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Important Questions for CBSE Class 10 Maths Chapter 11 - Areas Related to Circles

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Crucial Practice Problems for CBSE Class 10 Maths Chapter 11: Areas Related to Circles

Class 10 Maths Chapter 11 Important Questions Areas Related to Circles are given here based on the latest pattern of CBSE for 2024-2025. Students who are preparing for the board exams can practice Areas Related to Circles Important Questions to score full marks for the questions from this chapter. Along with the Important Questions for Class 10 Maths Chapter 11 related to Circles, we have also provided Extra Questions of Chapter 11 Class 10 Maths. Students can refer to the solutions whenever they get stuck while solving a problem. Also, in the end, some practice questions are provided for students to boost their preparation for the exam.

Areas Related to Circles chapter contains many formulas and concepts thus, it is important from the examination perspective, since most of the questions and objective-based questions will come in the exam from this chapter. Students can also refer to Chapter 11 Maths Class 10 Important Questions while preparing for their exam. NCERT Solution is also available at Vedantu to help the students in scoring full marks in the board exam.

Area related to circles includes the area of a circle, segment, sector, angle and length of a circle are provided in this pdf. In this chapter, we will discuss the concepts of the perimeter (also known as circumference) and area of a circle and apply this knowledge in finding the areas of two special parts of a circular region (or a circle) known as sector and segment. You can download Maths NCERT Solutions Class 10 and NCERT Solution for Class 10 Science to help you to revise the complete Syllabus and score more marks in your examinations.

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Study Important Questions for Class 10 Maths Chapter 11- Area Related to Circles

Very Short Answer Questions (1 Mark)

Unless stated otherwise, take π=227.

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.

Ans: Let us suppose R, radius of the circle whose circumference is equal to the circumference of the two circles with radius 19 cm and 9 cm respectively.

Thus, 2πR=2π(19)+2π(9)

R=19+9

R=28 cm

Radius of the circle, R=28 cm.

 

2.   The circumference of a circular field is 528 cm. Then its radius is

a.   42 cm

b.   84 cm

c.   72 cm

d.   56 cm

Ans: b. 84 cm

Circumference of a circle is given by, 2πR.

2πR=528

R=84 cm

 

3. The circumference of a circle exceeds its diameter by 180 cm. Then its radius is

a.   32 cm

b.   36 cm

c.   40 cm

d.   42 cm

Ans: d. 42 cm

Let the diameter of the circle be D.

D=2R.

Circumference of a circle is given by, 2πR.

2πR=D+180

2πR=2R+180

R=42 cm

 

4. Area of the sector of angle 60 of a circle with radius 10 cm is

a.   52521 cm2

b.   52821 cm2

c.   52421 cm2

d.   None of these

Ans: b. 52821 cm2

Area of the sector of a circle =(θ360)×πr2.

Area =(60360)×π(10)2

Area =52821 cm2.

 

5.   Area of a sector of angle P of a circle of radius R is

a.   P180×2πR

b.   P180×πR2

c.   P360×2πR

d.   P720×2πR2

Ans: d. P720×2πR2

Area of the sector of a circle =(θ360)×πr2.

Area =(P360)×π(R)2

Area =P720×2πR2.

 

6. If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then

a.   R1+R2=R

b.   R1+R2>R

c.   R1+R2<R

d.   None of these

Ans: a. R1+R2=R

Circumference of a circle is given by, 2πR.

Thus, according to the given condition, 2πR1+2πR2=2πR

R1+R2=R.

 

7. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is

a.   22:7

b.   14:11

c.   7:22

d.   11:14

Ans. b. 14:11

The perimeter of a circle is given by, 2πR.

The perimeter of a square is given by, 4a.

Given, 2πR=4a.

Ratio of area of circle to area of square is πR2a2

πR2a2=4πR2π2R2

Hence, the ratio of their areas is, 14:11.

 

8. The circumference of a circular field is 154 m. Then its radius is

a.   7 m

b.   14 m

c.   7.5 m

d.   28 cm

Ans. Circumference of a circle is given by, 2πR.

2πR=154

R=24.5 m.

