Class 10 - Maths - Areas Related to Circles

Exercise 12.1

Unless stated otherwise, use π = 22/7.

Question 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.

Answer:

We have,

Radius of circle-I R1 = 19 cm

Radius of circle-II R2 = 9 cm

Circumference of circle-I = 2πR1 = 2π * 19 cm

Circumference of circle-II = 2πR2 = 2π * 9 cm

Sum of the circumferences of circle-I and circle-II = 2π * 19 + 2π * 9

= 2π(19 + 9)

= (2π * 28) cm

Let R be the radius of the circle-III.

Circumference of circle-III = 2πR

According to the condition,

2πR = 2π * 28

=> R = (2π * 28)/2π

=> R = 28

Thus, the radius of the new circle = 28 cm.

Question 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.

Answer:

We have,

Radius of circle-I, r1 = 8 cm

So, Area of circle-I = πr12 = π(8)2 cm2

Area of circle-II = πr22 = π(6) 2 cm2

Let the radius of the circle-III be R

∴ Area of circle-III = πR2

Now, according to the condition,

πr12 + πr12 = πR2

=> π(8)2 + π(6)2 = πR2

=> π(82 + 62) = πR2

=> 64 + 36 = R2

=> 100 = R2

=> 102 = R2

=> R = 10

Thus, the radius of the new circle = 10 cm.

Question 3:

Fig. 12.3 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.

Answer: Diameter of the innermost region = 21 cm

Radius of the innermost (Gold Scoring) region = 21/2 = 10.5 cm

So, area of Gold region = π(10.5)2

= (22/7) * 10.5 * 10.5

= 22 * 1.5 * 10.5

= 346.50 cm2

Area of Red region = π(10.5 + 10.5)2 - π(10.5)2

= π(21)2 - π(10.5)2

= (22/7) * 21 * 21 - 346.50

= 22 * 21 * 3 - 346.50

= 1386 – 346.50

= 1039.50 cm2

Area of Blue region = π(21 + 10.5)2 - π(21)2

= π(31.5)2 - π(21)2

= π{(31.5)2 - (21)2}

= π(31.5 - 21)(31.5 + 21)

= (22/7) * 10.5 * 52.5

= 22 * 1.5 * 52.5

= 1732.50 cm2

Area of Black region = π(31.5 + 10.5)2 - π(31.5)2

= π(42)2 - π(31.5)2

= π{(42)2 - (31.5)2}

= π(42 – 31.5)(42 + 31.5)

= (22/7) * 10.5 * 73.5

= 22 * 1.5 * 73.5

= 2425.50 cm2

Area of White region = π(42 + 10.5)2 - π(42)2

= π(52.5)2 - π(42)2

= π{(52.5)2 - (42)2}

= π(52.5 – 42)(52.5 + 42)

= (22/7) * 10.5 * 94.5

= 22 * 1.5 * 94.5

= 3118.50 cm2

Question 4:

The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes

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

Answer:

Given diameter of the wheel = 80

So, the radius of the wheel r = 80/2 = 40

Now circumference of the wheel = 2πr

=2π * 40

= 80π

Now Speed of the car = 66 km/hour

= (66*1000 *100)/60

(1 km = 1000m and 1 m = 100 cm, 1 hour = 60 minutes)

= (66*1000 *10)/6                  (100 and 60 is divided by 10)

= 11*10000                             (66 and 6 is divided by 6)

= 110000 cm/minute

Distance travelled by car in 10 minutes = 110000*10 = 1100000 cm

Let the number of revolution of the wheel is n.

