exercise – 3...an umbrella has eight ribs which are equally spaced. assuming the umbrella to be a...

NCERT Textual Exercise (Solved)

70

Area Related to Circles

EXERCISE – 3.1 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.

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.

3. The following figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 min when the car is travelling at a speed of 66 km/h?

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) π units (C) 4 units (D) 7 units

Test Yourself – AR 1

1. Find the area of a circle whose circumference is 22 cm (π = 227

).

2. If the diameter of a semicircular protractor is 14 cm, find its perimeter.

3. The wheels of a car are of diameter 70 cm each. How many complete revolutions do each wheel make in 1 min when the car is travelling at a speed of 52.8 km/h?


71


4. The area of a circular field is 3218 27

m2. Find the cost of fencing the field

at the rate of `14 per metre (π = 227

)

5. In the adjoining figure, grass is planted on a circular strip of width 4 m. The radius of the inner edge of the grass strip is 12 m. If the cost of planting the grass is `3 per m2, then find the total cost of planting the grass on the whole

circular strip (π = 227

).

6. A car travels 1 km distance in which each wheel makes 450 complete revolutions. Find the radius of its wheels.

7. If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.

8. Two circles touch externally. The sum of their areas is 130 π cm2 and the distance between their centres is 14 cm. Find the radii of the circles.

9. The area of a circular road is 22176 sq. m. If the diameter of outer circle is

280 m, find the width of the road (π = 227

).

10. A man runs with the speed 15.84 km/h. If he completes 12 rounds around a

circular playground in an hour, find the area of the ground (π = 227

).

EXERCISE – 3.2

Unless stated otherwise, use π = 227

.

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

2. Find the area of a quadrant of a circle whose circumference is 22 cm. 3. The length of the minute hand of a clock is 14 cm. Find the area swept by

the minute hand in 5 min.


72


4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding (use π = 3.14).

(i) minor segment

A BX

O

10cm

(ii) major sector.

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 corresponding chord.

XA B

O 21 cm60°

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle (use π = 3.14 and 3 = 1.73 ).

XB C

A

60°15 cm

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle (use π = 3.14 and

3 = 1.73).

A B

X

O

12 cm120°

8. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope. Find the following:


73


(i) the area of that part of the field in which the horse can graze (ii) the increase in the grazing area if the rope were 10 m long instead

of 5 m (use π = 3.14).

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find the following:

(i) the total length of the silver wire required (ii) the area of each sector of the brooch

10. An umbrella has eight ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.


74


11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

12. To warn ships for underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned (use π = 3.14).

13. A round table cover has six equal designs. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ` 0.35 per cm2 (use

3 = 1.7).

BX

A

F

E

D

C28 cm

60°

O

14. Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is

(A) p

180 × 2πR (B)

p180

× πR2


75


(C) p

3602 R (D) p

7202 2 R

Test Yourself – AR 2 1. A sector is cut from a circle of radius 21 cm. The angle of the sector is 150°.

Find the length of its arc and area. 2. A pendulum swings through an angle of 30° and describes an arc of 17.6

cm in length. Find the length of the pendulum. 3. A minute hand of a clock is 14 cm long. Find the area on the face of the

clock described by minute hand between 9 p.m. and 9.45 p.m. 4. A chord 10 cm long is drawn in a circle of radius 50 cm. Find the following: (i) the area of the circle (ii) the area of the sector OPQ (iii) the area of the minor segment PXQ (iv) the area of the major segment PYQ 5. The length of the minute hand of a clock is 14 cm.

Find the area swept by the minute hand in (i) one minute and (ii) one hour.

6. A horse is tied to a corner of field which is in the shape of an equilateral triangle. If the length of the rope by which it is tied be 7 m, find the area of the field over which it can graze.

7. A sector of 57° cut out from a circle has an area 5 518

cm². Find the radius of the circle.

8. A chord AB of a circle of radius 15 cm makes an angle of 60° at its centre. Find the difference between the areas of major and minor segments.