 

9. The area of a circle is 394.24 cm2. Then the radius of the circle is

a.   11.4 cm

b.   11.3 cm

c.   11.2 cm

d.   11.1 cm

Ans. c. 11.2 cm

The area of a circle is given by, πR2.

πR2=394.24

R=11.2 cm.

 

10. If the perimeter and area of circle are numerically equal, then the radius of the circle is

a.   2 units

b.   π units

c.   4 units

d.   7 units

Ans. a. 2 units

Perimeter / Circumference of a circle is given by, 2πR.

Area of a circle is given by, πR2.

Given, 2πR=πR2.

R=2 units.

 

11. The radius of a circle is 7π cm, then the area of the circle is

a.   154 cm2

b.   49π cm2

c.   22 cm2

d.   49 cm2

Ans. d. 49cm2

Area of a circle is given by, πR2.

Area =π(49π)

Area =49cm2.


12. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is

a.   31 cm

b.   25 cm

c.   62 cm

d.   50 cm

Ans. d. 25 cm

Let D (D=2R) be the diameter of the circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm.

πR2=π(24)2+π(7)2

R2=576+49

R=25 cm

 

13. The circumference of a circle is 528 cm. Then its area is

a.   22,176 cm2

b.   22,576 cm2

c.   23,176 cm2

d.   24,576 cm2

Ans. a. 22,176 cm2

Circumference of a circle is given by, 2πR.

2πR=528 cm

R=84 cm

Thus, area =π(84)2

Area =22,176 cm2

 

Very Short Answer Questions (2 Marks)

1. 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.

Ans: Area of a circle is given by, πR2.

Let R be the radius of the circle whose area is equal to the sum of the areas of the two circles of radii 8 cm and 6 cm, respectively.

Thus, we have: πR2=π(8)2+π(6)2

R2=64+36

R2=100

R=10 cm.

 

2. Figure depicts an archery target marked with its five scoring area 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 the five scoring regions.

Ans: Given, diameter of the region representing Gold score as 21 cm and the width of other bands as 10.5 cm.

Thus, radius of the Gold scoring region, R=212cm.

Area of a circle is given by, πR2.

Area of the Gold scoring region, Agold=π(212)2.

Agold=346.5 cm2

Area of scoring region = Area of previous circle – Area of next circle.

Area of Red scoring region, Ared=π(212+10.5)2π(212)2

Ared=1386346.5

Ared=1039.5 cm2

Area of Blue scoring region, Ablue=π(212+10.5+10.5)2(1039.5+346.5)

Ablue=3118.51386

Ablue=1732.5 cm2

Area of Black scoring region, Ablack=π(212+10.5+10.5+10.5)2(1039.5+346.5+1732.5)

Ablack=55443118.5

Ablack=2425.5 cm2

Area of White scoring region, Awhite=π(212+10.5+10.5+10.5)2(2425.5+1039.5+346.5+1732.5)

Awhite=8662.55544

Awhite=3118.5 cm2

 

3. 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?

Ans: Given, diameter of car wheel =80 cm.

Radius of the car wheel =40 cm.

Distance covered by the wheel in one revolution is equal to the circumference of the wheel.

2πR=2×227×40

Thus, distance covered by the wheel in one revolution is equal to 17607 cm.

Given, the distance covered in 1 hour =66km=6600000cm

Hence, distance covered by the wheel in 10 minutes =660000060×10=1100000cm.

Number of revolutions =Total distanceDistance covered in one revolution

Number of revolutions =1100000×71760

Hence, number of revolutions =4375.

 

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

Ans: Given, Angle of the sector is 60.

Radius, r=6 cm

Area of the sector of a circle =(θ360)×πr2.

Area =(60360)×π(6)2

Area =1327 cm2.

 

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

Ans: Given, the circumference of the circle as 22 cm.

We know, the circumference of a circle is given by, 2πR.

Therefore, 2πR=22

R=3.5 cm

For the quadrant of a circle, θ=900

 Area of quadrant is given by, A=θ360×πr2

9003600×227×72×72=778cm2.

Area of the quadrant of the circle =9.625 cm2.

 

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

Ans: Given, length of the minute hand (radius) , r=14 cm and from the given data we interpret the angle in the clock, θ=30.

Area swept by the minutes hand in 5 minutes, A=θ360×πr2

A=30360×227×14×14=1543cm2

A=51.33 cm2.

