i) Area of sector OACBO =

2

360

r.

(ii) Perimeter of sector OACBO

=2

2

360

+

rr .

3. Minor sector : A sector of a circle is

called a minor sector if the minor arc of the

circle is a part of its boundary. In the figure

aboveminor sector is OACBO.

4. Major sector : A sector of a circle is

called a major sector, if the major arc of the

circle is a part of its boundary. In the above

figure, OADBO is the major sector.


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5. The sum of the arcs of major and minor

sectors of a circle is equal to the circumference

of the circle.

6. The sum of the areas of major and minor

sectors of a circle is equal to the area of the

circle.

7. The area of a sector is given by

A =1

,2

lr where l=180

r

.

8. Angle described by minute hand in

60 minutes = 360.

Angle described by minute hand in one

minute =360

6 .60

=

Thus, the minute hand rotates through an

angle of 6 in one minute.

9. Angle described by hour hand in 12 hours

= 360.

Angle described by hour hand in 1 hour

360

12

= = 30.Angle described by hour hand in one minute

30 1

60 2

= = .

Thus, hour hand rotates through 1

2

in

1 minute.10. A segment of a circle is the region

bounded by an arc and a chord, including the arc

and the chord.

11. Minor segment : If the boundary of a

segment is a minor arc of a circle, then the

corresponding segment is called a minor

segment. In the figure, segment PQR (the area

which is shaded) is a minor segment.

12. M aj or s eg me nt : A s egme nt

corresponding a major arc of a circle is known as

the major segment. In the figure, segment PQSP

is a major segment.

13. Area of minor segment PRQS

221 sin

360 2

rr

=

.

14. Area of major segment PQSP

= r2 area of minor segment PRQS.

TEXTBOOK'S EXERCISE 12.2

Unless stated otherwise, use 22

.7

=

Q.1. Find the area of a sector of a circle

with radius 6 cm if angle of the sector is 60.Sol. Radius r = 6 cm

Angle = 60

We know that, area of the sector

= 2

360r

= 2 2 260 22 132(6) cm cm .

360 7 7

=

Q.2. Find the area of a quadrant of a circle

whose circumference is 22 cm.

Sol. Let the radius of the circle be r cm.

As per condition, 2r = 22

22

2 227

=r 22 7

2 22

=

r r =

7

2 cm

For a quadrant of a circle, = 90

We know that, area of the sector = 2

360

r

So, area of given quadrant

=

2290 22 7 cm

360 7 2


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= 2 290 22 7 7 77

cm cm .

360 7 2 2 8

=

Q.3. The length of the minute hand of a

clock is 14 cm. Find the area swept by the

minute hand in 5 minutes.

Sol. r = 14 cm [Given]

Angle traced in 5 minutes =360

5 3060

=

We know that, area of sector = 2

360

r

So, area swept by the minute hand in 5 minutes

= 2 230 2 154

14 14 cm cm .360 7 3

2 =

Q.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 segment. (Use= 3.14)

Sol. Radius = r = 10 cm [Given]

= 90

O

10cm

10cm

A B

90

We know that, area of minor sector

= 2

360

r

So, area of the minor sector OAB

= 290

3.14 10 10 cm360

= 78.5 cm2

Area of OA OB

OAB2 =

= 2 210 10

cm 50 cm2

=

Area of the minor segment = Area of minor

sector OAB Area ofOAB

= 78.5 cm2 50 cm2 = 28.5 cm2

(ii) Area of major sector = Area of circle Area

of minor sector

=r2 78.5 = (3.14 10 10 78.5) cm2

= (314 78.5) cm2 = 235.5 cm2.

Q.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 corre-

sponding chord [Imp.]

Sol. Radius = r = 21 cm [Given]

Angle = = 60

O

21cm

21cm

A BM

(i) The length of the arc = 2360

r

= 60 22

2 21 cm 22 cm.360 7

=

(ii) Area of the sector formed by the arc

= 2 260 22 21 21 c m

360 360 7 =

r

= 231 cm2.

( i ii ) A r e a o f t he s eg me nt f or me d b y t he

corresponding chord

= Area of sector OAB Area ofOAB

= 231 cm2 area ofOAB (i)

Now, we have to find the area of triangle AOB.

Draw OM AB

InOMA andOMB,

OMA = OMB [Each = 90]

OA = OB [Radius of the circle]OM = OM [Common side]

OMA OMB

[RHS congruence criterion]

AM = BM [CPCT]

M is the mid-point of AB

and AOM = BOM [CPCT]

AOM =BOM =1

2 AOB


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So, AOM =BOM =1

2

60 = 30

In OMA,

OMcos30

OA =

3 OM

2 21=

OM = 21 3

cm2

sin 30 =AM

OA

1 AM

2 21= AM =

21cm

2

AB = 2AM =21

2 cm2

= 21 cm

So, area of1

OAB AB OM2

=

= 21 21 3 441 3

21 cm2 2 4

=

Area of segment formed by the corresponding

chord = Area of the sector formed by the arc Area

ofOAB

=441 3

2314

cm2.

Alternate method :

We know that area of minor segment

=

221

360 2

r

r sin

= (231 1

2 21 21 sin 60) cm2

=2441 3231 cm

4


subtends an angle of60 at the centre. Find the

areas of the corresponding minor and major

segments of the circle.

(Use = 3.14 and 3 = 1.73) [Imp.]