So, n * distance travelled in one minute = Distance travelled in 10 minutes

=> n * circumference of the wheel = Distance travelled in 10 minutes

=> n * 80π = 1100000

=> n = 1100000/80π

=> n = (110000)/8π                            (1100000 and 80 is divided by 10)

=> n = (110000*7)/(22*8)                  (π = 22/7)

=> n = (10000*7)/(2*8)                     (110000 and 22 is divided by 11)

=> n = (5000*7)/8                              (10000 and 2 is divided by 2)

=> n = 625*7                                      (5000 and 8 is divided by 8)

=> n = 4375

So, the number of revolution in 10 minutes at speed of 66 km/hour is 4375

Question 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

Answer:

We have:

Numerical area of the circle = Numerical circumference of the circle

=> πr2 = 2πr

=> πr2 – 2πr = 0

=> π(r2 – 2r) = 0

=> r2 – 2r = 0                        [Since π ≠ 0]

=> r(r – 2) = 0

=> r = 0 or r = 2

But r cannot be zero

So, r = 2 units.

Thus, the option (A) is the correct answer.

Exercise 12.2

Unless stated otherwise, use π = 22/7.

Question 1:

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

Answer: Given, radius of circle (r) = 6 cm

Angle of the sector (θ) = 60°

Now, area of sector = (θ/360°) * πr2

= (60°/360°) * (22/7) * 62

= (1/6) * (22/7) * 6 * 6

= (22/7) * 6

= 132/7 cm2

Question 2:

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

Answer: Let radius of circle be r cm.

Given, circumference of the circle = 22

=> 2πr = 22

=> 2 * (22/7) * r = 22

=> 2r/7 = 1

=> r = 7/2 cm

So, area of the quadrant (1/4)th of the circle = (θ/360) * πr2

= (90°/360°) * (22/7) * (7/2)2

= (1/4) * (22/7) * (7/2) * (7/2)

= (1/4) * 11 * (7/2)

= 77/8 cm2

Question 3:

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

Answer:

Length of minute hand = radius of the circle

=> r = 14 cm

Angle swept by the minute hand in 60 minutes = 360°

Angle swept by the minute hand in 5 minutes = (360°/60) * 5 = 6 * 5 = 30°

Now, area of the sector with r = 14 and θ = 30° is given as

= (θ/360°) * πr2

= (30°/360°) * (22/7) * 142

= (1/12) * (22/7) * 14 * 14

= (11 * 14)/3

= 154/3 cm2

Thus, the required area swept by the minute hand by 5 minutes = 154/3 cm2

Question 4:

A chord of a circle of radius 10 cm subtends a right angle at the centre.

Find the area of the corresponding:          (i) minor segment          (ii) major sector.        (Use π = 3.14)

Answer:

Length of the radius (r) = 10 cm

Sector angle θ = 90°

Area of the sector with θ = 90° and r = 10 cm is given as

= (θ/360°) * πr2

= (90°/360°) * 3.14 * 102

= (1/4) * 3.14 * 10 * 10

= 3.14 * 5 * 5

= 78.5 cm2

(i) Area of the minor segment = Area of minor sector – Area of ΔAOB

= 78.5 – (1/2) * 10 * 10

= 78.5 – 50

= 28.5 cm2

(ii) Area of major segment = Area of the circle] – Area of the minor segment

= πr2 – 78.5

= 3.14 * 10 * 10 – 78.5

= 314 – 78.5

= 235.5 cm2

Question 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

Answer: Given, radius = 21 cm and θ = 60°

(i) Circumference of the circle = 2πr

= 2 * (22/7) * 21

= 2 * 22 * 3

= 132 cm

Now, length of the arc APB = (60°/360°) * 132

= 132/6

= 22 cm

(ii) Area of the sector with angle 60° = (θ/360°) * πr2

= (60°/360°) * (22/7) * 212

= (1/6) * (22/7) * 21 * 21

= 11 * 21

= 231 cm2

(iii) Area of the segment APQ = Area of the sector AOB – Area of ΔAOB   ………….1

In ΔAOB, OA = OB = 21 cm

So, ∠A = ∠B = 60°       [∠O = 60°]