A

E

B DC

9. In the adjoining figure, PR and QS are diameters of the circle. If PR = 28 cm and PS = 14 3 , find the following:

(i) Area of ∆OPS (ii) Total area of shaded segment


76


P

Q

R

S

O

120°

10. In the adjoining figure, O is the centre and seg PQ is the chord of the circle. If OP = 12 cm and chord PQ is at a distance of 6 cm from the centre, find the area of the shaded region (π = 3.14, 3 = 1.73).

P

R

Q

O

12 6

M

EXERCISE – 3.3 1. Find the area of the shaded region, if PQ = 24 cm, PR = 7 cm and O is the

centre of the circle.

2. Find the area of the shaded region, if the radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.


77


3. Find the area of the shaded region in adjacent figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

A B

D C

P

14 cm 4. Find the area of the shaded region, where a circular arc of radius 6 cm has

been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

A B

C D

12 cm

X6 cm

O

5. From each corner of a square of side 4 cm, a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut in the middle. Find the area of the remaining portion of the square.

A B

D C4 cm

1cm

2cm

6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the design (shaded region).


78


7. In the given 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.

A B

D C

7 cm 7 cm

8. The following figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find the following:

(i) the distance around the track along its inner edge (ii) the area of the track

9. In the following figure, AB and CD are two diameters of the circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.


79


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. Find the area of the shaded region (use π = 3.14 and 3 = 1.73205).

A

B C

11. On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.

12. In the adjoining 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 following:

(i) Quadrant OACB (ii) Shaded region

D

A

O

2 cm

3.5 cmB

C

13. In the adjoining figure, square OABC is inscribed in quadrant OPBQ. If OA = 20 cm, find the area of the shaded region (use π = 3.14).

Q

C

O A P

B

20 cm


80


14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ∠AOB = 30°, find the area of the shaded region.

721

Q

C

A

D

XB

30°

15. 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.

B

14 cm

A C14 cm

PQ

16. Calculate the area of the designed region in the adjoining figure, common between the two quadrants of circles of radius 8 cm each.

Test Yourself – AR 3 1. A memento is made as shown in the figure. Its base PBCR is silver plated

from the front side at a rate of `20 per cm2. Find the total cost of the silver plating.

O

B C

RP3 cm

7 cm

3 cm

7 cm


81


2. In the adjoining 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 flower beds. Also, find the total area of the

lawn and the flower beds (use π = 227

).

A B

C D

O56 m

3. ABCD is a square. Region I is a semicircle on CD as diameter, region II and III are quadrants of circles with centres A and B respectively, each having

radius 7 cm (π = 227

). Find the area of the shaded region.

A B

P R

D CI

14 cm

II III

7 cm 7 cm

4. O is the centre and AOC is the diameter of the circle. Find the following: (i) The sum of the areas of the two designed segments made by the

chords AB and BC (ii) The area of the designed segment made by the chord PQ (iii) Total cost of making designs at a rate of `4.20 per cm2 (take π =

3.14).


82


4 cm

4 cm

P

Q

B

A

C

O

5. ABCD is a square of 10 cm and semicircles are drawn with each side of the square as diameter. Find the area of the shaded region (π = 3.14).

6. Find the area of the shaded region of the equilateral ∆ABC (take π = 3.14 and 3 = 1.73).

A2 260°

B C6

7. Find the area of the shaded region where ABCD is square of side 14 cm.A B

D C


83


8. A regular hexagon of side 6 cm is inscribed in a circle. Find the area of the region in the circle which is outside the hexagon (p = 3.14, 3 = 1.732).

PL

Q

M R

S 9. PLRM is a square inscribed in sector PQRS. PR = 40 cm. Find the area

of shaded portion (π = 3.14). 10. PQRS is square, PQ = 14 cm. Four circles are drawn with radius 7 cm and

every vertex as the centre. Find the area of the shaded portion.