Sol. Radius = r = 15 cm, Angle = = 60

O

15cm

15cm

A BM

Area of the minor sector = 2

360

r

= 2 260

3.14 (15) cm360

= 117.75 cm2 ....(i)

Area of :AOB

Draw OM AB

InOMA andOMB,

OA = OB [Radii of the same circle]


OM = OM [Common side]

OMA OMB[RHS congruence criterion]

AM = BM [CPCT]

AM = BM =1

2 AB

andOM =BOM [CPCT]

OM =BOM =1 1

AOM (60 )2 2

=

= 30

In OMA,

cos 30 =OM

OA

3 OM

2 15=

15 3

OM2

= cm

Also, sin 30 =AM

OA

1 AM

2 15=

AM =15

2 2AM = 15 AB = 15 cm

Area ofAOB =1

AB OM2

= 21 15 315 cm

2 2 = 2

225 3cm

4

=225 1.73

4

cm2 = 97.3125 cm2 (ii)

Area of corresponding minor segment of thecircle

= Area of the minor sector Area ofAOB

= (117.75 97.3125) cm2 = 20.4375 cm2

(iii)


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(ii) New length of the peg = 10 m

Area of the new sector

=

21R 3.14 10 10 90

360 360

=

m2

= 3.14 5 5 m2 = 78.50 m2.

Hence, increase in grazing area

= (78.50 19.625)m2 = 58.875 m2.

Q.9.A brooch is made with silver wire in the

form of a circle with diameter35 mm. The wire

is also used in making5 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.

Sol. (i) Diameter of circle = 35 mm

Radius of circle = 35

mm2

Number of diameters = 5

Length of 5 diameters = 35 5 mm = 175 mm

Circumference of circle = 2r

= 22 35

2 mm7 2

= 110 mm

The total length of the silver wire required= (110 + 175) mm = 285 mm

(ii) r =35 mm,2

=360 36

10 =

Area of each sector of the brooch

= 2

360

r =

236 22 35 35mm

360 7 2 2

= 2385

mm .4

Q.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.[V. Imp.]

Sol. Radius of the circle = 45 cm,

Number of ribs = 8

So, central angle = 360 458

= =

Area between the two consecutive ribs of the

umbrella = area of sector

= 2

360r

=

245 22 45 45 cm360 7

= 222275

cm .28

Q.11. A car has two wipers which do not

overlap. Each wiper has a blade of length 25cm

sweeping through an angle of 115. Find the

total area cleaned at each sweep of the blades.

Sol. Length of blade = r = 25 cm

Sweeping angle = = 115

We know that, area of sector =

2

360

r

Total area cleaned at each sweep of the blades

= 22

360r

= 2 2115 22

2 (25) cm360 7

= 2158125

cm .126

Q.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= 3.14).

Sol. Sector angle = = 80,

Radius = r = 16.5 km


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Area of the sea over which the ships are warned

= Area of sector

= 2

360r

=

2803.14 (16.5)

360

km2

= 189.97 km2.

Q.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) [2011 (T-II)]

Sol. Radius of the cover design = 28 cm

Number of equal designs = 6

Sector angle = 360

606

= =

Area of minor sector OAB

= 2

360r

=

2 260 2228 cm

360 7

= 21232

cm3

= 410.67 cm2 ....(i)

Area of AOB :Draw OM AB

InOMA andOMB,

OA = OB [Radii of the same circle]

OM = OM [Common]


OMA OMB [RHS congruence criterion] AM = BM [CPCT]

AM = BM =1

AB2

[Mis mid-point ofAB]

AOM BOM = [CPCT]

1AOM BOM AOB2 = =

1(60 ) 30

2= =

In OMA, OMcos30OA

=

3 OM

2 28= OM = 14 3cm

sin 30 =AM

OA

1 AM

2 28= AM = 14 cm

2AM = 28 cm AB = 28 cm

Area of1

AOB AB OM2

=

= 21

28 14 3 cm2

= 2196 3 cm

= 196 1.7 cm2 = 333.2 cm2

Area of minor segment

= Area of minor sector Area of AOB= (410.67 333.2) cm2 = 77.47 cm2

Area of one design = 77.47 cm2

Area of six designs = 77.47 6 cm 2

= 464.82 cm2 Cost of making the designs at the rate of

Rs. 0.35 per cm2

= Rs 464.82 0.35 = Rs. 162.68.

Q.14. Tic k th e c or re ct a n swer in th e

following :

Area of a sector angle p (in degrees) of a

circle with radius R is

(a) 2180

p

R (b) 2

180

pR

(c) 2360 R (d) 2

2720 RSol. (d) Angle of sector = =

Radius of circle = R

Area of sector = 2

R360

= 2R

360

p

=2 2

2 R 2 R 2(360) 720

p p =

Hence, the correct answer is (d).


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Q.1. Area of a quadrant of circle whose

circumference is 22 cm is :22

7

=

[2011 (T-II)](a) 3.5cm2 (b) 3.5cm(c) 9.625 cm2 (d) 17.25 cm2

Sol. (c) 2r = 22 r =7

2

Area of a quadrant90 22 7 7

360 7 2 2

=

2 277 cm 9.625 cm

8= =

Q.2. The minute hand of a clock is 21 cm

long. The distance moved by the tip of the minute

hand in 1 hour is : [2011 (T-II)](a) 21 cm (b) 42 cm(c) 10.5 cm (d) 7 cm

Sol. (b) Distance moved by the tip of the minute

hand in 1 hour 360

2 21360

=

= 42 cm

Q.3. The angle through which the minute

hand of the clock moves from 8 to 8 : 35 is :(a) 210 (b) 90(c) 60 (d) 45 [2011 (T-II)]

Sol. (a) Angle described by the minute hand in

35 minutes = 6 35 = 210.