=> AOB is an equilateral triangle,

Hence, AB = 21 cm

Draw OM ⊥ AB such that OM/OA = sin 60°

=> OM/OA = √3/2

=> OM = (√3/2) * 21

Now, area of triangle OAB = (1/2) * AB * OM

= (1/2) * 21 * (√3/2) * 21

= 441√3/4 cm2      ………….2

From equation 1 and 2, we get

Area of segment = (231 - 441√3/4) cm2

Question 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)

Answer: Here, radius (r) = 15 cm

Sector angle θ = 60°

Area of the sector with angle 60° = (θ/360°) * πr2

= (60°/360°) * 3.14 * 152

= (1/6) * 3.14 * 15 * 15

= 1.57 * 5 * 15

= 117.75 cm2

Since ∠O = 60° and OA = OB = 15 cm

∠AOB is an equilateral triangle.

=> AB = 15 cm and ∠A = 60°

Draw OM Ʇ AB

So, OM/OA = sin 60° = √3/2

=> OM = OA * √3/2

=> OM = 15 * √3/2

Now, ar(ΔAOB) = 1/2 * AB * OM

= 1/2 * 15 * 15 * √3/2

= 225√3/4

= (225 * 3.14)/4

= 20.4375 cm2

Now area of the minor segment = (Area of minor sector) – (ar ΔAOB)

= (117.75 – 97.3125) cm2

= 20.4375 cm2

Area of the major segment = Area of the circle – Area of the minor segment

= πr2 - 20.4375

= 3.14 * 15 * 15 - 20.4375

= 706.5 – 20.4375 cm

= 686.0625 cm2

Question 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)

Answer: Given, θ = 120° and r = 12 cm

Area of the sector = (θ/360°) * πr2

= (120°/360°) * 3.14 * 122

= (1/3) * 3.14 * 12 * 12

= 3.14 * 12 * 4

= 150.72 cm2     …………..1

Now, area of ΔAOB = 1/2 * AB * OM    ………..2

In ΔOAB, ∠O = 120°

=> ∠A + ∠B = 180° – 120°

=> ∠A + ∠B = 60°

Since, OB = OA = 12 cm

So, ∠A = ∠B = 30°

Again, OM/OA = sin 30° = 1/2

=> OM = OA * 1/2

=> OM = 12/2 = 6 cm

In right angle triangle AMO,

AM2 = 122 – 62

=> AM2 = 144 – 36

=> AM2 = 108

=> AM = √108

=> AM = 6√3

=> 2AM = 6√3

=> AM = 12√3

From equation 2, we get

Area of ΔAOB = 1/2 * AB * OM

= 1/2 * 12√3 * 6

= 36√3

= 36 * 1.73

= 62.28 cm2    ……….3

From equation 1 and 3, we get

Area of the minor segment = Area of minor segment – Area of ΔAOB

= 150.72 – 62.28 cm

= 88.44 cm2

Question 8:

A horse is tied to a peg at one corner of a square shaped

grass field of side 15 m by means of a 5 m long rope  (see Fig. 12.11).

Find:

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

(ii) the increase in the grazing area if the rope were 10 m long

instead of 5 m. (Use π = 3.14) Answer:

Here, Length of the rope = 5 m

Radius of the circular region grazed by the horse = 5 m

(i) Area of the circular portion grazed = (θ/360°) * πr2

= (90°/360°) * 3.14 * 52

= (1/4) * 3.14 * 5 * 5

= 78.5/4

= 19.625 m2

(ii) When length of the rope is increased to 10 m,

So, r = 10 m

Area of the circular region where θ = 90°.

= (θ/360°) * πr2

= (90°/360°) * 3.14 * 102

= (1/4) * 3.14 * 10 * 10

= 314/4   = 78.5 m2

So, increase in the grazing area = 78.5 – 19.625 = 58.875 m2

Question 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 as shown in Fig. 12.12.

