Question Bank In Mathematics Class X (Term–II)

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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 circlewith diameter d (d = 2r, where r is the radius) isgiven by C = d = 2r.

2. Perimeter of a semicircle with radius r= 2r + r = r( + 2).

3. Area of a circle with radius r is given byA = r2.

4. Area of a semicircle of radius r =2

2

πr.

5. Area of a ring whose outer and innerradii are R and r respectively

= (R2 – r2) = (R + r) (R – r)

6. If two circles touch internally, then thedistance between their centres is equal to thedifference of their radii.

7. If two circles touch externally, then thedistance between their centres is equal the sum oftheir radii.

8. The distance moved by a rotating wheelin one revolution is equal to the circumference ofthe wheel.

9. The number of revolutions completed bya rotating wheel in one minute

=Distance moved in one minute

.Circumference of the wheel

(Unless stated otherwise, use22

7π = )

Q.1. The radii of two circles are 19 cm and9 cm respectively. Find the radius of the circlewhich has circumference equal to the sum of thecircumferences 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 hascircumference equal to the sum of the circumferencesof 1st and 2nd circle.

So, 2πR = 2π (19) + 2π (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 and6 cm respectively. Find the radius of the circlehaving area equal to the sum of the areas of thetwo 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 =2rπ

Area of 1st circle = r12 =

2π × (8) cm2

Area of 2nd circle =22πr =

2π × (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 (Term–II)

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Q.3. Figure depicts an archery targetmarked with its five scoring areas from thecentre outwards as Gold, Red, Blue, Black andWhite. The diameter of the region representingGold score is 21 cm and each of the other bandsis 10.5 cm wide. Find the area of each of the fivescoring 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 andwhite regions be A1, A2, A3, A4 and A5 respectively.

A1 = r12 =

22

7× (10.5)2 cm2 = 346.5 cm2

A2 = 2 22 1( )π −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 2

22( ) [(31.5) (21) ] cm

7r rπ − = × −

=222

[992.25 441] cm7

× −

=22

7× 551.25 cm2 = 1732.5 cm2

A4 = 2 24 3( )π −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 25 4( )r rπ − =

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 diameter80 cm each. How many complete revolutionsdoes each wheel make in 10 minutes when thecar 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

= π = × ×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 minutes

Circumference of the wheel

=×

= =1100000 1100000 7

4375.1760 1760

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Q.5. Tick the correct answer in the followingand 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) π units(c) 4 units (c) 7 unitsSol. Let the radius of the circle be r units.

Its circumference = π2 r units

And its area = π 2r sq. units

As per condition : π = π 22 r r r = 2 units

Hence, the correct answer is (a).

OTHER IMPORTANT QUESTIONS

Q.1. If the diameter of the wheel of a cycleis 7 cm, then its area is :

(a) 77 cm2 (b) 22 cm2

(c)277

cm2

(d) 277 cmπ

Sol. (c) Here, diameter = 7 cm radius =7

cm.2

Area = π 2r = × ×22 7 7

7 2 2cm2 =

277cm

2.

Q.2. If a copper wire of length 88 cm is bentin the form of a circle, then the radius of thecircle is :

(a) 7 cm (b) 14 cm

(c) 21 cm (d) 22 cm

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

2 rπ = 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, thenthe distance travelled in 10 complete revolutionsis approximately :

(a) 20 m (b) 26 m(c) 30 m (d) 32 mSol. (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 thecircles 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 ratioof 4 : 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,

thenπ

= =π

21

22

A (4 ) 16

A 9(3 )

x

x

Q.7. If a steel wire in the form of a rhombusof side 11 cm 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 = π2 r

π =2 44r r =×

=×

44 77cm.

2 22

Q.8. The radius of the wheel of a cycle is25 cm. The number of revolutions it will take tocover a distance of 22 m is :

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

Sol. (b) Circumference of the wheel

= π ⋅2 25 cm = π50 cm = ×22

507

cm

No. of revolutions for covering 22 m

=× ×

=×

22 100 714

50 22.

Q.9. If the diameter of a semicircularprotractor is 14 cm, then the perimeter of theprotractor is :

(a) 26 cm (b) 14 cm(c) 28 cm (d) 36 cm [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

diameter7

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 acircle 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 = π 2 ,x New area = π = π2 2(2 ) 4x x .

Q.13. The length of a wire in the form of anequilateral triangle is 44 cm. If it is rebent intothe form of a circle, then area of the circle is :

(a) 484 cm2 (b) 176 cm2

(c) 154 cm2 (d) 44 cm2

Sol. (c) Perimeter of the triangle = circumference

of the circle

⇒ 44 = 2 rπ ⇒ r =44 7

7 cm2 22

×=

×

Area of the circle =2rπ =

22× 7 × 7

7cm2

= 154 cm2.

Q.14. If the diameter of a semi-circularprotractor 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 + rπ = + ×

2214 7

7cm = 36 cm.

Q.15. Find the circumference and area of thecircle 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 2227 7 cm 154 cm .

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Q.16. Find the radius of the circle which hascircumference of 100 cm.

Sol. C = 2r r =

C 100 700

222 442

7

= 15.9 cm.

Q.17. A wheel has diameter 84 cm. Find howmany complete revolutions must it make to cover792 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= ×

π= 300.

Q.18. The difference between circumferenceand diameter of a circle is 105 cm. Find theradius of the circle.

Sol. Circumference = 2 × × Radius

Diameter = 2 × Radius

According to the question,

2r – 2r = 105

2r( – 1) = 105

22

2 1 1057

r − =

15 105 72 105

7 2 15r 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 ringwhose inner circumference is 352 m and outercircumference is 396 m. Find the width of the

track.22

Use =7

π [2011 (T-II)]

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

metres and r metres respectively. Then,

2R = 396 and 2r = 352

22 22

2 R = 396 and 2 3527 7

× × × × =r

7 1 7 1

R 396 and = 35222 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 circumferenceand diameter of a circle is 135 cm. Find the

radius of the circle.22

Use =7

π [2011 (T-II)]

Sol. Circumference = 2 × × Radius

Diameter = 2 × Radius

According to the question,

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 acar rotate in a journey of 2002 m, if the radiusof 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= = × =

π .

