# Chapter 12 – Areas Related to Circles Questions and Answers: NCERT Solutions for Class 10 Maths

Class 10 Maths NCERT book solutions for Chapter 12 - Areas Related to Circles Questions and Answers.

Education Blogs Chapter 12 – Areas Related to Circles Questions and Answers: NCERT Solutions...

Class 10 Maths NCERT book solutions for Chapter 12 - Areas Related to Circles Questions and Answers.

The radius of the 1stcircle = 19 cm (given)

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

The radius of the 2ndcircle = 9 cm (given)

∴ Circumference of the 2ndcircle = 2π×9 = 18π cm

So,

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

Now, let the radius of the 3rdcircle = R

∴ The circumference of the 3rdcircle = 2πR

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

Hence, 56π = 2πR

Or, R = 28 cm.

Radius of 1stcircle = 8 cm (given)

∴ Area of 1stcircle = π(8)2= 64π

Radius of 2ndcircle = 6 cm (given)

∴ Area of 2ndcircle = π(6)2= 36π

So,

The sum of 1stand 2ndcircle will be = 64π+36π = 100π

Now, assume that the radius of 3rdcircle = R

∴ Area of the circle 3rdcircle = πR2

It is given that the area of the circle 3rdcircle = Area of 1stcircle + Area of 2ndcircle

Or, πR2= 100πcm2

R2= 100cm2

So, R = 10cm

The radius of 1stcircle, 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 2ndcircle, r2= 10.5cm+10.5cm = 21 cm

Thus,

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

= (π(21)2− 346.5) cm2

= 1386 − 346.5

= 1039.5 cm2

Similarly,

The radius of 3rdcircle, r3= 21 cm+10.5 cm = 31.5 cm

The radius of 4thcircle, r4= 31.5 cm+10.5 cm = 42 cm

The Radius of 5thcircle, r5= 42 cm+10.5 cm = 52.5 cm

For the area of nthregion,

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

∴ 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

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 = 11×105cm

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

=( 11×105)/80 π = 4375.

(A) 2 units

(B) π units

(C) 4 units

(D) 7 units

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.

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°)×πr2cm2

= (36/6)π cm2

= 6×22/7 cm2= 132/7 cm2

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 r2cm2

= (49/16) π cm2

= 77/8 cm2= 9.6 cm2

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°) × πr2cm2

= (1/12) × π142

= (49/3)×(22/7) cm2

= 154/3 cm2

(i) minor segment

(ii) major sector. (Use π = 3.14)

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

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

(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

Radius = 21 cm

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 r2cm2

= 441/6×22/7 cm2

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

(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×a2sq. Units.

Area of segment APB = 231-(√3/4)×(OA)2

= 231-(√3/4)×212

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

Radius = 15 cm

θ = 60°

So,

Area of sector OAPB = (60°/360°)×πr2cm2

= 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

Radius, r = 12 cm

Now, draw a perpendicular OD on chord AB and it will bisect chord AB.

So, AD = DB

= (120/360)×(22/7)×122

= 150.72 cm2

Consider the ΔAOB,

∠ OAB = 180°-(90°+60°) = 30°

Now, cos 30° = AD/OA

√3/2 = AD/12

Or, AD = 6√3 cm

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

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

= 150.72 cm2– 62.28 cm2= 88.44 cm2

(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)

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

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

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

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 = 285 mm

(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

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

6364.29/8 cm2

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

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

(Use π = 3.14)

Let O bet the position of Lighthouse.

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°)×πr2km2

= 189.97 km2

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)×a2sq. units

Here, a = OA

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

Area of sector ACB = (60°/360°)×πr2cm2

= 410.66 cm2

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

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

(A) p/180 × 2πR

(B) p/180 × π R2

(C) p/360 × 2πR

(D) p/720 × 2πR2

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.

Here, P is in the semi-circle and so,

P = 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

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

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

= 161.54 cm2

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°)×πr2cm2

= 68.44 cm2

Area of the sector OBD = (40°/360°)×πr2cm2

= (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

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

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

= 196 cm2-154 cm2

= 42 cm2

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= (√3/4)×122= 36√3 cm2

Area of the circle = πR2= (22/7)×62= 792/7 cm2

Area of the sector making angle 60° = (60°/360°) ×πr2cm2

= (1/6)×(22/7)× 62cm2= 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

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 = πR2cm2= (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

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

∴ AO = Radius of the circle = (2/3) AD

⇒ (2/3)AD = 32 cm

⇒ AD = 48 cm

In ΔADB,

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)2cm2= 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

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

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

= 196 cm2– 154 cm2

= 42 cm2

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.

Width of the track = 10 m

Distance between two parallel lines = 60 m

Length of parallel tracks = 106 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

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

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

Radius of the circles = 200/2 cm = 100 cm

Area of the sector = (60°/360°)×π r2cm2

= 1/6×3.14×(100)2cm2

= 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

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

(i) quadrant OACB,

(ii) shaded region.

Radius of the quadrant = 3.5 cm = 7/2 cm

(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

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

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

= 6.125 cm2

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)2cm2= 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

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°)×πR2cm2

= (1/12)×(22/7)×212cm2

= 231/2cm2

Area of the smaller circle = (30°/360°)×πr2cm2

= 1/12×22/7×72cm2

=77/6 cm2

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

= 616/6 cm2= 308/3cm2

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

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

Area of quadrant AECB = Area of quadrant AFCD = (¼)×22/7×82

= 352/7 cm2

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

= (352/7 -32)+(352/7- 32) cm2

= 2×(352/7-32) cm2

= 256/7 cm2

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