 

7. 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

Ans: Given, Radius, r =10 cm and the angle, θ=90.

Area of minor sector, A=θ360×πr2

A= 90360×3.14×10×10  = 78.5 cm2 

Area of ΔOAB, = 12 Base × Height.

12bh= 12×10×10  

50 cm2.

Area of minor segment can be obtained as,

Area of minor segment =  Area of minor sector   Area of OAB .

A=78.550 

A=28.5 cm2

ii. Major Segment

Ans: For major sector, radius = 10 cm and θ=36090=270

Area of major sector A= θ360×πr2 

 A=270360×3.14×10×10

A= 235.5 cm2 .

 

8. A horse is tied to a peg at one corner of a square shape grass field of side 15 m by means of a 5 m long rope (see figure). Find:

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

Ans: Area of a sector of a circle is given by, A=θ360×πr2

Area of the part of the field where the horse can graze is in the form of a quadrant with radius, r=5m.

For the quadrant of a circle, θ=900

Thus, the area is given as, A=90360×3.14×5×5

 A=19.625 m2.

ii. the increase in the grazing area if the rope were 10 m long instead of 5 m.

Ans: Area of quadrant with 10m rope, the radius would be, r=10m

Thus, the grazing area for 10 m long rope is given by,  A=90360×3.14×10×10 .

A=78.5 m2.

The increase in grazing area can be calculated by subtracting area of 5m rope from the area of 10m rope.

Hence, Increase in area =78.519.625=58.875 m2

 

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 figure. Find: 

i. The total length of the silver wire required. 

Ans: Given, diameter of silver wire circle as 35 mm 

Thus, radius is 352 mm.

Circumference of a circle is given by,  2πr.

Therefore, the circumference of the circle made by silver wire is,

 2×227 ×352 =110 mm   ……(1)

As the wire is used in making 5 diameters,

length of 5  diameters, L=35×5

L=175 mm   ……(2) 

Total length of the silver wire required would be the sum of circumference of the circle and the length of 5 diameters.

Hence, adding equation (1) and (2), we get 

110 + 175 = 285  mm

ii. The area of each sector of the brooch. 

Ans: Given, diameter of silver wire circle as 35 mm 

Thus, radius is 352 mm.

Angle of each sector,θ=36010=36.

The area of each sector of the brooch is given by, A=θ360×πr2

A=36360×227×352×352 

A=  3854mm2 

 

10. An umbrella has 8  ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of  radius 45 cm,  find  the area between  the  two  consecutive  ribs of  the umbrella. 

Ans: Let umbrella be a flat circle with radius,r= 45 cm.

Given, the umbrella has 8 ribs, thus the angle θ would be, θ=3608=45

Area between two consecutive ribs of the umbrella can be obtained using the formula, A=θ360×πr2.

A=45360×227×45×45 

A= 2227528 cm2

A=795.54 cm2.

 

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

Ans: Given, a car has two wipers with a blade length of 25 cm sweeping an angle of 115.

Let us suppose radius,  r = 25 cm and

Angle, θ=115.

The two wipers do not overlap.

The total area cleaned at each sweep of the blade will be,  2×(θ360×πr2)

2×(115360×227×25×25)=158125126cm2

Hence, area cleaned at each sweep is 1254.96 cm2.

 

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. (Useπ=3.14)

Ans: Given, a lighthouse spreads a red light over a sector of angle 80 to a distance of 16.5 km.

Let us suppose radius, r =16.5 km and

Angle, θ=80.

The area of sea over which the ships are warned is, A=θ360×πr2A=80360×3.14×16.5×16.5 

A=189.97 km2 

 

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

Ans: Area of shaded region ABDC can be determined by subtracting area of sector OBD from the area of sector OAC, i.e.,

Area of shaded region ABDC= Area of sector OAC Area of sector OBD

Area of sector is given by, A= θ360×πr2 .

Thus, Area of sector OAC=40360×227×(14)2 

Area of sector OBD=40360×227×(7)2  

Area of shaded region ABDC=40360×227×(14)240360×227×(7)2

Area=40360×227[(14)2(7)2]=40360×227×7×21

Area =1543 cm2.