Q.22. Two circles touch externally. The sumof their areas is 130 sq. cm and the distancebetween their centres is 14 cm. Find the radii ofthe circles. [Imp.]

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

r cm respectively. Let C1 and C2 be the centres of thegiven circles, then,

C1C2 = R + r

R + r = 14 …(i)

[C1C2 = 14 cm (given)]

Given that the sum of the areas of two circlesis 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 Radhika’s house there is a flowerpot. The sum of radii of circular top and bottomof the flower pot is 140 cm and the difference oftheir circumferences is 88 cm. Find thediameters 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 − = = =π ×

2r = 140 + 14 = 154

r =154

2cm = 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. Thecost of ploughing the field at Rs 1.50 per m2 isRs 5775. Find the cost of fencing the field at Rs8.50 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 = πr2

2 2223850 3850

7r rπ = ⇒ × =

7

3850 175 7 35 3522

r = × = × = ×

= 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 121sq cm. The same wire he bent in the form of acircle. 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 rπ = ⇒ × × = ⇒ =

Hence, area of the circle

2 227 7

7r

= π = × × cm2 = 154 cm2.

Q.26. The diameter of a circular pond is17.5 m. 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 24.5 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 distancebetween their centres is 6 cm. Find the radii ofthe 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]

π + π = π2 2R 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 r

R + r = 14 …(iii)

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

Hence, the radii of the given circles are 10 cmand 4 cm respectively.

Q.28. Ram Prakash walks around a circularpark of area 88704 m2. How long will he take towalk 10 rounds at the speed of 4.5 km hr–1?

Sol. Let r be the radius of the circular park,

Area = r2 = 88704 m2

= × =2 788704 28224

22r

r = 168 mCircumference of the circular field

222 2 168 m 1056 m.

7r= π = × × =

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 × 1000hours

176hours

75=

= 2 hours 20 minutes 48 seconds.

Q.29. In the given figure, OPQR is arhombus whose three vertices P, Q, R lie on acircle of radius 8 cm. Find the area of the shadedregion. [2011 (T-II)]

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

Let OQ and PR intersect at S

Since the diagonals of a rhombus bisect eachother 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 3 cm = 8 3 cm

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 is72 m and the area of the circular swimming tank

inscribed in it.22

.7

Take π =

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

272

5184 5184 12962

= − = − = 3888

⇒ 2 3888r =∴ Area of inscribed circular swimming tank

2 222(3888) m

7r= π = ×

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, thenits 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 4π cm2, then itsradius is :

(a) 1 cm (b) 2 cm(c) 3 cm (d) 4 cm

4. What will be the cost of polishing acircular table-top of radius 2 m at the rate of Rs7 per m2?

(a) Rs 44 (b) Rs 88

(c) Rs 49 (d) Rs 84

5. Find the circumference and area of acircle of radius 8.4 cm.

6. Find the area of a circle whosecircumference is 22 cm.

7. The radius of a wheel is 84 cm. Howmany revolutions will it make to go 52.8 km?

8. The difference between the area of acircle 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 whenwheel makes 1000 revolutions.

10. Sum of radii of two circles is 140 cm andthe difference of their circumferences is88 cm. Find the diameters of the circles.

11. A circular flower bed lies inside arectangular field of size 25 m × 18 m. The areaof the field excluding the flower bed is 296 m2.Find the diameter of the flower bed.

12. A larger wheel of diameter 50 cm isattached to a smaller wheel of diameter30 cm. Find the number of revolutions made bythe smaller wheel, when the larger one makes 15revolutions.

13. From an equilateral triangle of side 24cm, a circle of radius 7 cm is cut off. Find thearea of remaining portion of the triangle.

14. The short and long hand of a clock are4 cm and 6 cm respectively. Find the sum ofdistances travelled by their tips in one day.

12.2 AREAS OF SECTOR AND SEGMENTOF A CIRCLE

1. Length of an arc which subtends an angle

of θ° at the centre =2

360 180

r rπ θ° π θ°=° ° .

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

π θ°°

r.

(ii) Perimeter of sector OACBO

=2

2360

π θ°+°

rr .

3. Minor sector : A sector of a circle iscalled a minor sector if the minor arc of thecircle is a part of its boundary. In the figureabove minor sector is OACBO.

4. Major sector : A sector of a circle iscalled a major sector, if the major arc of thecircle is a part of its boundary. In the abovefigure, OADBO is the major sector.

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5. The sum of the arcs of major and minorsectors of a circle is equal to the circumferenceof the circle.

6. The sum of the areas of major and minorsectors of a circle is equal to the area of thecircle.

7. The area of a sector is given by

A =1

,2

lr where l =180

θ° × π °r

.

8. Angle described by minute hand in60 minutes = 360°.

Angle described by minute hand in one

minute =360

6 .60

° = °

Thus, the minute hand rotates through anangle 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 through1

2° in

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

bounded by an arc and a chord, including the arcand the chord.

11. Minor segment : If the boundary of asegment is a minor arc of a circle, then thecorresponding segment is called a minorsegment. In the figure, segment PQR (the areawhich is shaded) is a minor segment.

12. Major segment : A segmentcorresponding a major arc of a circle is known asthe major segment. In the figure, segment PQSPis a major segment.

13. Area of minor segment PRQS

221

sin360 2

rr

π θ°= − θ

°.

14. Area of major segment PQSP= π r2 – area of minor segment PRQS.


Unless stated otherwise, use22

.7

π =

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

θ× π

°

=2 2 260 22 132

(6) cm cm .360 7 7

°× × =

°

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

2cm

For a quadrant of a circle, = 90°

We know that, area of the sector = 2

360

θ× π

°r

So, area of given quadrant

=

2290 22 7

cm360 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 aclock is 14 cm. Find the area swept by theminute 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

θ× π

°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 cmsubtends a right angle at the centre. Find thearea 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

θ× π

°r

So, area of the minor sector OAB

= 2903.14 10 10 cm

360

°× × ×

°= 78.5 cm2

Area ofOA OB

OAB2

×∆ =

=2 210 10

cm 50 cm2

×=

Area of the minor segment = Area of minorsector OAB – Area of OAB

= 78.5 cm2 – 50 cm2 = 28.5 cm2

(ii) Area of major sector = Area of circle – Areaof 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 arcsubtends 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

θ× π

°r

=60 22

2 21 cm 22 cm.360 7

°× × × =

°(ii) Area of the sector formed by the arc

=2 260 22

21 21 cm360 360 7

θ °× π = × × ×

° °r

= 231 cm2.

