# Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

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## Solution:

The radius of the 1st circle = 19 cm (given)

∴ Circumference of the 1st circle = 2π×19 = 38π cm

The radius of the 2nd circle = 9 cm (given)

∴ Circumference of the 2nd circle = 2π×9 = 18π cm

So,

The sum of the circumference of two circles = 38π+18π = 56π cm

Now, let the radius of the 3rd circle = R

∴ The circumference of the 3rd circle = 2πR

It is given that sum of the circumference of two circles = circumference of the 3rd circle

Hence, 56π = 2πR

Or, R = 28 cm.

## Solution:

Radius of 1st circle = 8 cm (given)

∴ Area of 1st circle = π(8)2 = 64π

Radius of 2nd circle = 6 cm (given)

∴ Area of 2nd circle = π(6)2 = 36π

So,

The sum of 1st and 2nd circle will be = 64π+36π = 100π

Now, assume that the radius of 3rd circle = R

∴ Area of the circle 3rd circle = πR2

It is given that the area of the circle 3rd circle = Area of 1st circle + Area of 2nd circle

Or, πR2 = 100πcm2

Þ R2 = 100cm2 So, R = 10cm

## Solution:

The radius of 1st circle, r1 = 21/2 cm (as diameter D is given as 21 cm)

So, area of gold region = π r12 = π(10.5)2 = 346.5 cm2

Now, it is given that each of the other bands is 10.5 cm wide, So, the radius of 2nd circle, r2 = 10.5cm+10.5cm = 21 cm Thus,

∴ Area of red region = Area of 2nd circle − Area of gold region = (πr22−346.5) cm2

= (π(21)2 − 346.5) cm2

= 1386 − 346.5

= 1039.5 cm2

Similarly,

The radius of 3rd circle, r3 = 21 cm+10.5 cm = 31.5 cm The radius of 4th circle, r4 = 31.5 cm+10.5 cm = 42 cm The Radius of 5th circle, r5 = 42 cm+10.5 cm = 52.5 cm For the area of nth region,

A = Area of circle n – Area of circle (n-1)

∴ Area of blue region (n=3) = Area of third circle – Area of second circle

= π(31.5)2 – 1386 cm2

= 3118.5 – 1386 cm2

= 1732.5 cm2

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

∴ Area of black region (n=4) = Area of fourth circle – Area of third circle

= π(42)2 – 1386 cm2

= 5544 – 3118.5 cm2

= 2425.5 cm2

∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle

= π(52.5)2 – 5544 cm2

= 8662.5 – 5544 cm2

= 3118.5 cm2

## Solution:

The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm) So, the circumference of wheels = 2πr = 80 π cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm

It is given that the distance covered by the car in 1 hr = 66km Converting km into cm we get,

Distance covered by the car in 1hr = (66×105) cm

In 10 minutes, the distance covered will be = (66×105×10)/60 = 1100000 cm/s

∴ Distance covered by car = 11×105 cm

Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels)

=( 11×105)/80 π = 4375.

## Solution:

Since the perimeter of the circle = area of the circle,

2πr = πr2

Or, r = 2

So, option (A) is correct i.e. the radius of the circle is 2 units.

## Solution:

It is given that the angle of the sector is 60°

We know that the area of sector = (θ/360°)×πr2

∴ Area of the sector with angle 60° = (60°/360°)×πr2 cm2

= (36/6)π cm2

= 6×22/7 cm2 = 132/7 cm2

## Solution:

Circumference of the circle, C = 22 cm (given)

It should be noted that a quadrant of a circle is a sector which is making an angle of 90°. Let the radius of the circle = r

As C = 2πr = 22,

R = 22/2π cm = 7/2 cm

∴ Area of the quadrant = (θ/360°) × πr2 Here, θ = 90°

So, A = (90°/360°) × π r2 cm2

= (49/16) π cm2

= 77/8 cm2 = 9.6 cm2

## Solution:

Length of minute hand = radius of the clock (circle)

∴ Radius (r) of the circle = 14 cm (given)

Angle swept by minute hand in 60 minutes = 360°

So, the angle swept by the minute hand in 5 minutes = 360° × 5/60 = 30° We know,

Area of a sector = (θ/360°) × πr2

Now, area of the sector making an angle of 30° = (30°/360°) × πr2 cm2

= (1/12) × π142

= (49/3)×(22/7) cm2

= 154/3 cm2

## Solution:

Here AB be the chord which is subtending an angle 90° at the center O.