Hence, area of the shaded region is 51.33 cm2.

 

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

Ans: From the given figure,

Area of shaded region = Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC) 

Area of the shaded region =14×14[12×227(142)2+12×227(142)2]

Area of the shaded region =196227×7×7

Area of the shaded region is 42cm2.

 

15. Find the area of the shaded region in figure, 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. 

Ans: Area of the shaded region = Area of circle + Area of equilateral triangle OAB - Area of the region common to the circle and the triangle.

Area of a circle is given by, πR2.

Area of equilateral triangle is given by, 34(a)2 (a is the side)

Area of the shaded region =π(6)2+34(12)260360×π(6)2

Area of the shaded region =30π+363=(6607+363)cm2

Hence, the area of the shaded region is 156.63 cm2.

 

16. 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 of 2 cm is cut as shown in figure. Find the area of the remaining portion of the figure:

Ans: Given, a square of side 4 cm. Four quadrants of a circle of radius 1 cm are cut from each corners of the square.

Also, a circle of diameter 2 cm is cut.

Area of the remaining portion = Area of square – (4× Area of quadrant of radius 1 cm + Area of circle of diameter 2 cm).

Area of a square is given as (side)2.

Area of a circle is given by, πR2.

Similarly, area of quadrant is given by (θ360)×πr2.

Thus,

Area of the remaining portion =4×4[4×90360×227×(1)2+227×(22)2]

Area of the remaining portion =162×227=687 cm2.

Hence, area of the remaining portion is 9.71 cm2.

 

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

Ans: Given, circular table cover of radius, r=32cm and an equilateral triangle ABC is drawn in the middle.

Area of the design can be obtained by,

Area of design = Area of circle – Area of the equilateral triangle ABC.

Area of a circle is given by, πR2.

Area of equilateral triangle is given by, 34(a)2 (a is the side).

Thus, area of design =π(32)234(a)2   ……(1)

G is the centroid of the equilateral triangle. 

radius of the circumscribed circle =  23h cm 

So, we have 32=23h

h=48 cm

From the figure of equilateral triangle, a2=h2+(a2)2.

3a24=h2

a2=4×(48)23

a=3072

Substituting for a in equation (1), we get:

Area of design =π(32)234(3072)2

227×10247683=(2252877683)cm2

Thus, area of the design is 1888.08 cm2.

 

18. In figure ABCD is a square of side 14cm. 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. 

Ans: Given, a square of side 14 cm. With the corners of the square as centres four circles are drawn.

Thus, area of the shaded region = Area of the square – (4× Area of sector)

Area of a circle is given by, πR2 and area of sector is given by, (θ360)×πr2Area of the shaded region =14×144×90360×227×(142)2=196227×7×7

Hence,areaoftheshadedregionis42\text{ }c{{m}^{2}}$.

 

19. The given figure depicts a racing track whose left and right ends are semi-circular. The difference between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10m wide, find: 

i. The distance around the track along its inner edge, 

Ans: Given that, difference between the two inner parallel line segments is 60m and they are each 106m long and the track is 10m wide.

Distance around the track along its inner edge 

106+106+2×[180360×2×227×(602)]=212+60×227

212+13207=28047m

ii. The area of the track. 

Ans: Area of track can be calculated as,

106×10+106×10+2×[12×227(30+10)212×227(30)2]

=1060+1060+227[(40)2(30)2]

2120+227×700=4320m2

 

20. In figure, 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. 

Ans: Given, radius of the bigger circle, OA or R=7 cm.

Diameter of the smaller circle is the radius of the bigger circle.

Thus, Radius of the smaller circle, r=72 cm.

Area of the circle is given by, πR2 and

Area of the semicircle is given by, πr22.

ΔACB is made up of two right- angled triangles and we know, area of a right-angled triangle is given as  12×base×height.

Area of the shaded region can be obtained as:

Area of shaded region = Area of smaller circle + Area of semicircle ACB  Area of ΔACB.

Area of shaded region =227×(72)2+12×227×(7)2(7×72+7×72)=1332

Thus, area of the shaded region is 66.5 cm2.

 

21. 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 figure). Find the area of the shaded region. (Use π=3.14 and 3=1.73205)

Ans: Given, area of the equilateral triangle ABC=17320.5 cm2.