(iii) Area of the segment formed by thecorresponding chord

= Area of sector OAB – Area of OAB= 231 cm2 – area of OAB …(i)

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

Draw OM AB

In OMA and OMB,

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 ABand AOM = BOM [CPCT]

AOM = BOM =1

2AOB

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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 3 441 321 cm

2 2 4× × =

Area of segment formed by the correspondingchord = Area of the sector formed by the arc – Areaof OAB

=441 3

2314

−

cm2.

Alternate method :

We know that area of minor segment

=

221

360 2

π θ°−

°r

r sin

= (231 –1

2× 21 × 21 × sin 60°) cm2

=2441 3

231 cm4

−

Q.6. A chord of a circle of radius 15 cmsubtends an angle of 60° at the centre. Find theareas of the corresponding minor and majorsegments 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

θ× π

°r

=2 260

3.14 (15) cm360

°× ×

°= 117.75 cm2 ....(i)

Area of :AOB∆Draw OM AB

In OMA and OMB,

OA = OB [Radii of the same circle]


OM = OM [Common side]

∴ OMA OMB

[RHS congruence criterion]

∴ AM = BM [CPCT]

⇒ AM = BM =1

2AB

and OM = 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 of AOB =1

AB OM2

× ×

= 21 15 315 cm

2 2× × = 2225 3

cm4

=225 1.73

4

×cm2 = 97.3125 cm2 …(ii)

∴ Area of corresponding minor segment of thecircle

= Area of the minor sector – Area of AOB

= (117.75 – 97.3125) cm2 = 20.4375 cm2

…(iii)

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Area of circle = r2

= 3.14 × 15 × 15 cm2 …(iv)Area of the corresponding major segment of the

circle= Area of circle – Area of the corresponding

minor segment of the circle= (3.14 × 15 × 15) cm2 – 20.4375 cm2

= (706.5 – 20.4375) cm2 = 686.0625 cm2.

Q.7. A chord of a circle of radius 12 cmsubtends an angle of 120° at the centre. Find thearea of the corresponding segment of the circle.

(Use = 3.14 and 3 1.73= ) [2011 (T-II)]

Sol. Radius = r = 12 cm, Angle = = 120°

O

12cm

12cm

A BM

∴ Area of the sector = 2

360

θ× π

°r

= 21203.14 12 12 cm

360

°× × ×

°= 150.72 cm2 ....(i)

Area of AOB :

In OMA and OMB,OA = OB [Radii of the same circle]OMA = OMB = 90°OM = OM [Common]

OMA OMB [RHS congruence criterion]

∴ AM = BM [CPCT]

AM = BM =1

2AB

AOM = BOM [CPCT]

AOM = BOM

1 1BOA (120) 60

2 2= ∠ = × ° = °

∴ In OMA,

cos 60° =OM

OA

1 OM

2 12=

OM = 6 cm …(ii)

sin 60° =AM

OA

3 AM

2 12= AM = 6 3 cm

2AM = 12 3 AB = 12 3 cm …(iii)

Area of AOB∆ =1

AB OM2

× ×

= 2112 3 6 cm

2× × = 236 3 cm

= 36 × 1.73 cm2

= 62.28 cm2 …(iv)So, area of the corresponding segment of the

circle= Area of the corresponding sector of the circle

– Area of AOB

= (150.72 – 62.28) cm2 = 88.44 cm2.

Q.8. A horse is tied to a peg at one corner ofa square shaped grass field of side 15 m bymeans of a 5 m long rope (see figure). Find

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

(ii) the increase in the grazing area if the ropewere 10 m long instead of 5 m. (use = 3.14)

Sol. (i) Side of the square = 15 m

Length of peg = Radius of rope R = 5 m

Central angle = 90°

Area of sector =2R

360

π θ°

=23.14 5 5 90

m360

× × × °°

=23.14 25

m4

×

=278.5

m4

= 19.625 m2

Hence, area of the field in which the horse cangraze = 19.625 m2.

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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

π θ × × × °=

° °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 theform of a circle with diameter 35 mm. The wireis also used in making 5 diameters which dividethe circle into 10 equal sectors as shown infigure. 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 = 5Length of 5 diameters = 35 × 5 mm = 175 mmCircumference 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

3610

°= °

∴ Area of each sector of the brooch

=2

360

θ× π

°r =

236 22 35 35mm

360 7 2 2

°× × ×

°

=2385

mm .4

Q.10. An umbrella has 8 ribs which areequally spaced (see figure). Assuming umbrella

to be a flat circle of radius 45 cm, find the areabetween 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

°θ = = °

Area between the two consecutive ribs of theumbrella = area of sector

= 2

360r

θ× π

°= 245 22

45 45 cm360 7

°× × ×

°

=222275

cm .28

Q.11. A car has two wipers which do notoverlap. Each wiper has a blade of length 25 cmsweeping through an angle of 115°. Find thetotal area cleaned at each sweep of the blades.

Sol. Length of blade = r = 25 cm

Sweeping angle = = 115°


360

rπ θ°

Total area cleaned at each sweep of the blades

=22

360r

θ × × π °

= 2 2115 222 (25) cm

360 7

° × × × °

=2158125

cm .126

Q.12. To warn ships for underwater rocks, alighthouse spreads a red coloured light over asector of angle 80° to a distance of 16.5 km.Find the area of the sea over which the ships arewarned. (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

θ× π

°=

2803.14 (16.5)

360

°× ×

°km2

= 189.97 km2.

Q.13. A round table cover has six equaldesigns as shown in figure. If the radius of thecover is 28 cm, find the cost of making thedesigns 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

°θ = = °

Area of minor sector OAB

= 2

360r

θ× π

°=

2 260 2228 cm

360 7

= 21232cm

3= 410.67 cm2 ....(i)

Area of AOB :

Draw OM AB

In OMA and OMB,OA = OB [Radii of the same circle]OM = OM [Common]


∴ OMA OMB∆ ≅ ∆ [RHS congruence criterion]

∴ AM = BM [CPCT]

⇒ AM = BM =1

AB2

[M is mid-point of AB]

AOM BOM∠ = ∠ [CPCT]

⇒ 1AOM BOM AOB

2∠ = ∠ = ∠

1(60 ) 30

2= ° = °

In OMA,∆ OMcos30

OA° =

3 OM

2 28= OM = 14 3 cm

sin 30° =AM

OA

1 AM

2 28= AM = 14 cm

2AM = 28 cm AB = 28 cm

∴ Area of1

AOB AB OM2

∆ = × ×

= 2128 14 3 cm

2× × = 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 cm2

= 464.82 cm2

Cost of making the designs at the rate ofRs. 0.35 per cm2

= Rs 464.82 × 0.35 = Rs. 162.68.