Find:

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

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

Diameter of the circle = 35 mm

Radius (r) = 35/2 mm

(i) Circumference = 2πr

= 2 * (22/7) * (35/2)

= 22 * 5

= 110 mm

Length of 1 piece of wire used to make diameter to divide the circle into 10 equal sectors

= 35 mm

So, length of 5 pieces = 5 * 35 = 175 mm

Hence, total length of the silver wire = 110 + 175 mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,

Hence, Sector angle θ = 360°/10 = 36°

Now, area of each sector = (θ/360°) * πr2

= (36°/360°) * (22/7) * (35/2) * (35/2)

= (11 * 35)/4

= 385/4 mm2

Question 10:

An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle

of radius 45 cm, find the area between the two consecutive ribs of the umbrella. Answer:

Here, radius (r) = 45 cm

Since circle is divided in 8 equal parts,

Now, sector angle corresponding to each part θ = 360°/8 = 45°

Now, area of a sector(part) = (θ/360°) * πr2

= (45°/360°) * (22/7) * 45 * 45

= (1/8) * (22/7) * 45 * 45

= (1/4) * (11/7) * 45 * 45

= (11 * 45 * 45)/(4 * 7)

= 22275/28 mm2

Hence, the required area between two pins = 22275/28 mm2

Question 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.

Answer:

Here, radius (r) = 25 cm

Sector angle (θ) = 115°

Area cleaned by each sweep of the blades = (θ/360°) * πr2 * 2

[Since each sweep will have to and from movement]

= (115°/360°) * (22/7) * 25 * 25 * 2

= (23 * 11 * 25 * 25)/(18 * 7)

= 158125/126 mm2

Question 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)

Answer:

Here, Radius (r) = 16.5 km

Sector angle (θ) = 80°

Now, Area of the sea surface over which the ships are warned

= (θ/360°) * πr2

= (80°/360°) * 3.14 * 16.5 * 16.5

= (80°/360°) * 314/100 * 165/10 * 165/10

= (157 * 11 * 11)/100

= 18997/100

= 189.97 km2

Question 13:

A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm,

find the cost of making the designs at the rate of Rs 0.35 per cm2. (Use √3 = 1.7) Answer: Here, r = 28 cm

Since, the circle is divided into six equal sectors.

So, sector angle θ = 360°/6 = 60°

Area of the sector with angle 60° and radius 28 cm = (θ/360°) * πr2

= (60°/360°) * (22/7) * 28 * 28

= (1/6) * 22/7 * 28 * 4

= (44 * 28)/3

= 1232/3

= 410.67 cm2    ……………1

Now, area of 1 design = Area of segment APB

= Area of sector – Area of ΔAOB      …………..2

In ΔAOB, ∠AOB = 60°, OA = PN = 28 cm

So, ∠OAB = 60° and ∠OBA = 60°

=> ΔAOB is an equilateral triangle.

=> AB = AO = BO

=> AB = 28 cm

=> Draw OM ⊥AB

Now, in right ΔAOM, we have

OM/OA = sin 60° = √3/2

=> OM = OA * √3/2

=> OM = 28 * √3/2

=> OM = 14√3 cm

Area of ΔAOB = 1/2 * AB * OM

= 1/2 * 28 * 14√3

= 14 * 14 * 1.7

= 333.3 cm2   ………..3

Now, from (1), (2) and (3), we have:

Area of segment APQ = 410.67 – 333.2

= 77.47 cm2

=> Area of 1 design = 77.47 cm2

So, area of the 6 equal designs = 6 * 77.47

= 464.82 cm2

Cost of making the design at the rate of Rs. 0.35 per cm2 = Rs. 0.35 * 464.82

= Rs. 162.68

Question 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) (p/180) * πR2        (C) (p/360) * 2πR           (D) (p/720) * 2πR2

Answer:

Here, radius (r) = R

Angle of sector (θ) = p

Area of the sector = (θ/360°) * πr2

= (p/360°) * πR2

= (2/2) * (p/360°) * πR2

= (p/720°) * 2πR2

Hence, option (D) is the right answer.

Exercise 12.3

Unless stated otherwise, use π = 22/7.

Question 1:

Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. Answer:

Since O is the centre of the circle,

So, QOR is a diameter.