P Q

S R

88


NCERT Textual Exercises and Assignments

Exercise – 3.1 1. Let the radius of the bigger circle whose circumference is equal to the sum of the circumference

of two smaller circle be R. ∴ 2πR = 2πr1 + 2πr2

∴ 2πR = 2π × 19 + 2π × 9 ∴ 2pR = 2≠ ( 19 + 9 ) ∴ R = 28 ∴ Radius of the bigger circle = 28 cm. 2. Let the radius of the bigger circle be r. ∴ Area of the bigger circle = πr² ∴ As per the given condition, πr2 = π (8)2 + π (6)2

⇒ π πr2 2 28 6= +( ) ∴ r2 = 82 + 62

∴ r2 = 64 + 36 ∴ r2 = 100 ∴ r = 10 cm. ∴ Radius of the bigger circle = 10 cm. 3. Radius of the smallest circle (r1) = 10.5 cm ∴ r2 = 10.5 + 10.5 = 21 cm ∴ r3 = 21 + 10.5 = 31.5 cm ∴ r4 = 31.5 + 10.5 = 42 cm ∴ r5 = 42 + 10.5 = 52.5 cm ∴ A1 = πr1

2

= 227 × 10.5 × 10.5

∴ A1 = 346.5 cm2

89


∴ A2 = r r22

12

= π( r r22

12 )

= 227

(212 – 10.52)

= 227

× (21 + 10.5) (21 – 10.5)

= 227

× 31.5 × 10.5

∴ A2 = 1039.5 cm2 ∴ A3 = π ( r r3

222− )

= 227

× (31.52 – 212)

= 227 × (31.5 + 21) (31.5 – 21)

= 227 × 52.5 × 10.5

∴ A3 = 1732.5 cm2

∴ A4 = π ( r r42

32− )

= 227

× (422 – 31.52)

= 227 × (42 + 31.5) (42 – 31.5)

= 227 × 73.5 × 10.5

∴ A4 = 2425.5 cm2

∴ A5 = π (7 r r52

42− )

= 227 × (52.52 – 422)

= 227 × (52.5 + 42) (52.5 – 42)

= 227 × 94.5 × 10.5

∴ A5 = 3118.5 cm2

4. Radius of the circle = 802 = 40 cm.

Distance travelled in one revolution = circumference

90


= 2πr = 2 × π × 40 = 80π cm Distance travelled in 1 h = 66 km

∴ Distance travelled in 10 min = 6660

× 10

= 11 km = 11 × 1000 × 100 cm ... (1 km = 100000 cm)

∴ Total no. revolution =Total distance covered

Distance covered in one revolution

= 11 × 100000

80 × 227

= 11 × 1000 × 100 × 7

1760

= 77 × 1000 × 100

1760 ∴ Total number of revolutions = 4375. 5. Perimeter of a circle = Area of the circle ∴ 2πr = πr2 ∴ 2r = r2

∴ 2 = rr

2

∴ r = 2 units. ∴ Option (A) is correct.

Test Yourself – AR 1 1. 38.5 cm2 2. 58 cm 3. 400 revolutions 4. `2816 5. `1056 6. 35.35 cm 7. π : 2 8. 11 cm, 3 cm 9. 28 m 10. 0.1386 km2.

Exercise – 3.2 1. r = 6 cm q = 60°

91


∴ Area of the sector = θ π360

2× r

= 60360

227

6 6× × ×

∴ Area of the sector = 1327

cm2.

2. Circumference = 22 cm Circumference of the circle = 2πr

∴ 22 = 2 227

× × r

∴ 22 × 72 × 22 = r

∴ r = 72

cm

Area of the quadrant of a circle = 14

2πr

= 14

227

× 72

× 72

×

∴ Area of the quadrant of a circle = 778

cm2.

3. Length of the minute hand of a clock (r) = 14 cm Angle swept by minute hand in 60 min = 360°

∴ Angle swept by minute hand in 1 min =

3602× r

∴ Angle swept by minute hand in 5 min = 5 × 6° = 30° ∴ q = 30°


× 227

× 14 × 14

= 30360

× 227

× 14 × 14

∴ Area of the sector = 1543 cm2.