Q.14. Tick the correct answer in thefollowing :

Area of a sector angle p (in degrees) of acircle with radius R is

(a) 2180

× πp

R (b)2

180× π

pR

(c) 2360

× πp

R (d)22

720× π

pR

Sol. (d) Angle of sector = θ = p

Radius of circle = R

Area of sector =2R

360

θ× π =

2R360

p× π

=2 22 R 2 R

2(360) 720

p p× π = × π

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

π =

[2011 (T-II)](a) 3.5 cm2 (b) 3.5 cm(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 cm8

= =

Q.2. The minute hand of a clock is 21 cmlong. The distance moved by the tip of the minutehand 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 hour360

2 21360

°= × π ×

°= 42 cm

Q.3. The angle through which the minutehand 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 aclock is 6 cm. The area swept by minute hand in10 minutes is :

(a) 212 cmπ (b) 236 cmπ(c) 29 cmπ (d) 26 cmπ

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

660

°= °

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

Area of the sector =2

360

θ× π

°r

=2 2 260

6 cm 6 cm360

°× π × = π

°

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) 3 cm2 (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

°× π

°

= 2122 cm 3.66

6× = cm2

Q.6. If an arc subtends an angle of 45° at thecentre of the circle of radius a cm, then length ofthe arc is :

(a)6

πa cm (b)

3

πa cm

(c)4

πa cm (d)

2

πa cm

Sol. (c) Length of the arc = 2360

θ× π

°r

=45

2360

°× × π ×

°a cm =

4

πa cm

Q.7. The area of the sector which subtends anangle of 60° at the centre of a circle is4.4 cm2, then area of the circle is :

(a) 36 cm2 (b) 26.4 cm2

(c) 36.6 cm2 (d) 30 cm2

Sol. (b) Area of the sector =2

360

θ× π

°r

⇒ 4.4 =260

360× πr ⇒ 2rπ = 4.4 × 6

⇒ 2rπ = 26.4

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

2

3

πcm at the centre of the circle of radius 4 cm is :

(a) 30° (b) 45°(c) 60° (d) 90°

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Sol. (a) Length of arc = 2360

θ× π

°r

2

2 43 360

π θ= × π ×

° = 30°

Q.9. If a pendulum swings through an angleof 60° 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 = 2360

θ× π

°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 is3

97

cm2. The angle

subtended by the sector at the centre of the circleis :

(a) 60° (b) 120°(c) 45° (d) 90°

Sol. (b) Area of the sector =23

360

θ× π ×

°cm2

66 22

97 360 7

θ= × ×

°

=3 360

9

× °= 120°

Q.11. The perimeter of the sector of a circlewhose central angle is 45° and radius7 cm is :

(a) 39 cm (b) 19.5 cm(c) 35 cm (d) 17.5 cm

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

2 7360

°× π ×

°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 of7

60π

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

(a) 7 cm (b) 7 cm

(c) 14 cm (d)21

2cm

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

=2

360

θ× πr

7

60π = 26

360× πr [ for 1 minute is 6°]

r2 = 7 cm ⇒ r = 7 cm

Q.13. The length of the minute hand of aclock is 7 cm. Find the area swept by the minutehand 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 cm

360 7

= × × =

.

Q.14. The minute hand of a clock is 21 cm

long. Find the area swept by the minute hand onthe 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 21 cm

( )22 230 22

21 cm 5.5 cm .360 7

= × × =

Q.15. The perimeter of a sector of a circle ofradius 5.6 cm is 27.2 cm. Find the area of thesector. [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 cm5.6 + 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 is5

18of the

area of the circle. Find x. [2009]

Sol. Let the radius of the circle be r.

Area of the circle =2rπ

And, area of the sector OAPB =2

360

π°

r x

But,25

18πr =

2

360

π°

r x⇒ x =

360 5

18

° ×= 100°.

Q.17. A chord of a circle of radius 14 cmsubtends a right angle at the centre. What is thearea of the minor sector? [2008C]

O

AB

14cm

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

Area of minor sector =290

360

°× π

°r

=1 22

14 144 7

× × × cm2 = 154 cm2.

Q.18. What is the perimeter of a sector ofangle 45° of a circle with radius 7 cm?

22Use =

7

π [2008C]

Sol. The arc AB of length l of a sector of angle45° in a circle of radius 7 cm is given by

l = 2360

rθ°

× π°

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

Sol. BO = OD, r = 60 cm

= 360° – BOD∠ = 360° – 90° = 270°

(i) Area (table top) =2

360

θ× π

°r

=270

3.14 60 60360

°× × ×

°cm2

= 45 × 60 ×314

100cm2

= 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 cardboardwith radius 1.26 m, a sector with central angle40° 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

AB

40°

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

=2

360

θ× π

°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 and 3.5 cm are given. Find

the area of shaded region.22

Use =7

π [2011 (T-II)]

Sol. Let A1 and A2 be the areas of sectors OABand OCD respectively. Then,

A1 = area of a sector of angle 30° in a circle ofradius 7 cm.

A1

2 230 227 cm

360 7

= × ×

2Using : A =360

rθ × π

A1 =277

cm6

A2 = Area of a sector of angle 30° in acircle of radius 3.5 cm

A22 230 22

(3.5) cm360 7

= × ×

A22 21 22 7 7 77

cm cm12 7 2 2 24

= × × × =

Area of the shaded region = A1 – A2

277 77cm

6 24

= −

2 277 77(4 1) cm cm

24 8= × − =

= 9.625 cm2.

Q.22. The area of a sector is1

10that of the

complete circle. Find the angle of the sector ofthe 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

π θ°

r

According to the question,2

360

π θ°

r= 21

10πr

1

360 10

θ=

° =

360

10

° = 36°.