Q.4. The length of the minute hand of a

clock is 6cm. The area swept by minute hand in

10 minutes is :

(a) 212 cm (b) 236 cm

(c) 29 cm (d) 26 cm

Sol. (d) Angle swept in 1 minute = 360

660

=

So, angle swept in 10 minutes, = 10 6 = 60

Area of the sector = 2

360

r

= 2 2 260 6 cm 6 cm

360

=

Q.5. If the length of minute hand of a watch

is 7 cm, then the area swept by it between 9

a.m. to 9 : 10 a.m. is :

(a) 3cm2 (b) 3.5 cm2

(c) 3.6 cm2 (d) 4.2 cm2

Sol. (c) Angle swept in 10 minutes

=360

10 6060

=

Therefore, area of the sector

= 2 260

( 7) cm360

= 21

22 cm 3.666

= cm2

Q.6.If an arc subtends an angle of45 at the

centre of the circle of radius a cm, then length of

the arc is :

(a)6

a cm (b)

3

a cm

(c)4

a cm (d)

2

a cm

Sol. (c) Length of the arc = 2360

r

= 45

2360

a cm =

4

a cm

Q.7.The area of the sector which subtends an

angle of 60 a t th e c e n tr e o f a c ir c le is

4.4cm2, then area of the circle is :

(a) 36cm2 (b) 26.4cm2

(c) 36.6cm2 (d) 30cm2

Sol. (b) Area of the sector = 2

360

r

4.4 = 260

360 r 2r = 4.4 6

2r = 26.4

Q.8.The angle subtended by an arc of length

2

3

cm at the centre of the circle of radius4 cm is:

(a) 30 (b) 45

(c) 60 (d) 90


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Sol.(a) Length of arc = 2

360

r

2

2 43 360

=

= 30

Q.9. If a pendulum swings through an angle

of60 and describes an arc of 11 cm in length,

then length of the pendulum is :

(a) 12.5 cm (b) 11.5 cm(c) 10.5 cm (d) 9 cm

Sol. (c) Arc length = 2

360

r

11 =60 22

2360 7

r

66 7

44

= r

21

10.52

= =r cm

Q.10. The area of the sector cut off from the

circle of radius 3 cm is 39

7 cm2. The angle

subtended by the sector at the centre of the circle

is :

(a) 60 (b) 120

(c) 45 (d) 90

Sol.(b) Area of the sector = 23

360

cm2

66 22

97 360 7

=

= 3 360

9

= 120

Q.11. The perimeter of the sector of a circle

who s e c en tra l a ng le is 45 a nd r a diu s

7 cm is :

(a) 39cm (b) 19.5cm

(c) 35cm (d) 17.5cm

Sol.(b) The length of the arc =45

2 7360

cm

=1 22 11

2 7 cm cm8 7 2

=

Perimeter of the sector = 11

7 7 cm2

+ +

= 39

2cm = 19.5 cm

Q.12. A minute hand swepts an area of 7

60

cm2 in 1 minute. The length of the minute handis :

(a) 7 cm (b) 7 cm

(c) 14 cm (d) 21

2 cm

Sol. (a) Area of the sector swept out in 1 minute

= 2

360

r

7

60 = 2

6

360 r [ for 1 minute is 6]

r2 = 7 cm r = 7 cmQ.13. The length of the minute hand of a

clock is 7 cm. Find the area swept by the minute

hand from 6.00 p.m. to 6.10 pm. [2011 (T-II)]Sol.We have,

Angle described by the minute hand in one

minute = 6

Angle described by the minute hand in 10

minutes = (6 10) = 60

Area swept by the minute hand in 10 minutes

= Area of a sector of angle 60 in a circle of

radius 7 cm

2 2 260 22

(7) cm 25.66 cm360 7 = = .


long. Find the area swept by the minute hand on

the face of the clock from 7.00 a.m. to 7.05 a.m.

[2011 (T-II)]

Sol.We have,

Angle described by the minute hand in one

minute = 6

Angle described by the minute hand in 5

minutes = (6 5) = 30

Area swept by the minute hand in 5 minutes

= Area of a sector of angle 30 in a circle of

radius 21cm

( )2

2 230 22 21 cm 5.5 cm .360 7

= =

Q.15.The perimeter of a sector of a circle of

radius 5.6 cm is 27.2 cm. Find the area of the

sector. [2011 (T-II)]Sol.Let OAB be the given sector. Then,

Perimeter of sector OAB = 27.2 cm


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OA + OB + arc AB = 27.2 cm

5.6 + 5.6 + arc AB = 27.2

arc AB = (27.2 11.2) cm = 16 cm

Area of sector OAB

= 2 21 1

16 5.6 cm 44.8 cm2 2

= =l r

Q.16. In the given figure, O is the centre of

a circle. The area of sector OAPB is 5

18

of the

area of the circle. Find x. [2009]

Sol.Let the radius of the circle be r.

Area of the circle = 2r

And, area of the sector OAPB =

2

360

r x

But, 25

18r =

2

360

r x x =

360 5

18

= 100.


subtends a right angle at the centre. What is the

area of the minor sector? [2008C]

O

AB

14cm

Sol.Here, r = 14 cm and AOB 90 . =

Area of minor sector = 290

360

r

= 1 22

14 144 7

cm2 = 154 cm2.

Q.18. What is the perimeter of a sector of angle 45 o f a c irc le with r ad iu s 7 cm ?

22Use =

7

[2008C]

Sol.The arc AB of length l of a sector of angle

45 in a circle of radius 7 cm is given by

l = 2360

r

=45 22

2 7360 7

cm

= 5.5 cm

Perimeter of sector (OAB)

= OA + OB + arc AB = (7 + 7 + 5.5) cm

= 19.5 cm.

Q.19. In the figure, the shape of the top of atable in a restaurant is that of a sector of a

circle with centre O and 90 .BOD = If BO = OD = 60 cm, find

(i) the area of the top of the table

(ii) the perimeter of the table top.

[Take = 3.14] [2009, 2011 (T-II)]

Sol. BO = OD, r = 60 cm

= 360 BOD = 360 90 = 270

(i) Area (table top) = 2

360

r

= 270

3.14 60 60360

cm2

= 45 60 314

100 cm2

= 27 314 cm2 = 8478 cm2.

(ii) Perimeter (table top) = 2 2360

r r

=

270 314

2 60 cm 120360 100

+ cm

=90 314

cm 120 cm100

+

=2826 1200

10

+cm = 402.6 cm.