It is given that the radius (r) of the circle = 10 cm

(i) Area of minor sector = (90/360°)×πr2

= (¼)×(22/7)×102

Or, Area of minor sector = 78.5 cm2

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

Also, area of ΔAOB = ½×OB×OA

Here, OB and OA are the radii of the circle i.e. = 10 cm

So, area of ΔAOB = ½×10×10

= 50 cm2

Now, area of minor segment = area of minor sector – area of ΔAOB

= 78.5 – 50

= 28.5 cm2

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

= (3.14×102)-78.5

= 235.5 cm2

## Solution:

Given,

θ = 60°

(i)Length of an arc = θ/360°×Circumference(2πr)

∴ Length of an arc AB = (60°/360°)×2×(22/7)×21

= (1/6)×2×(22/7)×21

Or Arc AB Length = 22cm

(ii)It is given that the angle subtend by the arc = 60°

So, area of the sector making an angle of 60° = (60°/360°)×π r2 cm2

= 441/6×22/7 cm2

Or, the area of the sector formed by the arc APB is 231 cm2

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

(iii)Area of segment APB = Area of sector OAPB – Area of ΔOAB

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is

60°, ΔOAB is an equilateral triangle. So, its area will be √3/4×a2 sq. Units. Area of segment APB = 231-(√3/4)×(OA)2

= 231-(√3/4)×212

Or, Area of segment APB = [231-(441×√3)/4] cm2

## Solution:

Given,

θ = 60°

So,

Area of sector OAPB = (60°/360°)×πr2 cm2

= 225/6 πcm2

Now, ΔAOB is equilateral as two sides are the radii of the circle and hence equal and one angle is 60°

So, Area of ΔAOB = (√3/4) ×a2

Or, (√3/4) ×152

∴ Area of ΔAOB = 97.31 cm2

Now, area of minor segment APB = Area of OAPB – Area of ΔAOB Or, area of minor segment APB = ((225/6)π – 97.31) cm2 = 20.43 cm2 And,

Area of major segment = Area of circle – Area of segment APB

Or, area of major segment = (π×152) – 20.4 = 686.06 cm2

## Solution:

Now, draw a perpendicular OD on chord AB and it will bisect chord AB. So, AD = DB

Now, the area of the minor sector = (θ/360°)×πr2

= (120/360)×(22/7)×122

= 150.72 cm2

Consider the ΔAOB,

∠ OAB = 180°-(90°+60°) = 30°

We know OD bisects AB. So, AB = 2×AD = 12√3 cm

Now, sin 30° = OD/OA Or, ½ = OD/12

∴ OD = 6 cm

So, the area of ΔAOB = ½ × base × height Here, base = AB = 12√3 and

Height = OD = 6

So, area of ΔAOB = ½×12√3×6 = 36√3 cm = 62.28 cm2

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB

= 150.72 cm2– 62.28 cm2 = 88.44 cm2

## Solution:

As the horse is tied at one end of a square field, it will graze only a quarter (i.e. sector with θ =

90°) of the field with radius 5 m.

Here, the length of rope will be the radius of the circle i.e. r = 5 m It is also known that the side of square field = 15 m

(i) Area of circle = πr2 = 22/7 × 52 = 78.5 m2

Now, the area of the part of the field where the horse can graze = ¼ (the area of the circle) = 78.5/4 = 19.625 m2

(ii) If the rope is increased to 10 m,

Area of circle will be = πr2 =22/7×102 = 314 m2

Now, the area of the part of the field where the horse can graze = ¼ (the area of the circle)

= 314/4 = 78.5 m2

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

## Solution:

Diameter (D) = 35 mm

Total number of diameters to be considered= 5

Now, the total length of 5 diameters that would be required = 35×5 = 175 Circumference of the circle = 2πr