Q.23. The minute hand of a clock is 10 cmlong. Find the area of face of the clock describedby the minute hand between 9 a.m. and 9.35 a.m.

Sol. Angle described by the minute hand in oneminute = 6°

So angle described by the minute hand in35 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

10360 7

° × × ° cm2 = 183.3 cm2.

Q.24. A circular disc of 6 cm radius isdivided into 3 sectors with central angles 120°,150°, 90°. Find the ratio of the areas of threesectors.

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12.3 AREAS OF COMBINATIONS OFPLANE FIGURES

In our daily life we come acrosscombinations of plane figures for making

interesting designs such as flower beds, fabricdesigns, window designs, designs on table coversetc. In this section we will study the process ofcalculating 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 thecorresponding minor sector of a circle is equalto :

(a) area of the circle

(b)1

2area of the circle

(c)1

4area of the circle

(d)3

4area of the circle

2. The radius of a circle is 5 cm. The area ofthe sector formed by an arc of this circle oflength 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 circleof 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 thecentral angle of a semicircle?

(a) 0° (b) 90°

(c) 180° (d) 360°

5. Find the area of a sector of a circle withradius 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 majorsegment.

7. A chord of a circle of radius 28 cmsubtends 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 withcentral angle 90° is 25 cm. Find the area of theminor segment of the circle.

9. Find the area of shaded portions of thefollowing figures with given measurements :

(a) (b)

10. In a circle of radius 6 cm, a chord of10 cm makes an angle of 110° at the centre of thecircle. 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

π θ°

r

∴ Ratio of the areas of the three sectors

=2 2 2120 150 90

: :360 360 360

π × π × π ×r r r

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

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Hence, the area of the shaded region

is24523

cm .28

Q.2. Find the area of the shaded region inthe given figure, if radii of the two concentriccircles with centre O are 7 cm and 14 cmrespectively and AOC = 40°. [Imp.]

O

AB

D

C

40°

Sol. We know that, area of sector =2

360

π θ°

r

Area of the shaded region= Area of the sector OAC

– Area of the sector OBD

=2 2 240 40

(14) cm (7)360 360

° °π − π

° °cm2

=2 2 240

{(14) (7) } cm360

°π −

°

=2 21 22 154

(14 7)(14 7) cm cm .9 7 3

× − + =

Q.3. Find the area of the shaded region inthe given figure, if ABCD is a square of side14 cm and APD and BPC are semicircles. [Imp.]

A B

CD

P

Sol. Side of square = 14 cmDiameter of each semi-circle = 14 cm

Radius =14

cm 7 cm2

=

Area of square = (side)2 = 14 × 14 cm2

= 196 cm2 …(i)

Unless stated otherwise, use22

7π = .

Q.1. Find the area of the shaded region infigure, if PQ = 24 cm, PR = 7 cm and O is thecentre of the circle.

O

Q

PR

Sol. PQ = 24 cm

RPQ = 90° [Angle in a semi-circle is 90°]

PR = 7 cm

In QPR,∆RQ2 = PR2 + PQ2 [By Pythagoras Theorem]

= (7)2 + (24)2 = 49 + 576 = 625

RQ = 625 = 25 cm

Diameter of the circle = RQ = 25 cm

Radius of the circle (r) =25

2cm

Area of the semi-circle =21

2× πr cm2

=

21 25

2 2

π cm2

=2 2625 625 22

cm cm8 8 7

π = × =26875

cm28

…(i)

Area of1

RPQ PQ PR2

∆ = × ×

2124 7 cm

2= × × = 84 cm2

∴ Area of the shaded region= Area of the semi-circle

– Area of right triangle RPQ

=2 26875 6875 2352

84 cm cm28 28

− − =

=24523

cm28

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Area of semi-circle APD =21

2πr

2 21 227 7 cm 77 cm

2 7= × × × = …(ii)

Area of semi-circle BPC

= 2 21 1 227 7 cm

2 2 7π = × × ×r

= 77 cm2

Area of the shaded region = Area of the squareABCD – (Area of semi-circle APD + Area of semi-circle BPC)

= (196 – 154) cm2 = 42 cm2.

Q.4. Find the area of the shaded region inthe given figure, where a circular arc of radius 6cm has been drawn with vertex O of anequilateral triangle OAB of side 12 cm as centre.

[2011 (T-II)]

Sol. Radius of circle = r = 6 cmSide of equilateral triangle = 12 cmArea of shaded region= Area of circle + Area of equilateral triangle

OAB – Sectorial area common to the circle and thetriangle

= 2 2 2 2 2 23 60(6) cm (12) cm (6) cm

4 360

°π + − π

°= 2 2 236 cm 36 3 cm 6 cmπ + − π= 2 230 cm 36 3 cmπ +

2 22230 cm 36 3 cm

7= × +

=2660

36 3 cm7

+ .

Q.5. From each corner of a square of side4 cm a quadrant of a circle of radius 1 cm is cutand also a circle of diameter 2 cm is cut asshown in figure. Find the area of the remainingportion 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 + Areaof a circle]

= (4 × 4) cm2 –

2290

4 1360

° 2 × π( ) + π ° 2 cm2

= (16 – 2) cm2 =22

16 – 27

× cm2

=44

16 –7

cm2 =

112 – 44

7cm2 =

68

7cm2 .

Q.6. In a circular table cover of radius32 cm, a design is formed leaving an equilateraltriangle ABC in the middle as shown in figure.Find the area of the design (shaded region).

[Imp.]

Sol. Radius of table cover = 32 cmArea of equilateral ABC

2 23 3(side)

4 4a= = [a = side]

Area of the design (shaded region)= Area of the circular table cover

– Area of the equilateral triangle ABC

= 2 23(32)

4aπ − …(i)

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Let h be the height of ABC. Since the centre ofthe circle coincides with the centroid of theequilateral 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 =2

2

4

ah +

2

2

4

aa − =

22 23

4

ah h⇒ =

2

2 4

3

ha =

24(48)

3= = 3072

From equation (i),

∴ Required area =2 2 3

(32) cm4

π − 3072 cm2

=2 222

(1024)cm 768 3 cm7

−

=222528

768 3 cm7

− .