Also, radius of each circle is half the length of the side of the equilateral triangle.

Area of the equilateral triangle is given by, 34×(side)2.

Let side of the equilateral triangle ABC be a.

34×a2=17320.5

a2=40000

a=200  cm

radius of circle, r=100 cm

Area of the shaded region can be obtained as:

Area of shaded region = Area of the equilateral triangle ABC  3× Area of the sector.

We know, area of the sector of a circle =(θ360)×πr2.

Thus,

Area of shaded region =17320.53×(60360)×3.14×(100)2

17320.515700=1620.5cm2

Hence, area of the shaded region is 1620.5 cm2.

 

22. On a square handkerchief, nine circular designs each of radius 7cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Ans: Given, radius of circular design is 7 cm.

Observing the figure, the side of the square is 6×7=42 cm.

The handkerchief consists of 9 circular designs.

Thus, area of the shaded region = Area of square ABCD  9× Area of circular design.

Area of shaded region =(42×42)9×π×(7)2

17641386=378 cm2

Hence, area of the shaded region is 378 cm2.

 

23. In figure, OACB is a quadrant of a circle with centre O and radius 3.5cm. If OD = 2 cm, find the area of the: 

i. quadrant OACB 

Ans: Given radius, r=3.5 cm and

OD=2 cm.

Area of the quadrant of a circle =(θ360)×πr2.

Area of the quadrant OACB=90360×227×(3.5)2

 14×227×3510×3510= 778cm2 

Area of the quadrant OACB is 9.625 cm2.

ii. shaded region 

Ans: Given radius, r=3.5 cm and

OD=2 cm.

Area of the shaded region = Area of the quadrant OACB  Area of the triangle BOD.

Area of the shaded region =9.62512×3.5×2=6.125 cm2

Hence, area of the shaded region is 6.125 cm2.

 

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

Ans: Given, a square OABC inscribed in a quadrant OPBQ and

OA=20 cm.

OA=AB=BC=OC.

Taking the ΔOAB into consideration and applying Pythagoras Theorem,

OB2=OA2+AB2

OB2=(20)2+(20)2

OB=800

OB=202 cm

Area of the shaded region = Area of quadrant OPBQ - Area of square OABC.

We know, area of the quadrant of a circle =(θ360)×πr2.

Area of shaded region =90360×3.14(202)220×20

14×3.14×800400=228 cm2

Hence, area of the shaded region is 228 cm2.

 

25. AB and CD respectively are the arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If AOB=30. Find the area of the shaded region.

Ans: Given, two arcs AB and CD of two concentric circles with radii 21 cm and 7 cm respectively.

Also, AOB=30

We know, area of the quadrant of a circle =(θ360)×πr2.

Area of the shaded region = Area of arc AB - Area of arc CD.

Area of shaded region =(30360)×227×(21)2(30360)×227×(7)2

(30360)×227×[(21)2(7)2]=3083 cm2

Hence, area of the shaded region is 102.67 cm2.

 

26. In figure, 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.

Ans: Given, a quadrant ABC of a circle of radius 14 cm.

Let us take, ΔABC and apply Pythagoras Theorem.

AB2+AC2=BC2

(14)2+(14)2=BC2

BC=392

BC=142 cm.

Radius of the semicircle =1422=72 cm.

We know, area of the quadrant of a circle =(θ360)×πr2.

Area of the shaded region BPCQ= Area of BCQB - (Area of BACP - Area of ΔABC).

Area of shaded region =180360×227(72)2[90360×227×(14)214×142] 

12×227×98(14×227×19698)=98 cm2

Hence, area of the shaded region is 98 cm2.

 

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

Ans: Given, two quadrants each of radii 8 cm which has a design in the common region.

Let us consider the figure as:

Applying Pythagoras Theorem in the right- angled triangle ABC, AC2=AB2+BC2

 AC2= (8)2+ (8)2= 2(8)2 

  AC=128=82  cm 

Draw a perpendicular, BMAC. 