Or, C = πD = 22/7×35 = 110

Area of the circle = πr2

Or, A = (22/7)×(35/2)2 = 1925/2 mm2

(i) Total length of silver wire required = Circumference of the circle + Length of 5 diameter

= 110+175 = 185 mm

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

(ii) Total Number of sectors in the brooch = 10

So, the area of each sector = total area of the circle/number of sectors

∴ Area of each sector = (1925/2)×1/10 = 385/4 mm2

## Solution:

The radius (r) of the umbrella when flat = 45 cm

So, the area of the circle (A) = πr2 = (22/7)×(45)2 =6364.29 cm2 Total number of ribs (n) = 8

∴ The area between the two consecutive ribs of the umbrella = A/n

Þ 6364.29/8 cm2

Or, The area between the two consecutive ribs of the umbrella = 795.5 cm2

## Solution:

Given,

Radius (r) = 25 cm Sector angle (θ) = 115° Since there are 2 blades,

The total area of the sector made by wiper = 2×(θ/360°)×π r2

= 2×(115/360)×(22/7)×252

= 2×158125/252 cm2

= 158125/126 = 1254.96 cm2

### 12.   To warn ships for underwater rocks, a lighthouse spreads a red colored 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)

Solution:

Let O bet the position of Lighthouse.

Here the radius will be the distance over which light spreads. Given, radius (r) = 16.5 km

Sector angle (θ) = 80°

Now, the total area of the sea over which the ships are warned = Area made by the sector

Or, Area of sector = (θ/360°)×πr2

= (80°/360°)×πr2 km2

= 189.97 km2

## Solution:

Total number of equal designs = 6

angle AOB= 360°/6 = 60°

Radius of the cover = 28 cm

Cost of making design = ₹ 0.35 per cm2

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is

60°, ΔAOB is an equilateral triangle. So, its area will be (√3/4)×a2 sq. units Here, a = OA

∴ Area of equilateral ΔAOB = (√3/4)×282 = 333.2 cm2

Area of sector ACB = (60°/360°)×πr2 cm2

= 410.66 cm2

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

So, area of a single design = area of sector ACB – area of ΔAOB

= 410.66 cm2 – 333.2 cm2 = 77.46 cm2

∴ Area of 6 designs = 6×77.46 cm2 = 464.76 cm2

So, total cost of making design = 464.76 cm2 ×Rs.0.35 per cm2

= Rs. 162.66

## Solution:

The area of a sector = (θ/360°)×πr2 Given, θ = p

So, area of sector = p/360×πR2

Multiplying and dividing by 2 simultaneously,

= (p/360)×2/2×πR2

= (2p/720)×2πR2

So, option (D) is correct.

## Solution:

Here, ÐP is in the semi-circle and so, ÐP = 90°

So, it can be concluded that QR is hypotenuse of the circle and is equal to

the diameter of the circle.

∴ QR = D

Using Pythagorean theorem, QR2 = PR2+PQ2

Or, QR2 = 72+242

Þ QR= 25 cm = Diameter Hence, the radius of the circle = 25/2 cm Now, the area of the semicircle = (πR2)/2 = (22/7)×(25/2)×(25/2)/2 cm2

= 13750/56 cm2 = 245.54 cm2

Also, area of the ΔPQR = ½×PR×PQ

=(½)×7×24 cm2

= 84 cm2

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

Hence, the area of the shaded region = 245.54 cm2-84 cm2

= 161.54 cm2

## Solution:

Given,

Angle made by sector = 40°,

Radius the inner circle = r = 7 cm, and Radius of the outer circle = R = 14 cm We know,

Area of the sector = (θ/360°)×πr2

So, Area of OAC = (40°/360°)×πr2 cm2

= 68.44 cm2

Area of the sector OBD = (40°/360°)×πr2 cm2

= (1/9)×(22/7)×72 = 17.11 cm2

Now, area of the shaded region ABDC = Area of OAC – Area of the OBD

= 68.44 cm2 – 17.11 cm2 = 51.33 cm2

## Solution:

Side of the square ABCD (as given) = 14 cm So, Area of ABCD = a2

= 14×14 cm2 = 196 cm2

We know that the side of the square = diameter of the circle = 14 cm So, side of the square = diameter of the semicircle = 14 cm

∴ Radius of the semicircle = 7 cm Now, area of the semicircle = (πR2)/2

= (22/7×7×7)/2 cm2 =

= 77 cm2

∴ Area of two semicircles = 2×77 cm2 = 154 cm2

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

Hence, area of the shaded region = Area of the Square – Area of two semicircles

= 196 cm2 -154 cm2

= 42 cm2

## Solution:

It is given that OAB is an equilateral triangle having each angle as 60° Area of the sector is common in both.