Q.7. In figure, ABCD is a square of side14 cm. With centres A, B, C and D, four circlesare drawn such that each circle touchesexternally 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

rπ θ°

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

=2 2 290

14 14 cm 4 7 cm360

°× − × π ×

°

=2 2 222

196 cm (7) cm7

−

= (196 – 154) cm2 = 42 cm2.

Q.8. Figure below depicts a racing trackwhose left and right ends are semicircular. Thedistance 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 itsinner edge

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

Sol. (i) Length of each parallel line segment= 106 m

Distance between two inner parallel linesegments = 60 m

Width of track = 10 mThe distance around the track along its inner edge

= 106 m + 106 m + 260

m2

π

= (212 + 60 π ) 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 1

2 (30 10) (30) m2 2

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

=2 222

2120m 700m7

+ ×

= (2120 + 2200) m2 = 4320 m2.

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Q.9. In figure, AB and CD are two diametersof a circle (with centre O) perpendicular to eachother and OD is the diameter of the smallercircle. If OA = 7 cm, find the area of the shadedregion. [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 1

cm (7) cm 14 7cm2 2 2

π + π − × ×

=2 2 249 49

cm cm 49cm4 2

ππ × + −

=249 3 22

cm4 7

× ××

– 49 cm2

=249 66 49 28

cm28

× − ×=

23234 1372cm

28

−

=21862

cm28

= 66.5 cm2.

Q.10. The area of an equilateral triangleABC is 17320.5 cm2. With each vertex of thetriangle as centre, a circle is drawn with radiusequal to half the length of the side of the triangle(see figure). Find the area of the shaded region.

(Use π = 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 equilateraltriangle ABC be a cm. Then,

Area of triangle = 2 23cm

4a

As per condition, 2317320.5

4a =

a2 =17320.5 4

3

×= 10000 × 4 = 40000

a = 40000 = 200 cm

Area of each sector

=

22 260 200

cm 10000 cm360 2 6

° π π = × °

Area of each shaded region = Area of theequilateral triangle ABC – 3 × (Area of each sector)

=2 217320.5 cm (10000) cm

2

π−

= 17320.5 cm2 – 3.14 × 5000 cm2

= (17320.5 – 15700) cm2 = 1620.5 cm2.

Q.11. On a square handkerchief, ninecircular designs each of radius 7 cm are made(see figure). Find the area of the remainingportion of the handkerchief.

Sol. Radius of each circle = 7 cmDiameter of each circle = 14 cmSide of square = 42 cm

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Area of the remaining portion of the handkerchief

= Area of the square ABCD – Area of ninecircular designs

= (42 × 42) cm2 – 2 29 (7) cmπ

=2 2 222

1764 cm 9 (7) cm7

− × ×

= (1764 – 1386) cm2 = 378 cm2.

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


360

rπ θ°

(i) Area of the quadrant OACB

=2 290

(3.5) cm360

°π

°

=21 22 35 35

cm4 7 10 10

× × × =77

8cm2.

(ii) Area of the shaded region = Area of thequadrant OACB – Area of OBD

=2 2 277 OB OD 77 3.5 2

cm cm cm8 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

O A P

B

Sol. In AOB,

OB = 2 2OA AB+[Using Pythagoras Theorem]

= 2 2OA + OA

= 2OA 2(20)= cm = 20 2 cm

Area of the shaded region = Area of the quadrantOPBQ – Area of the square OABC

=2 2 290

(20 2) cm 20 20 cm360

°π − ×

°= 200 cm2 – 400 cm2

= 200 × 3.14 cm2 – 400 cm2

= 628 cm2 – 400 cm2 = 228 cm2.

Q.14. AB and CD are respectively arcs oftwo concentric circles of radii 21 cm and 7 cmand centre O (see figure). If AOB = 30°, findthe area of the shaded region. [2011, (T-II)]

A B

C D

O

30°7cm

21cm

Sol. Radius of sector OBA = r1 = 21 cmRadius of sector ODC = r2 = 7 cmArea of the shaded region

= Area of the sector OAB

– Area of the sector OCD

=2 2

1 2

360 360

r rπ θ π θ−

° °

=2 2 2 230 30

(21) cm (7) cm360 360

° °π − π

° °

=2 21 22 1 22

21 21cm 7 7cm12 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 acircle of radius 14 cm and a semicircle is drawnwith BC as diameter. Find the area of the shadedregion. [2008, 2011 (T-II)]

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B

AC

Sol. In BAC, using Pythagoras Theorem

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

BC = 14 2 cm Radius of circle = 14 cm

Area of1

BAC AB AC2

∆ = × ×

=21

14 14 cm2

× × = 98 cm2 …(i)

Area of sector ABC =2

360

rπ θ°

=22 14 14 90

7 360

× ××

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

=14 2

cm2

= 7 2 cm

Area of the semi-circle

=21

2rπ =

21 227 2 7 2 cm

2 7× × ×

= 154 cm2

Required area = Area of semi-circle with BC asdiameter – [Area of sector ABC – Area of BAC]

= 154 cm2 – [154 – 98] cm2 = 98 cm2.

Q.16. Calculate the area of the designedregion 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 andradius 8 cm

=2 22 2 90

360 360

r rπ θ π × °=

° °=

2 22 8 8 90

7 360

× × × ××

cm2

=2704

cm7

Since the designed area is common to both thesectors.

Therefore, area of design = Area of both sectors– Area of square

= 2 2704 25664 cm cm

7 7

− = = 36.57 cm2.

Q.1. In the given figure, if O is the centre ofthe circle AB CD and OB = 3 cm, then the areaof the shaded portion is :

O

C

A B

D

(a)246

7cm (b)

236

7cm

(c)226

7cm (d) none of these

Sol. (b) Area of the shaded portion

=2 2 21 1

3 cm 6 3 cm2 2

π × − × ×

=2 2 211 4 36

9 9 cm 9 cm cm7 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 theremaining position is :

A

B C

D

14 cm

(a) 24 cm2 (b) 48 cm2

(c) 42 cm2 (d) 56 cm2

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

− π × = − × ×

= (196 – 154) cm2 = 42 cm2.