Then AM=MC=  12AC 

 12×82 = 42 cm 

From the figure the right triangle AMB, 

(AB)2 = (AM)2+(BM)2 [Pythagoras theorem] 

(8)2=(42)2+BM2

BM2=6432=32

BM =42  cm

Area of  ΔABC =12  × AC × BM 

 82×422 = 32 cm2 

We know, area of the quadrant of a circle =(θ360)×πr2.

Area of half shaded region = Area of quadrant ABC - Area of ΔABC.

Area of half shaded region =90360×227×(8)232

 16×22732 = 1287 cm2 

Thus, area of designed region can be obtained as, twice the area of half shaded region.

2 ×1287  = 2567 cm2 

Hence, area of the designed region is 36.57 cm2.

 

28. Find the circumference of a circle of diameter 14cm. 

Ans: Given, diameter of circle, d=14 cm

Circumference of a circle is given by, 2πr

Since, d=14 cm

r=142=7 cm

r=7cm

Thus, the circumference of the circle is, C=2πr

C=2×227×7

C=44 cm

The circumference of the circle is 44 cm.

 

29. The diameter of a circular pond is 17.5m. It is surrounded by a path of width 3.5m. Find the area of the path. 

Ans: Given, diameter of circular pond, d=17.5 m.

Radius, r=17.52=8.25 m

It is surrounded by a path of width 3.5 m.

Outer radius R is the sum of circular pond radius with the width of the surrounded path.

R=8.75+3.5=12.25 m

Equation for area of path A  is given by π[R2r2].

A=π[(R+r)(Rr)]

A=227[(12.25+8.75)(12.258.75)]

A=227×21×3.50

A=231 m2

Hence, area of the path is  231 m2.

 

30. Find the area of the shaded region where ABCD is a square of side 14 cm.

Ans: Given, side of square, a=14 cm.

Area of the shaded region can be calculated by subtracting area of 4 circles from the area of square ABCD, i.e.,

Area of shaded region = Area of square ABCD  4×Area of circle.

Area of the given square ABCD is given by, A=14cm×14cm

A=196cm2

From the figure, we observe that the diameter of each circle is half the side of the square.

Thus, diameter of circle =142=7 cm

  Radius of each circle =72cm

Area of the circle =πr2  

227×(72)2cm2   

Area of shaded region =196 cm2(4×38.5) cm2 

(196154)cm2=42 cm2

Hence, area of the shaded region is 42 cm2.

 

31. The radius of a circle is 20cm. Three more concentric circles are drawn inside it in such a manner that it is divided into four parts of equal area. Find the radius of the largest of the three concentric circles. 

Ans: Given radius of the circle, r=20 cm

Area of a circle is given by, A=πr2

Substituting for radius, r.

A=π(20)2 

A=400πcm2

As the circle is divided into four parts of equal area, area of each circle would be,

14400πcm2=100πcm2

Let R be the radius of the largest of the three circles and its area is 400π100π=300πcm.

Thus the radius, R can be calculated as

πR2=300π

 R2=300

R=300 =103 cm

Radius of the largest of the three concentric circles is 103 cm.

 

32. OACB is a quadrant of a circle with centre O and radius 7 cm. If OD = 4cm, then find area of shaded region. 

Ans: Given radius r=7 cm and OD=4 cm

Area of quadrant is given by, A=θ360×πr2

From the figure it is clear that, angle θ=90

Area of quadrant  OACB=90360π(7)2

14×227×49=772cm2

Area of ΔOAD=12(7×4)

Area of shaded region = Area of quadrant OACB Area of ΔAOB

77212(7×4)=492cm2

Thus, area of the shaded region is 24.5 cm2.

 

33. A pendulum swings through on angle of 30and describes an arc 8.8 cm in length. Find the length of pendulum. 

Ans: Given, pendulum swings at angle 30 , with an arc of length 8.8 cm.

Sketching the figure, we have:

In the below given figure let r be the length of pendulum and $\angle

AOB={{30}^{{}^\circ }}then,\angle AOB=\dfrac{\pi }{{{180}^{{}^\circ }}}\times {{30}^{{}^\circ }}$

AOB=π6

Use θ=lr to determine the length of the pendulum.

π6=8.8r

 r=8.8×6π

r=16.8cm

The length of the pendulum is 16.8 cm.