Radius of the circle = 6 cm. Side of the triangle = 12 cm.

Area of the equilateral triangle = (√3/4) (OA)2= (√3/40×122 = 36√3 cm2 Area of the circle = πR2 = (22/7)×62 = 792/7 cm2

Area of the sector making angle 60° = (60°/360°) ×πr2 cm2

= (1/6)×(22/7)× 62 cm2 = 132/7 cm2

Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector

= 36√3 cm2 +792/7 cm2-132/7 cm2

= (36√3+660/7) cm2

## Solution:

Side of the square = 4 cm Radius of the circle = 1 cm

Four quadrant of a circle are cut from corner and one circle of radius are cut from middle. Area of square = (side)2= 42 = 16 cm2

Area of the quadrant = (πR2)/4 cm2 = (22/7)×(12)/4 = 11/14 cm2

∴ Total area of the 4 quadrants = 4 ×(11/14) cm2 = 22/7 cm2

Area of the circle = πR2 cm2 = (22/7×12) = 22/7 cm2

Area of the shaded region = Area of square – (Area of the 4 quadrants + Area of the circle)

= 16 cm2-(22/7) cm2+(22/7) cm2

= 68/7 cm2

## Solution:

Radius of the circle = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.

⇒ BD = AB/2

Since, AD is the median of the triangle

By Pythagoras theorem, AB2 = AD2 +BD2

⇒ AB2 = 482+(AB/2)2

⇒ AB2 = 2304+AB2/4

⇒ 3/4 (AB2)= 2304

⇒ AB2 = 3072

⇒ AB= 32√3 cm

Area of ΔADB = √3/4 ×(32√3)2 cm2 = 768√3 cm2

Area of circle = πR2 = (22/7)×32×32 = 22528/7 cm2

Area of the design = Area of circle – Area of ΔADB

= (22528/7 – 768√3) cm2

## Solution:

Side of square = 14 cm

Four quadrants are included in the four sides of the square.

∴ Radius of the circles = 14/2 cm = 7 cm Area of the square ABCD = 142 = 196 cm2

Area of the quadrant = (πR2)/4 cm2 = (22/7) ×72/4 cm2

= 77/2 cm2 Total area of the quadrant = 4×77/2 cm2 = 154cm2

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

Area of the shaded region = Area of the square ABCD – Area of the quadrant

= 196 cm2 – 154 cm2

= 42 cm2

## Solution:

Width of the track = 10 m

Distance between two parallel lines = 60 m

Length of parallel tracks = 106 m

DE = CF = 60 m

Radius of inner semicircle, r = OD = O’C

= 60/2 m = 30 m Radius of outer semicircle, R = OA = O’B

= 30+10 m = 40 m Also, AB = CD = EF = GH = 106 m

Distance around the track along its inner edge = CD+EF+2×(Circumference of inner semicircle)

= 106+106+(2×πr) m = 212+(2×22/7×30) m

= 212+1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)

= (AB×CD)+(EF×GH)+2×(πr2/2) -2×(πR2/2) m2

= (106×10)+(106×10)+2×π/2(r2-R2) m2

= 2120+22/7×70×10 m2

= 4320 m2

## Solution:

Radius of larger circle, R = 7 cm Radius of smaller circle, r = 7/2 cm Height of ΔBCA = OC = 7 cm

Base of ΔBCA = AB = 14 cm

Area of ΔBCA = 1/2 × AB × OC = (½)×7×14 = 49 cm2

Area of larger circle = πR2 = (22/7)×72 = 154 cm2 Area of larger semicircle = 154/2 cm2 = 77 cm2

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

Area of the shaded region = Area of larger circle – Area of triangle – Area of larger semicircle + Area of smaller circle

Area of the shaded region = (154-49-77+77/2) cm2

= 133/2 cm2 = 66.5 cm2

## Solution:

ABC is an equilateral triangle.