Q.3. In the given figure, area of the portionACBDA is :

C

A

B

O7 cm

D

2 cm

(a) 35.5 cm2 (b) 31.5 cm2

(c) 25.5 cm2 (d) 21.5 cm2

Sol. (b) Area of the portion ACBDA

= Area of sector – Area of AOD∆

=290 1

7 7 7 2 cm360 2

° × π × × − × × °

=2 263

cm 31.5 cm2

=

Q.4. In the given figure, 4 quadrants each ofradius 1 cm are cut from the square of side 4 cm.The area of remaining portion is :

(a) 12 cm2 (b) 13 cm2

(c) 13.4 cm2 (d) 12.8 cm2

1 cm 1 cm

1 cm1 cm

1 cm

1 cm

1 cm

1 cm4 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 oftwo semi-circles and a circle of radius2 m as shown below, then total area of the flowerbed is :

14m

2m

20m

(a) 51m2 (b) 49 m2

(c) 55m2 (d) 53m2

Sol. (d) Total area of the flower bed

= 22 2 2 21

7 m 2 m2

π × + π ×

= ( ) 2 249 4 m 53 mπ + π = π .

Q.6. Four cows are tethered at four cornersof a square field of side 12 m. If each cow cangraze the maximum area, then the total areagrazed by them is :

(a) 6m2 (b) 24m2

(c) 36m2 (d) 30m2

Sol. (c) Total area grazed by 4 cows= 4(Area of quadrant of radius 6 m)

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=1 24 64

× π × = 236 mπ

Q.7. In the given figure, SR is the diameterof length 12 cm and SP = PQ = QR, then thearea of the shaded region is :

S RP

Q

(a) 25 cm2 (b) 20 cm2

(c) 18 cm2 (d) 24 cm2

Sol. (b) Area of the shaded region

= Area of the circle – 2 (Area of semi-circleof diameter 8 cm)

= (2 21

6 2 42

π × − × × π × ) cm2

=2 2(6 4)(6 4) cm 20 cm

Q.8. The area of the shaded portion in thefollowing figure is :

(a) 13 cm2 (b) 7 cm2

(c) 16 cm2 (d) 25 cm2

8 cm

6cm

Sol. (d) Area of the shaded portion= 2(Area of semi-circle of diameter 8 cm)

+ 2(Area of semi-circle of diameter 6 cm)

=

2 22 21 8 1 6

2 cm 2 cm2 2 2 2

× π + × π

=2 2(16 9 ) cm 25 cmπ + π = π

Q.9. In the figure, PQ = 24 cm, PR = 7 cmand O is the centre of the circle. Find the area of

shaded region (take 3.14π = )

[2009, 2008 C, 2011(T-II)]

Sol. QPR 90∠ = ° [Angle in a semi-circle]

QR2 = PR2 + PQ2 [Pythagoras theorem]= 49 + 576 = 625

QR = 25 cm

Radius of the circle, r =25

cm2

= 12.5 cm.Area of the shaded region = Area of the semi-

circle – area of the triangle

=2 1

PQ RP2 2

r

=21

[3.14 12.5 12.5 24 7] cm2

× × − ×

=21

[490.625 168] cm2

−

= 21322.625 cm

2× = 161.31 cm2.

Q.10. In the figure, find the perimeter ofshaded region where ADC, AEB and BFC aresemi-circles on diameters AC, AB and BCrespectively. [2008]

Sol. AC = (2.8 + 1.4) cm = 4.2 cm

Perimeter of the shaded region

= Length of the semi-circle ADC + length of thesemi-circle BFC + length of the semi-circle AEB.

=AC BC AB

2 2 2

=4.2 1.4 2.8

cm2 2 2

π + +

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=22

[2.1 0.7 1.4] cm7

+ +

224.2 cm 13.2 cm

7= × = .

Q.11. Find the area of the shaded region inthe figure, where ABCD is a square ofside 14 cm. [2008, 2011(T-II)]

Sol. Diameter of each circle =14

2cm = 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) × × π × cm2

=22

196 – 4 3.5 3.57

× × × cm2

= (196 –154) cm2 = 42 cm2.

Q.12. In the figure, ABC is a right-angledtriangle right-angled at A. Semi-circles aredrawn on AB, AC and BC as diameters. Find thearea 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

π =

25

2 2

π sq units

=π25

8sq units

Area of the semi-circle with AC as diameter

=

2AC

2 2

π

222

π= × sq units = 2 sq units

Area of the semi-circle with AB as diameter

2AB

2 2

π = =

23

2 2

sq units

=π9

8sq units

Area of ABC =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 withdiameter AC – [area of the semi-circle with diameterBC – area of ABC]

=9 25

2 68 8

π π + π − − sq units

=9 16 25

68

π + π − π + 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 thelength of the side of the triangle. Find the areaof triangle not included in the circle.

[Take 3 = 1.73] [2009]

60°

60° 60°

Sol. Let each side of the equilateral triangle bex cm.

Then, 2349 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 angle60°

=2

27 603 cm

360

π × × ° °

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

[2011 (T-II)]

Sol. In ∆ ABC, BC2 = AB2 + AC2

AC 100 36 8⇒ = − = cm

OP AB and OQ AC

[Radius through the point of contact isperpendicular 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 = CQ4 + r = AC – AQ [From (i)]

4 + r = 8 – r

2r = 4 ⇒ r = 2 cm

Area of the shaded region

= Area of ABC – 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 threesemi-circles, A, B and C having diameters 3 cmeach, and another semi-circle E having a circleD with diameter 4.5 cm are shown. Calculate :

(i) the area of the shaded region.(ii) the cost of painting the shaded region at

the rate of 25 paise per cm2, to the nearest rupee.[HOTS]

Sol. (i) Area of the shaded region (i.e., area E, Band F) = Area of semi-circle with radius 4.5 cm –(area of semi-circle A + area of semi-circle C) + (areaof semi-circle B) – (area of circle with diameter4.5 cm).

= π × × 2 21(4.5) cm

2

− × π × + × π × 2 2 21 1

(1.5) (1.5) cm2 2

+ × π × − π

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

2

( )π π= π + 22 2 2 2 2(4.5) cm – (1.5) cm 1.5 cm

2 2

− π 2 2(2.25) cm

= π − −

2 22(4.5) (1.5)

(2.25)2 2

cm2

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= π × − × − 21 1

20.25 2.25 5.0625 cm2 2

= π − −[10.125 1.125 5.0625] cm2

= π × 23.9375 cm

= × =2 2223.94 cm 12.375 cm

7.