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

Now, in right angle triangle RPQ,

RQ2 = PQ2 + PR2

=> RQ2 = 242 + 72

=> RQ2 = 756 + 49

=> RQ2 = 625

=> RQ = √625

=> RQ = 25

Area of ΔRPQ = (1/2) * RQ * RP

= (1/2) * 24 * 7

= 12 * 7

= 84 cm2

Now, area of semi-circle = πr2/2

= (1/2) * (22/7) * (25/2) * (25/2)

= (11 * 625)/(7 * 4)

= 6875/28

= 245.54 cm2

So, area of the shaded portion = 245.54 – 84 = 161.54 cm2

Question 2:

Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠ AOC = 40°. Answer: Radius of the outer circle = 14 cm

Here, θ = 40°

So, area of the sector BOD = (40°/360°) * (22/7) * 7 * 7

= (1/9) * 22 * 7

= 154/9 cm2

Now, area of the shaded region = Area of sector AOC – Area of sector BOD

= (40°/360°) * (22/7) * 14 * 14 – 154/9

= (1/9) * 22 * 14 * 2 – 154/9

= 616/9 – 154/9

= (616 - 154)/9

= 462/9

= 154/3 cm2

Question 3:

Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles. Answer: Side of the square = 14 cm

Area of the square ABCD = 14 * 14 = 196 cm2

Now, diameter of the circle = (Side of the square) = 14 cm

Radius of each of the circle = 14/2 = 7 cm

Area of the semi-circle APB = πr2/2

= (1/2) * (22/7) * 7 * 7

= 11 * 7

= 77 cm2

Area of the semi-circle BPC = πr2/2

= (1/2) * (22/7) * 7 * 7

= 11 * 7

= 77 cm2

Now, area the shaded region = Area of the square] – [Area of semi-circle APD + Area of semi-circle BPC]

= 196 – [77 + 77]

= 196 – 154

= 42 cm2

Question 4:

Find the area of the shaded region in Fig. 12.22, 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. Answer: We know that each angle of equilateral triangle is 60°.

Area of sector OCDE = (60°/360°) * πr2

= (1/6) * (22/7) * 6 * 6

= 132/7 cm2

Area of equilateral triangle OAB = (√3/4) * (12)2

= (√3/4) * 12 * 12

= √3 * 3 * 12

= 36√3 cm2

Area of circle = πr2

= (22/7) * 6 * 6

= 792/7 cm2

Now, area of shaded region = area of circle + area of ΔOAB - area of sector OCDE

= 792/7 + 36√3 – 132/7

= (792 - 132)/7 + 36√3

= (36√3 + 660/7) cm2

Question 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 as shown in Fig. 12.23.

Find the area of the remaining portion of the square. Answer: Radius of each quadrant = 1 cm

Area of each quadrant = (90°/360°) * πr2

= (1/4) * (22/7) * 1 * 1

= (1/1) * (11/7)

= 11/14 cm2

Area of square = (side)2 = 42 = 16 cm2

Area of circle = πr2 = π * 1 * 1 = π = 22/7 cm2

Now, area of shaded region = area of square – area of circle – area of 4 quadrant

= 16 - 22/7 – 4 * 11/14

= 16 - 22/7 – 22/7

= 16 – 44/7

= (112 - 44)/7

= 68/7 cm2

Question 6:

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral

triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design (shaded region). Answer: Radius of circle = 32

Centroid O divides median AD into 2 : 1, therefore AO : OD = 2 : 1

=> AO/OD = 2/1

=> 32/OD = 2/1

=> OD = 32/2

=> OD = 16 cm

Therefore, AD = 32 + 16 = 48 cm

In ΔABD,

AB2 = AD2 + BD2

=> AB2 = 482 + (AB/2)2

=> AB2 = 482 + AB2/4

=> AB2 - AB2/4 = 482

=> 3AB2/4 = 482

Take square root on both sides, we get

=> √3AB/2 = 48

=> AB = (48 * 2)/√3

=> AB = (16 * √3 * √3 * 2)/√3

=> AB = 32√3

Area of equilateral triangle ABC = (√3/4) * (32√3)2

= 768√3 cm2

Area of circle = πr2 = (22/7) * 32 * 32 = 22528/7 cm2

Now, area of design = area of circle – area of equilateral triangle ABC

= (22528/7 - 768√3) cm2

Question 7:

In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such

that each circle touch externally two of the remaining three circles. Find the area of the shaded region. Answer: The circles drawn taking A, B, C and D from quadrants of radius 7 cm in square.

Radius of each quadrants = 7 cm

Area of each quadrant = (90°/360°) * πr2

= (1/4) * (22/7) * 7 * 7

= (1/2) * 11 * 7

= 77/2 cm2

Area of square = (side)2 = 142 = 196 cm2

Now, area of shaded region = area of square – area of 4 quadrants

= 196 – 4 * 77/2

= 196 – 77 * 2

= 196 – 154

= 42 cm2

Question 8:

Fig. 12.26 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 :

(i) the distance around the track along its inner edge

(ii) the area of the track. Answer: (i) The distance around the track along its inner edge

= AB + arc BED + CD + arc DFA

= 106 + 1/2 * 2πr + 106 + 1/2 * 2πr

= 106 + 1/2 * 2 * 22/7 * 30 + 106 + 1/2 * 2 * 22/7 * 30

= 212 + 2 * 22/7 * 30

= 212 + 1320/7

= 2804/7 m2

(ii) area of track = (Area of GHIJ – area of ABCD) + (area of semi-circle HKI – area of semicircle

BEC) + area of semi-circle GLJ – area of semi-circle AFD

= (106 * 80 – 106 * 60) + (1/2) * 22/7 * [402 – 302] + (1/2) * 22/7 * [402 – 302]

= 106(80 - 60) + (1/2) * 22/7 * 700 + (1/2) * 22/7 * 700

= 2120 + 22/7 * 700

= 2120 + 2200

= 4320 m2

Question 9:

In Fig. 12.27, AB and CD are two diameters of a 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. Answer: Radius of smaller circle = 7/2 cm

Area of smaller circle = πr2 = (22/7) * 7/2 * 7/2 = 77/2 cm2

Radius of larger circle = 7 cm

Area of smaller circle AECFB = (1/2) * πr2 = (1/2) * (22/7) * 7 * 7 = 77 cm2

Area of triangle ACB = (1/2) * AB * OC = (1/2) * 14 * 7 = 49 cm2

Area of shaded region = Area of smaller circle + Area of larger circle - Area of ΔACB

= 77/2 + 77 – 49

= 38.5 + 28

= 66.5 cm2

Question 10:

The area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of the triangle as centre, a

circle is drawn with radius equal to half the length of the side of the triangle (see Fig. 12.28). Find the

area of the shaded region. (Use π = 3.14 and √3 = 1.73205) Answer:

Let each side of equilateral triangle be a cm Given, area of an equilateral triangle ABC = 17320.5

=> (√3/4) * a2 = 17320.5

=> a2 = 17320.5/(√3/4)

=> a2 = (17320.5 * 4)/√3

=> a2 = (17320.5 * 4)/1.73205

=> a2 = 10000 * 4

=> a2 = 40000

=> a = 200 cm

Area of sector ADEF = (θ/360°) * πr2

= (60°/360°) * 3.14 * 100 * 100

= (1/6) * 314 * 100

= (1/3) * 157 * 100

= 15700/3 cm2

Now, area of shaded region = Area of equilateral triangle ABC - Area of three sector

= 17320.5 - 3 * 15700/3    = 17320.5 – 15700       = 1620.5 cm2

Question 11:

On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29).