Q.20. From a circular piece of cardboard

with radius 1.26 m, a sector with central angle

40 has been removed. Find

(i) Area of the portion removed

l

A B

45

O

7 cm


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(ii) Area of the remaining portion

(iii) Perimeter of the sector removed.

O

A

B

40

Sol.

(i) Area of the portion removed = Area of the sector

= 2

360

r =

40 221.26 1.26

360 7

m2

= 0.5544 m2.(ii) Area of remaining portion

= Area of the circle Area of the sector

= 22

1.26 1.26 0.55447

m2

= (4.9896 0.5544) m2 = 4.4352 m2.

(iii) Perimeter of the sector = 2r+ Length of the

arc = 2 1.26 m + 40 22

2 1.26360 7

m

= (2.52 + 0.88) m = 3.40 m.

Q.21.In the figure, sectors of two concentriccircles of radii 7 cm and3.5 cm are given. Find

the area of shaded region.22

Use =7

[2011 (T-II)]

Sol. Let A1

and A2

be the areas of sectors OAB

and OCD respectively. Then,

A1

= area of a sector of angle 30 in a circle of

radius 7 cm.

A1

2 230 22 7 cm360 7

=

2Using : A =360

r

A1

= 277

cm6

A2

= Area of a sector of angle 30 in acircle of radius 3.5 cm

A2

2 230 22 (3.5) cm360 7

=

A2

2 21 22 7 7 77cm cm12 7 2 2 24

= =

Area of the shaded region = A1

A2

277 77 cm6 24

=

2 277 77(4 1) cm cm

24 8= =

= 9.625 cm2.

Q.22. The area of a sector is 1

10that of the

complete circle. Find the angle of the sector of the circle. [Imp.]

Sol. Let the radius of the circle be r.

Area of circle = r2

Let the angle of the sector be. Then, area of the

sector =

2

360

r


2

360

r

= 21

10 r

1

360 10

=

=

360

10

= 36.


long. Find the area of face of the clock described

by the minute hand between9 a.m. and9.35a.m.Sol. Angle described by the minute hand in one

minute = 6

So angle described by the minute hand in

35 minutes = (6 35) = 210

Area swept by the minute hand in 35 minutes= Area of a sector of angle 210 in a circle of

radius 10 cm

= 2210 22 10

360 7

cm2 = 183.3 cm2.

Q.24. A circular disc of 6 cm radius is

divided into 3 sectors with central angles 120,

150, 90. Find the ratio of the areas of three

sectors.


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12.3 AREAS OF COMBINATIONS OF

PLANE FIGURES

In o ur d ai ly l if e we co me a cr os s

combinations of plane figures for making

interesting designs such as flower beds, fabric

designs, window designs, designs on table covers

etc. In this section we will study the process of

calculating areas of combination of figures.

PRACTICE EXERCISE 12.2A

Choose the correct option (Q. 1 4)

1.The sum of areas of a major sector and the

corresponding minor sector of a circle is equalto :

(a) area of the circle

(b)1

2 area of the circle

(c) 1


(d) 3


2.The radius of a circle is 5 cm. The area of

the sector formed by an arc of this circle of

length 9 cm is :

(a) 45 cm2 (b) 22.5 cm2

(c) 67.5 cm2 (d) 2.25 cm2

3. The circumference of a sector of a circle

of radius 7 cm and central angle 45 is :

(a) 19.5 cm (b) 39 cm

(c) 14 cm (d) 7 cm

4. What is the supplementary angle of the

central angle of a semicircle?(a) 0 (b) 90

(c) 180 (d) 360

5. Find the area of a sector of a circle with

radius 6 cm, if angle of the sector is 60.

6. A chord 10 cm long is drawn in a circle

whose radius is 5 2 cm. Find the area of major

segment.

7. A chord of a circle of radius 28 cm

subtends an angle 45 at the centre of the circle.

Find the area of the minor segment.

8. The perimeter of a sector of a circle with

central angle 90 is 25 cm. Find the area of the

minor segment of the circle.

9. Find the area of shaded portions of the

following figures with given measurements :

(a) (b)

10. In a circle of radius 6 cm, a chord of

10 cm makes an angle of 110 at the centre of the

circle. Find the length of the arc and area of thesector so formed.

90

120

150

Sol. Given r = 6 cmArea of the circle =r2 sq. cm

Area of sector with central angle =

2

360

r

Ratio of the areas of the three sectors

= 2 2 2120 150 90

: :360 360 360

r r r

= 120 : 150 : 90 = 4 : 5 : 3.


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Area of semi-circle APD = 21

2

r

2 21 22 7 7 cm 77 cm2 7

= = (ii)

Area of semi-circle BPC

= 2 21 1 22

7 7 cm2 2 7

= r

= 77 cm2

Area of the shaded region = Area of the square

ABCD (Area of semi-circle APD + Area of semi-

circle BPC)

= (196 154) cm2 = 42 cm2.

Q.4. Find the area of the shaded region in

the given figure, where a circular arc of radius6

c m h as b ee n d ra wn w it h v er te x O o f a n

equilateral triangle OAB of side12 cm as centre.

[2011 (T-II)]

Sol. Radius of circle = r = 6 cm

Side of equilateral triangle = 12 cm

Area of shaded region

= Area of circle + Area of equilateral triangle

OAB Sectorial area common to the circle and the

triangle

= 2 2 2 2 2 23 60

(6) cm (12) cm (6) cm4 360

+

= 2 2 236 cm 36 3 cm 6 cm +

= 2 230 cm 36 3 cm +

2 22230 cm 36 3 cm

7= +

= 2660 36 3 cm

7

+ .

Q.5. From each corner of a square of side

4cm a quadrant of a circle of radius 1cm is cut

and also a circle of diameter 2 cm is cut as

shown in figure. Find the area of the remaining

portion of the square.

A

D

B

C

Sol. Side of Square = 4 cm

Radius of quadrant of a circle = 1 cm

Area of the remaining portion of the square =

Area of the square [4 Area of a quadrant + Area

of a circle]

= (4 4) cm2

22904 1

360

2 ( ) + 2 cm2

= (16 2) cm2 =22

16 27

cm2

=44

16 7

cm

2 =112 44

7 cm2 =

68

7 cm2 .