∴ ∠ A = ∠ B = ∠ C = 60°

There are three sectors each making 60°.

Area of ΔABC = 17320.5 cm2

⇒ √3/4 ×(side)2 = 17320.5

⇒ (side)2 =17320.5×4/1.73205

⇒ (side)2 = 4×104

⇒ side = 200 cm

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

Radius of the circles = 200/2 cm = 100 cm Area of the sector = (60°/360°)×π r2 cm2

= 1/6×3.14×(100)2 cm2

= 15700/3cm2

Area of 3 sectors = 3×15700/3 = 15700 cm2

Thus, area of the shaded region = Area of equilateral triangle ABC – Area of 3 sectors

= 17320.5-15700 cm2 = 1620.5 cm2

## Solution:

Number of circular designs = 9 Radius of the circular design = 7 cm

There are three circles in one side of square handkerchief.

∴ Side of the square = 3×diameter of circle = 3×14 = 42 cm Area of the square = 42×42 cm2 = 1764 cm2

Area of the circle = π r2 = (22/7)×7×7 = 154 cm2 Total area of the design = 9×154 = 1386 cm2

Area of the remaining portion of the handkerchief = Area of the square – Total area of the design                                                                        = 1764 – 1386 = 378 cm2

## Solution:

(i) Area of quadrant OACB = (πR2)/4 cm2

= (22/7)×(7/2)×(7/2)/4 cm2

= 77/8 cm2

(ii) Area of triangle BOD = (½)×(7/2)×2 cm2

= 7/2 cm2

### YOU ARE READING: Areas Related to Circles Chapter 12 NCERT Solutions For Class 10 CBSE Mathematics

Area of shaded region = Area of quadrant – Area of triangle BOD

= (77/8)-(7/2) cm2 = 49/8 cm2

= 6.125 cm2

## Solution:

Side of square = OA = AB = 20 cm Radius of the quadrant = OB OAB is right angled triangle

By Pythagoras theorem in ΔOAB,

OB2 = AB2+OA2

⇒ OB2 = 202 +202

⇒ OB2 = 400+400

⇒ OB2 = 800

⇒ OB= 20√2 cm

Area of the quadrant = (πR2)/4 cm2 = (3.14/4)×(20√2)2 cm2 = 628cm2 Area of the square = 20×20 = 400 cm2

Area of the shaded region = Area of the quadrant – Area of the square

= 628-400 cm2 = 228cm2

## Solution:

Radius of the larger circle, R = 21 cm Radius of the smaller circle, r = 7 cm

Angle made by sectors of both concentric circles = 30° Area of the larger sector = (30°/360°)×πR2 cm2

= (1/12)×(22/7)×212 cm2

= 231/2cm2

Area of the smaller circle = (30°/360°)×πr2 cm2

= 1/12×22/7×72 cm2

=77/6 cm2

Area of the shaded region = (231/2) – (77/6) cm2

= 616/6 cm2 = 308/3cm2

## Solution:

Radius of the quadrant ABC of circle = 14 cm AB = AC = 14 cm

BC is diameter of semicircle. ABC is right angled triangle.

By Pythagoras theorem in ΔABC,

BC2 = AB2 +AC2

⇒ BC2 = 142 +142

⇒ BC = 14√2 cm

Radius of semicircle = 14√2/2 cm = 7√2 cm Area of ΔABC =( ½)×14×14 = 98 cm2

Area of quadrant = (¼)×(22/7)×(14×14) = 154 cm2

Area of the semicircle = (½)×(22/7)×7√2×7√2 = 154 cm2

Area of the shaded region =Area of the semicircle + Area of ΔABC – Area of quadrant

= 154 +98-154 cm2 = 98cm2

## Solution:

AB = BC = CD = AD = 8 cm

Area of ΔABC = Area of ΔADC = (½)×8×8 = 32 cm2

= 352/7 cm2

Area of shaded region = (Area of quadrant AECB – Area of ΔABC) = (Area of quadrant AFCD –