4. r = 10 cm q = 90°

Area of ∆OAB = 12

× AO × OB

= 12

×10 ×10

92


= 50 cm²

∴ Area of the sector (O-AXB) =

360

2 r

A BX

O

10cm

= 90360

× 3.14 ×10 ×10

= 78.5 cm2 ∴ Area of the minor segment = A (O-AXB) – A(∆AOB) = 78.5 – 50 ∴ Area of the minor segment = 28.5 cm2. ∴ Area of the circle = πr2

= 3.14 × 10 × 10 = 314 cm2 ∴ Area of the major sector = area of circle – area of minor sector = 314 – 78.5 ∴ Area of the major sector = 235.5 cm2.

5. l (arc AXB) =

3602 r

XA B

O 21 cm60°

= 60360

× 2 × 227

× 21

∴ Length of the arc = 22 cm

A (O-AXB) =

3602 r

= 60

360 × 22

7× 21 × 21

∴ Area of the sector = 231 cm2. In ∆OAB, OA = OB = radius of same circle ∠A = ∠B = x° ( angles opposite to equal sides of an isosceles triangle are equal) In ∆OAB, ∴ ∠A + ∠B + ∠O = 180° ∴ x + x + 60° = 180° ∴ 2x = 120° ∴ x = 60° ∴ ∆OAB is an equilateral triangle

∴ A (∆OAB) = 34

× (side)2 ( sides of ∆OAB = radius)

= 34

× 21 × 21

93


= 441

43

cm2

∴ Area of the minor segment = A (O-AXB) – A (∆AOB)

∴ Area of the minor segment = 231 – 44143 cm².

6. In ∆ABC, AB = AC = radius of same circle

XB C

A

60°15 cm

∠B = ∠C = x° In ∆ABC, ∴ ∠A + ∠B + ∠C = 180° ∴ 60° + x + x = 180° ∴ 2x = 120° ∴ x = 60° ∴ ∆ABC is an equilateral triangle

∴ A (∆ABC) = 34

× (side)²

= 3

4 × (15)2

=225 3

4cm2

= 225 ×1.73

4

= 97.3125 cm2

∴ A (A-BXC) =

360 r²

= 60

3603 14 15 2 .

= 16

× 3.14 × 15 × 15

= 117.75 cm2

∴ Area of the minor segment = A (A-BXC) – A (∆ABC) = 117.75 – 97.3125 ∴ Area of the minor segment = 20.4375 cm2

∴ Area of circle = πr2

= 3.14 × 15 × 15 = 706.5 cm2 ∴ Area of major segment = area of circle – area of minor segment

94


= 706.5 – 20.4375 ∴ Area of the major segment = 686.0625 cm2. 7. Draw OL ⊥ AB ∆AOL ≅ ∆BOL ..... (RHS rule) ∴ ∠AOL = ∠BOL =

12

× ∠AOB

A B

X

O

12 cm120°

= 12

× 120°

= 60° In ∆AOL,

sin 60° = ALAO

∴ 32

AL12

=

∴ AL = 12 3

2 ∴ AL = 6 3 cm

∴ AB = 2AL ... (AL = BL) ..... (perpendicular drawn from the centre to the chord bisects it)

= 2 × 6 3

= 12 3 cm

∴ cos 60° = OLOA

∴ 12 12= OL

∴ OL = 122

∴ OL = 6 cm

∴ A (∆OAB) = 12

× b × h

= 12

× AB × OL

= 12

×12 3 × 6

= 36 3

95


= 36 × 1.73 = 62.28 cm2

∴ A (O-AXB) =

360 r²

= 120360

× 3.14 ×12 ×12

= 452 16

3.