Q.6. In a circular table cover of radius

32 cm, a design is formed leaving an equilateral

triangle ABC in the middle as shown in figure.

Find the area of the design (shaded region).[Imp.]

Sol. Radius of table cover = 32 cm

Area of equilateralABC

2 23 3(side)4 4

a= = [a = side]

Area of the design (shaded region)

= Area of the circular table cover

Area of the equilateral triangle ABC

= 2 23(32)

4a (i)


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Leth be the height ofABC. Since the centre of

t h e c i rc l e c o i n ci d es w it h t h e c e n t roi d o f t h e

equilateral triangle.

Radius of the circle = 2

3h

As per condition, 2

323

h= h = 48 cm

Using Pythagoras theorem in ABD,

a2 =

22

2

ah

+ a2 =

22

4

ah +

22

4

aa =

22 23

4

ah h =

22 4

3

ha =

24(48)

3= = 3072

From equation (i),

Required area = 2 2 3

(32) cm4

3072 cm2

= 2 222

(1024)cm 768 3cm7

= 222528768 3 cm

7

.

Q.7. In figure, ABCD is a square of side

14 cm. With centres A, B, C and D, four circles

are drawn such that each circle touches

externally two of the remaining three circles.Find the area of the shaded region. [Imp.]

A

D

B

C

Sol. Side of square = 14 cmRadius of each circle = 7 cm

Area of square ABCD = (side)2

= 14 14 cm2 = 196 cm2

Area of sector =

2

360

r

Area of shaded region = Area of the square of side14 cm 4 [Area of a sector of central angle 90]

= 2 2 29014 14 cm 4 7 cm

360

= 2 2 222

196 cm (7) cm7

= (196 154) cm2 = 42 cm2.

Q.8. Figure below depicts a racing track

whose left and right ends are semicircular. The

distance between the two inner parallel linesegments is 60 m and they are each 106 m long.

If the track is 10 m wide, find :

(i) the distance around the track along its

inner edge

(ii) the area of the track. [2011 (T-II)]

Sol. (i) Length of each parallel line segment

= 106 m

D i st a nc e b e tw ee n t w o i n ne r p a ral l el l i ne

segments = 60 m

Width of track = 10 m

The distance around the track along its inner edge

= 106 m + 106 m + 2 60

m2

= (212 + 60 ) m =22

212 60 m7

+

= 1320 2804

212 m m7 7 + = (ii) Area of the track

= (106 10) m2 + (106 10) m2

+ 2 2 21 12 (30 10) (30) m

2 2

+ = 2 2 2 2 21060 m 1060 m [(40) (30) ] m+ +

= 2 2222120 m 700 m

7+

= (2120 + 2200) m2 = 4320 m2.


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Q.9.In figure, AB and CD are two diameters

of a circle (with centre O)perpendicular to eachother and OD is the diameter of the smaller

circle. If OA = 7 cm, find the area of the shaded

region. [2010, 2011(T-II)]

O

B

D C

A

Sol. Diameter of large circle = 14 cm

Radius of large circle = r = 7 cm

Diameter of smaller circle = 7 cm

Radius of smaller circle =7

cm2

Area of1

ABC Base Altitude2

=

Area of the shaded region

= Area of small circle + Area of semi-circle ACB

Area of triangle ABC

=

22 2 2 27 1 1cm (7) cm 14 7 cm

2 2 2

+

= 2 2 249 49

cm cm 49 cm4 2

+

= 249 3 22

cm4 7

49 cm2

= 249 66 49 28 cm

28

=

23234 1372 cm28

= 21862

cm28

= 66.5 cm2.

Q.10. The area of an equilateral triangle

ABC is 17320.5 cm2. With each vertex of the

triangle as centre, a circle is drawn with radius

equal to half the length of the side of the triangle

(see figure). Find the area of the shaded region.

(Use = 3.14 and 3 = 1.73205).[2011 (T-II)]

A

B C

Sol. Area of equilateral triangle = 17320.5 cm2

Let the length of the side of the equilateral

triangle ABC be a cm. Then,

Area of triangle = 2 23 cm

4a

As per condition, 23 17320.54

a =

a2 =17320.5 4

3

= 10000 4 = 40000

a = 40000 = 200 cm

Area of each sector

=

22 260 200 cm 10000 cm

360 2 6

=

Area of each shaded region = Area of the

equilateral triangle ABC 3 (Area of each sector)

= 2 217320.5 cm (10000) cm

2

= 17320.5 cm2 3.14 5000 cm2

= (17320.5 15700) cm2 = 1620.5 cm2.

Q.11. On a square handkerchief, nine

circular designs each of radius 7 cm are made

(see figure). Find the area of the remaining

portion of the handkerchief.

Sol. Radius of each circle = 7 cm

Diameter of each circle = 14 cm

Side of square = 42 cm


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Area of the remaining portion of the handkerchief

= Area of the square ABCD Area of nine

circular designs

= (42 42) cm2 2 29 (7) cm

= 2 2 222

1764 cm 9 (7) cm7

= (1764 1386) cm2 = 378 cm2.

Q.12. In figure, OACB is a quadrant of a

circle with centre O and radius 3.5 cm. If OD

= 2 cm, find the area of the

(i) quadrant OACB,

(ii) shaded region. [2011 (T-II)]

A

D

OB

C

Sol. Radius of quadrant, r = 3.5 cm

Angle of sector = 90

We know that, area of sector =

2

360

r

(i) Area of the quadrant OACB

= 2 290 (3.5) cm

360

= 21 22 35 35 cm

4 7 10 10 =

77

8 cm2.

(ii) Area of the shaded region = Area of the

quadrant OACB Area ofOBD

= 2 2 277 OB OD 77 3.5 2cm cm cm

8 2 8 2

=277 35

cm8 10

=2 277 7 49

cm cm8 2 8

= .