 

34. The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280. The field is to be ploughed at the rate of Rs.0.50per m2. Find the cost of ploughing the field. (Take π=227)

Ans: As per the question the cost of fencing a circular field at the rate of Rs24, per meter is Rs 5280.

With Rs. 24, the length of fencing =1 metre 

for Rs.5280, the length fencing=124×5280=220meters

Perimeter i.e., circumference of the field =220meters 

Let r be the radius of the field.

Circumference is given by, 2πR 

2πr=220

r=220×72×22=35m

Area of the field, A=πr2.

π(35)2=1225πm2

Rate = Rs.0.50 per m2 

Total cost of ploughing the field

Rs.(1225π×0.50)=Rs.1225×22×17×2  

Rs.(175×11)=Rs.1925

Hence, the cost of ploughing the field is Rs.1925.

 

35. Find the difference between the area of regular hexagonal plot each of whose side 72 m and the area of the circular swimming take in scribed in it. (Take π=227)

Ans: Given, side of hexagonal plot =72m

Area of equilateral triangle is given byA=34(side)2

AΔAOB=34(72)2=12963m2

Area of hexagonal plot =6× Area of triangle AOB 

6×12963=13468.032 m2

Now,OC2=OA2AC2

(72)2(722)2 =51841296=3888

OC2=3888

OC=3888=62.35 m

Area of circular region, A=πr2 

227×(62.35)2=12,219.42 m2.

Then the difference is, 13468.032 m212,219.42 m2=1248.603 m2.

 

36. In the given figure areas have been drawn of radius 21cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

Ans: Given, areas drawn with radius 21 cm at vertices, A,B,C and D.

Thus, required area is the area of the circle with radius 21 cm.

A=π(21)2=227×21×21 

A=1386cm2

 

37. A wheel has diameter 84cm,  find how many complete revolutions it must make to complete 792 meters. 

Ans: Given, diameter of wheel, d=2r = 84 cm.

Distance covered in one revolution = circumference

Circumference of the circle is given by 2πr.

Distance covered in one revolution =2πr=π(2r)=227×84

22×12=264 cm

Thus, for distance covered264cm, number of revolutions=1

For distance covered 792m/ 79200 cm, number of revolutions=1264×79200=300 .

Number of revolutions to complete 792 m is 300.

 

38. The given figure is a sector of a circle of radius 10.5cm. Find the perimeter of the sector. (Take π=227

Ans: Given radius, r=10.5 cm

We know that circumference, i.e., perimeter of a sector of angle P of a circle with radius R is given by, P360×2πR+2R.

Required perimeter can be calculated as,

60360×2×227×(10.5)+2(10.5)=16×447×212+21 

=32cm

Thus, the perimeter of a sector is 32 cm.

 

39. A car had 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. 

Ans: Given, radius of each wiper= 25 cm and sweeping angle, θ=115.

Since, area of sector is given by, A=θ360×πr2

Total area cleaned at each sweep of the blades =2(115360×227×25×25)

230×22×5×2572×7=230×11×12536×7

115×11×12518×7=1254.96 cm2

Thus, the total area cleaned at each sweep of the blades is 1254.96 cm2

 

40. In the given figure arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaped region. 

Ans: Given, arc with radius 14 cm each.

Area of sector is given by, A=θ360×πr2

From the figure area of shaded region is,A=3×60360×π(14)2=227×(14)22

A=11×2×14=308cm2

Thus, area of shaded region is 308 cm2.

 

41. The radii of two circles are 19cm and 9cm respectively. Find the radius of the circle which has its circumference equal to the sum of the circumference of the two circles.

Ans: Given, radii of two circles as 19 cm and 9 cm.

Circumference of the circle is given by C=2πr.

C1 = circumference of the 1st circle

2π(19)= 38π cm 

C2 = circumference of the 2nd circle

2π(19)=18π cm 

Circumference of the required circle = sum of circumferences of two circles

2πr=38π+18π cm 

2πr=56π

r=28cm

Thus, the radius of the required circle is 28 cm.

 

42. A car travels 0.99km distance in which each wheel makes 450 complete revolutions. Find the radius of its wheel 

Ans: Given, distance travelled by a wheel in 450 complete revolutions =0.99 km = 990m

Distance travelled in one revolution =990450=115.