= 150.72 cm2

∴ Area of the minor segment = A(O-AXB) – A (∆OAB) = 150.72 – 62.28 ∴ Area of the minor segment = 88.44 cm2. 8. (i) r = 5 cm q = 90°

∴ A (A – MXN) =

360 r²

= 90360

× 3.14 × 5 × 5

= 78 5

4.

= 19.625 m2

∴ Area of the grazed part = 19.625 m2. (ii) If r = 10 cm

∴ Area of sector = 360

r²

= 90

360 × 3.14 × 10 × 10

= 3144

= 78.5 m2

∴ Increase in the grazing area = 78.5 – 19.625 Increase in the grazing area = 58.875 m2.

9. Central angle = 36010

= 36° (i) Total length of the wire = Circumference + 5 (diameters) = 2πr + 5 (2r)

96


= 2 × 227

2 × 352

352

5

35m = 110 + 175 = 285 mm ∴ Total length of the wire = 285 mm.

(ii) Area of each sector =

360 r²

= 36360

× 227

× 352

× 352

11 5

..... (q of each sector = 36010

= 360)

= 1

102

× 5511

× 352

Area of each sector = 3854

mm2.

10. Angle between two ribs = 3608

∴ q = 45° r = 45 cm

∴ Area between the consecutive ribs = area of sector

=

360 r²

= 45

360× 22

7× 45 × 45

= 184

× 2211

7× 45 × 45


cm2.

11. r = 25 cm q = 115°

A

O

B

X

115°25 cm

∴ A (O – AXB) =

360 r²

= 115360

×227

× 25 × 25

= 2336

×117

× 25 × 25

97


∴ Total area = 2 ×2336

×117

× 25 × 25

= 23 × 11 × 625

126

∴ Total area = 158125126

cm2.

12. r = 16.5 km 16.5 cm

80°

A B

X

O q = 80°

∴ A (O - AXB) =

360 r²

= 80360

× 3.14 ×16.5 ×16.5

Area of the sector = 189.97 km2.

13. Central angle = 360

6 = 60°

∴ A (∆OAB) = 3

4 × (side)2

BX

A

F

E

D

C28 cm

60°

O

= 3

4 × (28)2

= 3 × 28 × 7

= 1.7 × 28 × 7 = 333.2 cm2

∴ A (O - AXB) =

3602 r

= 60

360×

227

× 284

× 28

= 16

× 22 × 4 × 28

= 2464

6 = 410.67 cm2

∴ Area of the minor segment = A (O - AXB) – A (∆ AOB) = 410.67 – 333.2 = 77.47 cm2

∴ Area of six segments = 6 × 77.47 = 464.82 cm2

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Cost of 1 cm2 = `0.35 ∴ Total cost = 464.82 × 0.35 Total cost = `162.68. 14. Area of sector of angle P of a circle with radius R

= p

3602× Rπ

= 2

2 360

2 p R

= p

720 × 2πR2

∴ Option (D) is correct.

Test Yourself – AR 2 1. 55 cm, 577.5 cm2 2. 33.6 cm

3. 462 cm2 4. (i) 11007

cm2 (ii) 2757

cm2

(iii) 1007

cm2 (iv) 10007

cm2

5. 10.27 cm2 , 616.20 cm2 6. 25.67 cm2

7. 3.26 cm 8. 686.1 cm2 – 20.4 cm2 = 665.7 cm2

9. 84.77 cm2, 241.12 cm2 10. 88.44 cm2

Exercise – 3.3 1. ∠QPR = 90° ... (diameter always subtends a right angle)

∴ A (∆PQR) = 12

× PR × QP

= 12

20 × 7 × 24

= 84 cm2

In ∆PQR, QP2 + PR2 = QR2 ... (by Pythagoras theorem) ∴ (24)2 + (7)2 = QR2

∴ QR2 = 576 + 49 ∴ QR2 = 625 ∴ QR = 25 cm

∴ Radius = 252 cm

99


∴ Area of semicircle = 12

2≠ r

= 12

× 2211

7× 25

2× 25

2

= 6875

28cm2

∴ Area of the shaded portion = area of semicircle – A (∆PQR)

= 6875

28 – 84

= 6875 235228−

∴ Area of the shaded portion = 452328

cm2.