Q.13. In the figure, a square OABC isinscribed in a quadrant OPBQ. If OA = 20 cm,

find the area of the shaded region.

(Use= 3.14).Q

C

OA P

B

Sol. In AOB,

OB = 2 2

OA AB+ [Using Pythagoras Theorem]

= 2 2OA + OA

= 2OA 2 (20)= cm = 20 2c mArea of the shaded region = Area of the quadrant

OPBQ Area of the square OABC

= 2 2 290(20 2 ) cm 20 20 cm

360

= 200 cm2 400 cm2

= 200 3.14 cm2 400 cm2

= 628 cm2 400 cm2 = 228 cm2.

Q.14. AB and CD are respectively arcs of

two concentric circles of radii 21 cm and7 cm

and centre O (see figure). If AOB = 30, find

the area of the shaded region. [2011, (T-II)]

A B

C D

O

307 cm

21cm

Sol. Radius of sector OBA = r1

= 21 cm

Radius of sector ODC = r2

= 7 cm

Area of the shaded region

= Area of the sector OAB

Area of the sector OCD

=

2 21 2

360 360

r r

= 2 2 2 230 30

(21) cm (7) cm360 360

=

2 21 22 1 22

21 21cm 7 7 cm12 7 12 7

= 2 2231 77

cm cm2 6

= 2693 77

cm6

= 2 2616 308

cm cm .6 3

=

Q.15. In the figure, ABC is a quadrant of a

circle of radius 14 cm and a semicircle is drawn

with BC as diameter. Find the area of the shaded

region. [2008, 2011 (T-II)]


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B

AC

Sol. InBAC, using Pythagoras Theorem

BC2 = AB2 + AC2 = (14)2 + (14)2 = 2(14)2

BC = 14 2 cm

Radius of circle = 14 cm

Area of 1

BAC AB AC

2

=

= 21

14 14 cm2

= 98 cm2 (i)

Area of sector ABC =

2

360

r

=22 14 14 90

7 360

= 154 cm2 (ii)Radius of semi-circle with BC as diameter

= 14 2

cm2

= 7 2 c m

Area of the semi-circle

= 21

2r = 2

1 227 2 7 2 cm

2 7

= 154 cm2

Required area = Area of semi-circle with BC as

diameter [Area of sector ABC Area ofBAC]

= 154 cm2 [154 98] cm2 = 98 cm2.

Q.16. Calculate the area of the designed

region in figure common between the twoquadrants of circles of radius 8 cm each.

[2011 (T-II)]

Sol. Side of the square = 8 cm

Area of the square = 64 cm2.

Area of two quadrants with centres B and D and

radius 8 cm

=

2 22 2 90

360 360

r r =

=

2 22 8 8 90

7 360

cm2

= 2704

cm7

Since the designed area is common to both the

sectors.

Therefore, area of design = Area of both sectors

Area of square

= 2 2704 256

64 cm cm7 7

= = 36.57 cm2.

Q.1. In the given figure, if O is the centre of

the circle ABCD and OB =3 cm, then the areaof the shaded portion is :

O

C

A B

D

(a)

246

7 cm (b)

236

7 cm

(c) 226

7cm (d) none of these

Sol. (b) Area of the shaded portion

= 2 2 21 13 cm 6 3 cm

2 2

= 2 2 211 4 369 9 cm 9 cm cm

7 7 7

= =



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Q.2. In a square handkerchief of side 14 cm,

a circular design of radius 7 cm is made asshown in the following figure. The area of the

remaining position is :

A

B C

D

14 cm

(a) 24cm2 (b) 48cm2

(c) 42cm2 (d) 56cm2

Sol. (c) Area of the remaining portion

= Area of whole handkerchief

Area of the circle of radius 7 cm

= 2 2 2 2 2 222

14 cm 7 cm 196 cm 7 7 cm7

=

= (196 154) cm2 = 42 cm2.

Q.3. In the given figure, area of the portion

ACBDA is :

C

A

B

O7 cm

D

2 cm

(a) 35.5 cm2 (b) 31.5cm2

(c) 25.5cm2 (d) 21.5cm2

Sol. (b) Area of the portion ACBDA

= Area of sector Area of AOD

= 290 17 7 7 2 cm

360 2

= 2 263

cm 31.5 cm2

=

Q.4. In the given figure, 4 quadrants each of

radius1 cm are cut from the square of side 4 cm.

The area of remaining portion is :

(a) 12cm2 (b) 13cm2

(c) 13.4cm2 (d) 12.8cm2

1 cm 1 cm

1 cm1 cm

1 cm

1 cm

1 cm

1 cm

4 cm

4 cm

Sol. (d) Area of the remaining portion

= Area of square 4 [Area of quadrant]

= (4 4 12) cm2

= (16 ) cm2 = (16 3.14) cm2 = 12.86 cm2

Q.5. If a park has flower bed in the shape of

two s em i-c ir cle s a n d a c irc le o f r a diu s

2m as shown below, then total area of the flowerbed is :

14 m

2m

20 m

(a) 51 m2 (b) 49m2

(c) 55 m2 (d) 53 m2

Sol. (d) Total area of the flower bed

= 2 2 2 2 21 7 m 2 m

2

+

= ( ) 2 249 4 m 53 m + = .

Q.6. Four cows are tethered at four corners

of a square field of side 12 m. If each cow can

graze the maximum area, then the total area

grazed by them is :

(a) 6 m2 (b) 24 m2(c) 36 m2 (d) 30 m2

Sol. (c) Total area grazed by 4 cows

= 4(Area of quadrant of radius 6 m)


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= 22

[2.1 0.7 1.4] cm

7

+ +

224.2 cm 13.2 cm

7= = .