Find the area of the remaining portion of the handkerchief. Answer: Radius of circle = 7 cm

Area of one circular design = πr2 = (22/7) * 7 * 7 = 22 * 7 = 154 cm2

Side of square = 42 cm

Area of square = (side)2 = 422 = 1764 cm2

Now, area of the remaining portion = area of square – area of 9 circular designs

= 1764 – 9 * 154

= 1764 – 1386

= 378 cm2

Question 12:

In Fig. 12.30, OACB is a quadrant of a circle with centre O

and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) quadrant OACB,                                (ii) shaded region. Answer:

(i) Radius of quadrant = 1 cm

Area of quadrant = (θ/360°) * πr2

= (90°/360°) * 22/7 * 3.5 * 3.5

= (1/4) * 22 * 0.5 * 3.5

= (1/4) * 11 * 3.5

= 38.5/4

= 385/40

= 77/8 cm2

Area of ΔOBD = (1/2) * OB * OD = (1/2) * 3.5 * 2 = 3.5 = 35/10 = 7/2 cm2

(ii) Area of shaded region = Area of quadrant - Area of ΔOBD

= 77/8 – 7/2

= (77 – 7 * 4)/8

= (77 – 28)/8

= 49/8 cm2

Question 13:

In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ.

If OA = 20 cm, find the area of the shaded region. (Use π = 3.14) Answer: In ΔOAB,

OB2 = OA2 + AB2

=> OB2 = 202 + 202

=> OB2 = 400 + 400

=> OB2 = 800

=> OB = √800

=> OB = 20√2

Radius of quadrant = 20√2 cm

Area of quadrant = (θ/360°) * πr2

= (90°/360°) * 3.14 * 20√2 * 20√2

= (1/4) * 3.14 * 20√2 * 20√2

= 3.14 * 20√2 * 5√2

= 628 cm2

Area of square = (side)2 = 202 = 400 cm2

Area of shaded region = Area of quadrant - Area of square

= 628 – 400

= 228 cm2

Question 14:

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32).

If ∠ AOB = 30°, find the area of the shaded region. Answer: Area of shaded region = Area of sector OAEB - Area of sector OCFD

= (30°/360°) * π * (21)2 - (30°/360°) * π * 72

= (1/12) * π * 441 - (1/12) * π * 49

= (1/12) * π * (441 – 49)

= (1/12) * 22/7 * 392

= 308/3 cm2

Question 15:

In Fig. 12.33, 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. Answer: Radius of sector = 14 cm

Area of sector = (θ/360°) * πr2

= (90°/360°) * 22/7 * 14 * 14

= (1/4) * 22 * 2 * 14

= 11 * 14

= 154 cm2

In ΔABC,

BC2 = AC2 + AB2

=> BC2 = 142 + 142

=> BC2 = 196 + 196

=> BC2 = 392

=> BC = √392

=> BC = 14√2

Therefore, the diameter of semi-circle = 14√2 cm

Radius of semi-circle = 14√2/2 = 7√2 cm

Area of semi-circle = (1/2) * πr2 = (1/2) * (22/7) * 7√2 * 7√2 = 11 * √2 * 7√2 = 154 cm2

Area of ΔABC = (1/2) * AC * AB = (1/2) * 14 * 14 = 7 * 14 = 98 cm2

Area of shaded region = Area of ΔABC + Area of semi-circle - Area of quadrant

= 98 + 154 – 154

= 98 cm2

Question 16:

Calculate the area of the designed region in Fig. 12.34 common between the two quadrants

of circles of radius 8 cm each. Answer: Area of sector DAFC = (θ/360°) * πr2

= (90°/360°) * 22/7 * 8 * 8

= (1/4) * 22/7 * 8 * 8

= 352/7 cm2

Area of ΔADC = (1/2) * DC * AD = (1/2) * 8 * 8 = 8 * 4 = 32 cm2

Area of segment = Area of sector DAFC - Area of ΔADC

= 352/7 – 32

= (352 – 32 * 7)/7

= (352 - 224)/7

= 128/7 cm2

Area of shaded region = Area of two segment

= 2 * 128/7

= 256/7 cm2

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