2. OB = OD = 7 cm OA = OC = 14 cm

A(O - AYC) = θ

360 × π(14)2

A(O - BXD) = θ

360 × π(7)2

∴ Area of the shaded portion = A (O - AYC) − A(O - BXD)

= θ

360 × π(14)2 – θ

360 × π(7)2

= 40

360 × π(142 − 72 )

= 19

× 227

× 14 + 7 14 �7

= 193

× 227

× 217

× 7


cm2.

3. Area of square = (Side)2 A B

D C

P

14 cm

= (14)2

= 196 cm2

∴ Area of semicircle = 12

2πr

∴ Area of two semicircles = 2 × 12× 22

7 × 7 × 7

100


= 154 cm2

∴ Area of the shaded portion = area of square – area of two semicircles = 196 – 154 ∴ Area of the shaded portion = 42 cm2. 4. Area of circle = πr2

A B

C D

12 cm

X6 cm

O = 227

× 6 × 6

= 7927

cm2

Area of ∆AOB = 34

× (side)2

= 34× 12

3× 12

= 36 3 cm2

A (O - CXD) = θ

360 × πr2

= 60

360227

× × 6 × 6

= 132

7 cm2

∴ Area of the shaded portion = area of circle + A (∆AOB) − A(O-CXD)

= 7927

1327

36 3

∴ Area of the shaded portion = 6607+ 36 3

cm2.

5. Side of the square = 4 cm A B

D C4 cm

1cm

2cm

∴ Area of the square = (side)2

= (4)2

= 16 cm2

Radius of the middle circle = 22

= 1cm ∴ Area of circle = πr2

= 227

× (1)2

= 227

cm2

∴ Area of one sector = θ360

× πr2

101


∴ Area of four sectors = 4 × θ

360 × πr2

∴ Area of four sectors = 4 ×90360

× 227

× (1)2

= 227 cm2

∴ Area of the shaded portion = (Area of square) − (area of 4 sectors + area of circle)

= 16 – 227

227

= 16 – 447

= 112 447−


cm2.

6. Let O be the centre of the circle. ∴ Radius of circle, OB = 32 cm = OC ∠BAC = 60° ( ∆ABC is equilateral triangle) ⇒ ∠BOC = 2∠BAC (The angle subtended by the arc at the centre is twice the angle subtended

by the arc on the circle.) ⇒ ∠BOC = 2 × 60° = 120° Draw a perpendicular from O on BC as D.

A

B CD

O

60°32 cm

∆BOD ≅ ∆DOC (By RHS) ⇒ ∠BOD = ∠DOC (by cpct) ⇒ ∠BOD = ½∠BOC = ½ × 120° = 60° ⇒ ∠BOD = 60° In ∆BOD,

sin 60° = dBDOB

⇒ 3

2= BD

32 ⇒ BD = 16 3 ⇒ BC = 2BD = 2 × 16 3 32 3=

⇒ AB = BC = 32 3 cm = AC

⇒ Sides of ∆ABC = 32 3 cm

102


∴ A (∆ABC) = 34

× (side)2

= 3

4 32 3

2

= 3

432 32 3

8× × ×

= 768 3 cm2

Area of circle = πr2

= 227

× 32 × 32

= 22528

7 cm2

∴ Area of the design = Area of circle – A (∆ABC)

∴ Area of the design = 225287

768 3−−

cm².