Q.11. Find the area of the shaded region in

t he f ig ure , w he re A BC D i s a s qu are o f

side 14 cm. [2008, 2011(T-II)]

Sol. Diameter of each circle =14

2 cm = 7 cm

Radius of each circle =7

2cm = 3.5 cm

Area of the shaded region= area of the square 4 area of a circle

=

214 14 4 (3.5) cm

2

=22

196 4 3.5 3.57

cm2

= (196 154) cm2 = 42 cm2.

Q.12. In the figure, ABC is a right-angled

triangle right-angled at A. Semi-circles are

drawn on AB, AC and BC as diameters. Find the

area of the shaded region. [2008, 2011(T-II)]

Sol. BC2 = AB2 + AC2 [Pythagoras theorem]

BC = 9 16 units = 5 units

Area of the semi-circle with BC as diameter

=

2BC

2 2

=

25

2 2

sq units

=25

8

sq units

Area of the semi-circle with AC as diameter

=

2AC

2 2

222

= sq units = 2 sq units

Area of the semi-circle with AB as diameter

2AB

2 2

= =

23

2 2

sq units=

98

sq units

Area ofABC =1

2

AB AC

=1

2 3 4 sq units = 6 sq units

Area of the shaded region = area of the semi-

circle with diameter AB + area of the semi-circle with

diameter AC [area of the semi-circle with diameter

BC area ofABC]

=9 25

2 68 8

+ sq units

=

9 16 25

68

+

+ sq units = 6 sq units.

Q.13. The area of an equilateral triangle is

249 3 cm . Taking each angular point as centre,

circles are drawn with radius equal to half the

length of the side of the triangle. Find the area

of triangle not included in the circle.

[Take 3 = 1.73] [2009]

60

60 60

Sol. Let each side of the equilateral triangle bexcm.

Then, 23 49 3

4x =

x2 = 4 49 x = 2 7 = 14 cm Radius of each circle = 7 cm


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Area of the three sectors each with central angle

60

=2

27 603 cm360

= 222 7 73 cm

7 6

= 77 cm2

Required area = Area of the shaded region

= Area of the triangle Area of the three sectors= (49 1.73 77) cm2

= (84.77 77) cm2 = 7.77 cm2.

Q.14. In the figure, ABC is a right triangleright angled at A. Find the area of shaded region

if AB = 6 cm, BC = 10 cm and O is the centreof the incircle of ABC. (take = 3.14)

[2011 (T-II)]

Sol. In ABC, BC2 = AB2 + AC2

AC 100 36 8 = = cmOP AB and OQ AC

[ Ra d iu s t hr o ug h t he p o in t o f c on t ac t i sperpendicular to the tangent]

And OP = OQ = r

Hence, APOQ is a square.

BP = BR [Tangents drawn from an externalpoint are equal]

AB AP = BC CR

6 r = 10 CR

CR = 4 + r ... (i)

Also, CR = CQ

4 + r = AC AQ [From (i)]

4 + r = 8 r

2r= 4 r= 2 cmArea of the shaded region

= Area ofABC Area of the circle

=1

2 AB AC 22

=1

6 8 3 14 42

. cm2

= (24 12.56) cm2 = 11.44 cm2.

Q.15. In the figure below, there are three

semi-circles, A, B and C having diameters 3 cm

each, and another semi-circle E having a circle

D with diameter 4.5cm are shown. Calculate :

(i) the area of the shaded region.

(ii) the cost of painting the shaded region at

the rate of25 paise per cm2, to the nearest rupee.[HOTS]

Sol. (i) Area of the shaded region (i.e., area E, B

and F) = Area of semi-circle with radius 4.5 cm

(area of semi-circle A + area of semi-circle C) + (area

of semi-circle B) (area of circle with diameter

4.5 cm).

= 2 21

(4.5) cm2

+ 2 2 21 1(1.5) (1.5) cm

2 2

+

2 2 2 21 (1.5) cm (2.25) cm2

( )

= + 22 2 2 2 2(4.5) cm (1.5) cm 1.5 cm2 2

2 2(2.25) cm

=

2 22(4.5) (1.5) (2.25)

2 2cm2


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Sol. Here, OA = R= 21 m and OC = r= 14 m.

Area of the flower bed (i.e., shaded portion)

= area of the quadrant of a circle of radius R

area of the quadrant of a circle of radius r.

= = 2 2 2 21 1R (R )4 4 4

r r

= 2 2 21 22

(21) (14) m4 7

= 1 22

4 7 [(21 + 14) (21 14)] m2

=11

14 35 7 m2 = 192.5 m2.

Q.19. Find the area of the shaded region.

Sol. Here, the radius of the bigger semi-circle

= 14 cm

Area of the bigger semi-circle

= = 2 21 1 22

(14)2 2 7

r cm2 = 2308 cm

Radius of each of the smaller semi-circles = 7 cm

Area of 2 smaller semi-circles

= 21

2 2r

= = 2 2 21 222 (7) cm 154 cm

2 7

The area of the shaded region

= (308 + 154) cm2 = 462 cm2.

Q.20. In an equilateral triangle of side24cm, a circle is inscribed touching its side.Find the area of the remaining portion of thetriangle.

Sol. Area of the equilateral ABC with side24 cm

= 23

(side)4

= =2 2 23

(24) cm 144 3 cm4

(i)

Let rbe the radius of inscribed circle, then

Area ofAFO

1 1AF area of ABC

2 6r= =

=1 144 3

122 6

r

= =144 3

4 336

r

Also, area of the inscribed circle = r2

= = 2 2 222 22

(4 3) cm 4 4 3 cm7 7

= 150.85 cm2

Required shaded area

= Area ofABC area of inscribed circle

2(144 3 150.85) cm

= 2

(144 1.732 150.85) cm= (249.408 150.85) cm2 = 98.558 cm2.