(ii) The cost of painting the shaded region at the

rate of 25 paise per cm2 =×12.375 25

Rs100

= Rs 3.

Q.16. A child draws the figure of anaeroplane as shown. Here, the wings ABCD andFGHI are parallelograms, the tail DEF is anisosceles triangle, the cockpit CKI is a semi-circle and CDFI is a square. In the given figure,

BP ⊥ CD, HQ ⊥ FI and EL ⊥ DF. If CD = 8cm, BP = HQ = 4 cm and DE = EF = 5 cm, findthe area of the whole figure. Take = 3.14.

[HOTS]

Sol. Area of the given figure = (Area ofparallelogram ABCD + Area of parallelogram FGHI +Area of square DFIC + area of semi-circle CKI + areaof isosceles DEF)

= (DC × PB) + FI × HQ + (CD)2 +1

2× × (4)2

+ × × − 2 21

DF 4 DE DF4

= × + × + + × ×2 1 22

(8 4) (8 4) (8) 162 7

+ × × × − 2 21

8 4 (5) 84

cm2

= [64 + 64 + + × −176

2 100 647

] cm2

= [128 + 25.14 + 2 × 36 ] cm2

= [128 + 25.14 + 12] cm2 = 165.14 cm2.

Q.17. PQRS is a diameter of a circle ofradius 6 cm. The lengths PQ, QR and RS areequal. Semi-circles are drawn on PQ and QS asdiameters as shown in figure. Find the perimeterand area of the shaded portion.

Sol.

Diameter PS = 12 cm

PQ = QR = RS =12

3cm = 4 cm

QS = QR + RS = (4 + 4) cm = 8 cm(i) Required perimeter = (length of semi-circle

of radius 6 cm + length of semi-circle of radius4 cm + length of semi-circle of radius 2 cm)

= ( π × 6 + π × 4 + π × 2) cm = 12 π cm

(ii) Required area = (Area of semi-circle withdiameter PS + Area of semi-circle with diameter PQ– Area of semi-circle with diameter QS)

= × π × + × π × − × π ×

2 2 21 12 1 4 1 8

2 2 2 2 2 2

= × π × + × π × − × π ×

2 2 221 12 1 4 1 8

cm2 2 2 2 2 2

= π + −2 2 21(6 2 4 )

2cm2

= π = × × =2 2 21 1 22 264(24) cm 24 cm cm

2 2 7 7= 37.71 cm2.

Q.18. ABCD is a flower bed. If OA = 21 mand OC = 14 m, find the area of the bed.

22.

7Take

π =

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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.

π= π − π = −2 2 2 21 1

R (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

= × π = × ×2 21 1 22(14)

2 2 7r cm2 = 2308 cm

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

∴ Area of 2 smaller semi-circles = × π

212

2r

= × × × = 2 2 21 22

2 (7) cm 154 cm2 7

The area of the shaded region

= (308 + 154) cm2 = 462 cm2.

Q.20. In an equilateral triangle of side24 cm, 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 cm

4…(i)

Let r be the radius of inscribed circle, then

Area of AFO

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 cm

7 7

= 150.85 cm2

Required shaded area

= Area of ABC – 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 arectangle, with two semi-circles on its smallersides as diameters, added to its outside. If thesides of the rectangle are 36 m and 24.5 m, find

the area of the play ground.22

.7

π = Take

[HOTS]

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Sol. Length of the rectangle ABCD = 36 mBreadth 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)

= × π × = × π ×2 2 21 1(12.25) m

2 2r

= × × 2 21 22(12.25) m

2 7

= 235.8125 m2

Area of semicircle (II) = Area of semicircle (I)= 235.8125 m2

Area of playground = Area of semi-circleI + 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 mlong ropes at the three corners of a triangularfield having sides 20 m, 34 m and 42 m. Find thearea of the plot which can be grazed by thehorses. Also, find the area of the plot whichremains ungrazed. [HOTS]

Sol. Let A = 1, B = 2 and C = 3The area which can be grazed by three horses =

(Area of sector with central angle 1 and radius 7 cm+ Area of sector with central angle 2 and radius 7 cm

+ Area of sector with central angle 3θ and radius

7 cm.)

π θπ θ π θ= + +

° ° °

22 231 2

360 360 360

rr r

π= θ + θ + θ

°

2

1 2 3( )360

r π= × °

°

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 of ABC

= − − −( ) ( ) ( )s s a s b s c

= × × × =2 248 28 14 6 m 336 m

Hence, area grazed by the horses = 77 m2 andungrazed area = (336 – 77) m2 = 259 m2.

PRACTICE EXERCISE 12.3 A

1. A park is in the form of a rectangle120 m × 100 m. In the centre of the park, thereis a circular lawn as shown in the figure below.The area of the park excluding the lawn is 8700m2. Find the radius of the circular lawn.

22Use

7

π =

lawn

120 m

100 m

2. Find the area of the shaded region in the

34

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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 bedshave been shown on two sides of a square lawnABCD of side 56 m. If the centre of eachcircular flower bed is the point of intersection Oof the diagonals of the square lawn, find the sumof the areas of the lawn and the flower beds.

[2011 (T-II)]

5. A horse is placed for grazing inside arectangular field 70 m by 52 m. It is tethered toone corner by a rope 21 m long. On how much

area can it graze? How much area is leftungrazed?

70 m

52 m21 m

21 m

6. The area of a circle inscribed in anequi l ateral tri angle i s 154 cm2. Find the

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 cornersof a square field of side 50 m such that each cangraze the maximum unshared area. What area

will be left ungrazed? (Take 3.14π = )

A

C

B

D

50 m

8. Find the area of the region ABCDEFAshown in the given figure below, given thatABDE is a square of side 10 cm, BCD is asemi-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. Mark POQ = (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 C2 with 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 thefollowing table.

Circle Angle of the No. of equal Area of one360

θ°

× r2

sector = sectors in the circle sector

C1 45° 81

8× r2

45

360

°× r2 =

1

8r2

C2 60° 61

6× r2

60

360

°°× r2 =

1

6r2

C3 90° 41

4× r2

90

360

°°× r2 =

1

4r2

C4 120° 31

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

θ°

× 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 cmπ


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

Question Bank In Mathematics Class X (Term–II)

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