7. Side of the square = 14 cm

A B

D C

7 cm 7 cm ∴ Area of the square = (side)2

= (14)2

= 196 cm2

∴ Radius of the sector = 142

= 7 cm

∴ Area of one sector = θ360

×πr2

∴ Area of four sectors = 4 ×90360

× 227

× 7 × 7 = 154 cm2

∴ Area of the shaded portion = area of square − area of 4 sectors = 196 – 154 ∴ Area of the shaded portion = 42 cm2. 8. Length of ABCD = 106 m Length of EFGH = 106 m ∴ (EF) = (HG) = 106 m There are two semicircles GF and EH which make one

complete circle of radius = 602

= 30 m

103


∴ Circumference of circle = 2πr

= 2 × 227

× 30

= 13207

m

∴ The distance around the track

along its inner edge = 106 + 106 + 1320

7

= 212 + 1320

71484 + 1320

7=

The distance around the track along its inner edge = 28047

m.

∴ Area of the EDCF = × b = 106 × 10 = 1060 m2

∴ A (EDCF) = 1060 m2

Similarly A (ABGH) = 1060 m2

∴ Area of the region between two semicircles GF and BC

= 12 × π (r1

2– r22)

= 12

227

× (402 – 302 ) .... ( r1 =802

= 40 m , r2 = 602

= 30 m)

= 12× 22

7 × (40 + 30 )(40 – 30 )

= 12

227

× × 70 × 10

= 1100 m2

Similarly, area of the region between two semicircles EH and AD = 1100 m2

∴ Area of the track = 1100 + 1100 + 1060 + 1060 ∴ Area of the track = 4320 m2.

9. Radius of smaller circle =72 cm

∴ Area of smaller circle = πr2

= 227× ×7

272

= 772

cm2 ∠BOC = 90°

104


∴ BO = OC = 7 cm

∴ A (∆ BOC) = 12

× BO × OC

= 12 × 7 × 7

= 492

cm2

∴ A(O - BXC) = θ360

×πr2

= 90360

227

× × ×7 7

= 772

cm2

∴ Area of segment BXC = A (O - BXC) − A (∆ BOC)

= 772− 49

2

= 282

= 14 cm2

∴ Area of the other segment AYC = 14 cm2

∴ Area of the shaded region = area of smaller circle + area of segment BXC + area of segment AYC

= 772

+ 14 + 14

= 772 + 28

= 77 + 56

2

= 133

2 cm2

∴ Area of the shaded region = 66.5 cm2.

A

B C

10. ∴ A (∆ABC) = 3

4 × (side)2

∴ 17320.5 = 34

× (side)2

∴ 4 ×17320.53

= (side)2

105


∴ 4 ×17320.51. 73205

= (side)2

∴ 4× 173205

10173205100000

= (side)²

∴ 4 ×173205 ×100000

173205 ×10 = (side)2

∴ 40000 = (side)2

∴ Side = 200 cm

∴ Radius = 2002

= 100 cm

∴ Area of one sector = 360

2 r

∴ Area of three sectors = 3 × 60

360 × 3.14 × 100 × 100

= 15700 cm2

∴ Area of the shaded portion = A (∆ABC) – area of three sectors = 17320.5 – 15700 ∴ Area of the shaded portion = 1620.5 cm2.

A B

D C

11. Radius of one circle = 7 cm ∴ Area of nine circles = 9πr²

= 9 × 227

× 7 × 7

= 1386 cm2

∴ Side of square = 7 + 7 + 7 + 7 + 7 + 7 = 42 cm

∴ Area of square = (side)2

= (42)2 = 1764 cm2

∴ Area of the remaining portion = A (ABCD) − area of nine Circles = 1764 – 1386 ∴ Area of the remaining portion = 378 cm2.

12. ∴ A (∆BOD) = 12 × BO × OD

= 12 × 3.5 × 2

106


= 3.5 cm2

q = 90°

D

A

O

2 cm

3.5 cmB

C ∴ A(O - BCA) =

3602× r

= 90360

227

× × 3.5 × 3.5

= 14

× 227

× 3.50.5

× 3.5 = 38 5

4.