Q.21. A playground has the shape of a

rectangle, with two semi-circles on its smaller

sides as diameters, added to its outside. If the

sides of the rectangle are36 m and24.5 m, find

the area of the play ground. 22

.7

= Take

[HOTS]


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Sol.Length of the rectangle ABCD = 36 m

Breadth of rectangle ABCD = 24.5 m

Area of rectangle ABCD

= 36 24.5 m2 = 882 m2

Radius of semi-circle (I) = 24.5

m = 12.25 m.2

Area of semi-circle (I)

= = 2 2 21 1 (12.25) m2 2

r

= 2 21 22

(12.25) m2 7

= 235.8125 m2

Area of semicircle (II) = Area of semicircle (I)

= 235.8125 m2

Area of playground = Area of semi-circle

I + area of semi-circle II + area of rectangle ABCD

= (235.8125 + 235.8125 + 882) m2

= 1353.625 m2.

Q.22. Three horses are tethered with 7 m

long ropes at the three corners of a triangularfield having sides20 m, 34 m and42 m. Find the

area of the plot which can be grazed by the

horses. Also, find the area of the plot whichremains ungrazed. [HOTS]

Sol. LetA =1, B =

2and C =

3The area which can be grazed by three horses =

(Area of sector with central angle1

and radius 7 cm+ Area of sector with central angle

2and radius 7 cm

+ Area of sector with central angle3

and radius

7 cm.)

= + +

22 231 2

360 360 360

rr r

= + +

2

1 2 3( )360

r =

2

180360

r

[ Sum of three angles of a = 180]

2 222 7 7 180m 77 m

7 360

= =

Sides of plot ABC are a = 20 m, b = 34 m and

c = 42 m.

Semi-perimeter (s) = + + =20 34 42

m 48 m2

Area of triangular plot = Area ofABC

= ( ) ( ) ( )s s a s b s c

= =2 248 28 14 6 m 336 mHence, area grazed by the horses = 77 m2 and

ungrazed area = (336 77) m2 = 259 m2.

PRACTICE EXERCISE 12.3 A

1. A park is in the form of a rectangle

120 m 100 m. In the centre of the park, there

is a circular lawn as shown in the figure below.

The area of the park excluding the lawn is 8700

m2. Find the radius of the circular lawn.

22Use

7

=

lawn

120 m

100 m

2. Find the area of the shaded region in the


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figure below, if AB = 12 cm, BC = 5 cm.

(Take = 3.14)

D

A

C

B

O

3. Four equal circles, each of radius 5 cm,

touch each other, as shown in the figure below.

Find the area included between them.(Take = 3.14).

4. In the figure, two circular flower beds

have been shown on two sides of a square lawn

ABCD of side 56 m. If the centre of each

circular flower bed is the point of intersection O

of the diagonals of the square lawn, find the sum

of the areas of the lawn and the flower beds.

[2011 (T-II)]

5. A horse is placed for grazing inside a

rectangular field 70 m by 52 m. It is tethered to

one corner by a rope 21 m long. On how much

area can it graze? How much area is left

ungrazed?70 m

52 m

21 m

21 m

6. The area of a circle inscribed in an

equi l ateral tri angl e i s 154 cm2. F i n d t h e

perimeter of the triangle. (Take 3 1.73= ).A

B C

O

aa

D

h

r

a/2

7. Four cows are tethered at the four corners

of a square field of side 50 m such that each can

graze the maximum unshared area. What area

will be left ungrazed? (Take 3.14 = )

A

C

B

D

50 m

8. Find the area of the region ABCDEFA

shown in the given figure below, given that

ABDE is a square of side 10 cm, BCD is a

semi-circle with BD as diameter, EF = 8 cm, AF

= 6 cm and AFE = 90. (Take = 3.14)

10 cm

10 cm

10 cm

E

A

D

B

C

8cm

F

6cm


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B. FORMATIVE ASSESSMENT

ACTIVITY

Objective : To derive the formula for area of sector of a circle.

Materials Required :

Glaze paper, geometry box, A pair of scissors. Fevistick etc.

Procedure :

1. Draw some circles of any radius (say 3 cm) on glaze paper.

Cut these out and paste them on a drawing sheet.

2. Mark two points P and Q on circumference. Join OP and OQ.The region OPQ is called the sector of a circle. MarkPOQ =

(Figure 1). POQ is the angle of the sector.

3. Now on other circles, make different sectors of 45, 60, 90

and 120 (Figure 2).

4. Circle C1 with sector of 45, circle C2with sectors of 60,

circle C3 with sector of 90 and circle C4 with sector of 120.

Figure 2(a) Figure 2(b) Figure 2(c) Figure 2(d)

5. To calculate the area of sectors of C1, C2, C3 and C4, record your observations in the

following table.

Circle Angle of the No. of equal Area of one360

r2

sector = sectors in the circle sector

C1 45 8 1

8 r2

45

360

r2 =

1

8r2

C2 60 6 1

6 r2

60

360

r2 =

1

6r2

C3 90 4 1

4 r2

90

360

r2 =

1

4r2

C4 120 3 1

3 r2

120

360

r2 =

1

3r2

Figure 1


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Observation : We see from the above table that area of a sector =360

r2.

ANSWERS

Practice Exercise 12.1A

1. (b) 2. (a) 3. (b) 4. (b) 5. 52.8 cm, 221.76 cm2 6. 38.5 cm2 7. 10,000 revolutions 8. 2.8 m

9. 4.4 km 10. 154 cm, 126 cm 11.14 m 12. 25 13. 95.40 cm2 14. 954.56 cm or 340 cm


1. (a) 2. (b) 3. (a) 4. (a) 5. 18.85 6. 142.75 cm2 7. 30.85 cm2 8. 14 cm2 9. (a) 20.32 cm2

(b) 14 cm2 10. 11.51 cm, 34.5 cm2


1. 32.4 m 2. 72.665 cm2 3. 21.5 cm2 4. 4032 cm2 5. 346.5 m2, 3293.5 m2 6. 14 3 cm

7. 537.5 cm2 8. 115.28 cm2

Documents

Question Bankc Mathematics