Area of quadrant OACB = 9.625 cm2

∴ Area of the shaded portion = A(O - BCA) − A (∆ BOD) = 9.625 – 3.5 ∴ Area of the shaded portion = 6.125 cm2. 13. By Pythagoras theorem,

Q

C

O A P

B

20 cm

OA2 + AB2 = OB2 ∴ (20)2 + (20)2 = OB2

∴ OB2 = 400 + 400 ∴ OB2 = 800

∴ OB = 20 2 cm

∴ Radius = 20 2 cm

∴ A(O - PBQ) =

360

2× r

= 90360

× 3.14 × 20 2 × 20 2

= 14

× 3.14 × 800

= 200 × 3.14 = 628 cm2

∴ Area of square ΟABC = (side)2

= (20)2

= 400cm2

∴ Area of the shaded portion = A(O - PBQ) – A (OABC) = 628 – 400 ∴ Area of the shaded portion = 228 cm2.

14. ∴ A(O - AXB) = θ

360 × π (21)2

∴ A(O - CYD) = θ360

× π (7)2

107

Area Related to Circles ∴ Area of the shaded portion = A(O - AXB) − A(O - CYD)

= θ

360 × π (21)2 – θ360

× π (7)2

721

Q

C

A

D

XB

30° = 30360

21 72 2 ( )

= 1

12× 22

7 × (21 + 7) (21 – 7)

= 1123

× 2271

× 2841

× 14= 22 ×143

Area of the shaded portion = 3083

cm2.

15. In ∆ABC, AB2 + AC2 = BC2 ... (by Pythagoras theorem) ∴ (14)2 + (14)2 = BC2

∴ BC2 = 196 + 196 ∴ BC2 = 2 × 196 ∴ BC = 14 2 cm

∴ Radius of semicircle = 14 2

2

B

14 cm

A C14 cm

PQ = 7 2 cm

∴ ar (∆ABC) = 12

× AC × AB

= 12 × 14 × 14

= 98 cm2 ∴ Area of sector A - BPC=

θ360

× πr²

= 90360

× 227

× 14 × 14

= 154 cm2

∴ Area of minor segment BPC = ar(A - BPC) − ar (∆ABC) = 154 – 98 = 56 cm2

∴ Area of semicircle with BC as diameter = 12

2πr

=12

× 227

× 7 2×7 2

= 154 cm2

108

Area Related to Circles ∴ Area of the shaded portion = area of semicircle – area of minor segment BPC = 154 – 56 ∴ Area of the shaded portion = 98 cm2. Alternate Method In ∆ABC, AB2 + AC2 = BC2 (By Pythagoras theorem) ⇒ 142 + 142 = BC2

⇒ BC =14 2 cm

Radius of semicircle = 14 2

27 2= cm

Area of ∆ABC = 12× ×AC AB

= 12

14 147× × = 98 cm2

Area of semicircle = πr2

2

= 12

227

7 2 7 211

× × × = 154 cm2

Area of sector ABPC =

3602 r

=90

36022

714 14

4

112

2

× × × = 154 cm2

Area of shaded region = (area of ∆ABC + area of semicircle) – area of sector = (98 + 154) – 154 = 98 cm2

16. ∴ A (∆ABD) = 12

× AD × AB

= 12 × 8 × 8 = 32 cm2

∴ Area of sector A-BXD = θ360

× πr2

= 90360

× 227 × 8 × 8

= 3527

cm2

∴ Area of the minor segment BXD = A (A-BXD) − A (∆ABD)

= 3527

– 32

109


= 352 224

7−

= 1287

cm2

Similarly, area of the minor segment BYD = 128

7cm2

∴ Area of the shaded portion = 2 × 128

7


cm2.

Test Yourself – AR 3 1. `230 2. 896 m2, 4032 m2

3. 42 cm2 4. (i) 4.56 cm2 (ii) 2.28 cm2 (iii) `28.73 (approx.)

5. 43 cm2 6. 36.97 cm2

7. 42 cm2 8. 19.512 cm2

9. 456 cm2 10. 504 cm2

exercise – 3...an umbrella has eight ribs which are equally spaced. assuming the umbrella to be a...

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