Let r be the radius of the wheel.

Circumference of the circular field is given by, 2πr.

2πr=115

2×227r=115

r=7×10020=35m.

 

43. A sector is cut from a circle of diameter 21cm. If the angle of the sector is 150 find its area. 

Ans: Given, diameter, d= 21cm 

Thus, radius can be obtained as,  r=  212cm 

Angle of sector =150 

Area of the sector is given by,  A=θ360×πr2.

Hence, area  of the sector is, A=512×227×212×212=5×11×214×2 

A=144.38 cm2.

 

44. In the given figure AOBCA represent a quadrant of area  9.625cm2.  Calculate the area of the shaded portion. 

Ans: Given, area of quadrant 9.625 cm2.

From the figure, OD=2 cm and OA=3.5 cm.

Area of the quadrant of a circle =(θ360)×πr2.

Area of the shaded region = Area of the quadrant OACB  Area of the triangle BOD.

Area of the shaded region =9.62512×3.5×2=6.125 cm2

Hence, the area of the shaded region is 6.125 cm2.

 

Short Answer Questions (3 Marks)

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

Ans: Given PQ=24 cm and PR=7 cm

Since QR is a diameter passing through the centre O of the circle 

Angle of semi-circle, RPQ=90

QR2=PR2+PQ2

72+242=625=252

QR=25 cm

Diameter of the circle = 25cm 

Thus radius, r=252cm

Area of the semi-circle=12πr2

Area =12×227×252×252=11×62528cm2

Also, area of  ΔPQR=12PR×PQ

12×7×24=84cm2

From the given figure,

 Area of the shaded region = Area of the semi-circle with O as centre and OQ as radius – area of ΔPQR

Hence, area of the shaded region can be calculated as,

Area of the shaded region =11×6252884=11×62584×2828

6875235228=452328

Area of the shaded region  =161.54cm2.


2. The perimeter of a sector of a circle of radius 5.7 m is 27.2 m.Calculate:

(i). The length of arc of the sector in cm.

Ans: Given radius, r=5.7 m and perimeter =27.2 m. 

Then OA=OB=5.7m 

Now, 

OA+OB+arcAB=27.2m

5.7m+5.7m+arcAB=27.2 m

arcAB=27.211.4=15.8 m

The length of the arc of the sector is 15.8 m.

ii. The area of the sector in cm2 correct to the nearest cm2 

Ans: Area of sector OACB=(12×radius×arc)

(12×5.7×15.8)=45.03cm2

Area of sector correct to nearest cm2=45cm2

 

3. Find the area of the shaded region in the given figure where ABCD is a square of side 10cm  and semi-circles are drawn with each side of the square as diameter. [π=3.14]

Ans: Given, side of square ABCD =10 cm                                                     

Area of square ABCD =(side)2=(10)2=100 cm2                           

Given that semicircle is drawn with side of square as diameter.   

Diameter of semicircle =10 cm.                                                                  

Radius of semicircle =5 cm                                                                      

Area of semi-circle =12πr2                                                                       

Area =12×3.14×(5)2=39.25 cm2                                                                 

Area of 4 semi-circles – Area of shaded region = Area of square ABCD  

Area of shaded region = Area of 4 semi-circles - Area of square ABCD 

Area of shaded region =4×39.25100=57cm2

 

4. In the given figure ΔABC  is an equilateral triangle inscribed in a circle of radius 4 cm and centre O. Show that the area of the shaded region is 43(4π33)cm2.

Ans: Given, ΔABC as an equilateral triangle inscribed in circle with radius, r=4 cm.

In  ΔOBD, Let BD=a,OB=4cm

32=a4

a=432

a=23 cm

BC=2a=2×23=43 cm

OD4=cos60OD=4×12=2 cm

Area of shaded region = Area of sector OBPC – Area of ΔOBC

120360×π×4212×43×2=43[4π33]cm2

 

5. The radii of two circles are 8cmand 6cmrespectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles. 

Ans: Given, radii of two circles, r1=8 cm and r2=6 cm.

Area of a circle is given by πr2

Let A1 = Area of the first circle

A1=π(8)2=64πcm2 

A2=Area of the second circle

A2=π(6)2=36πcm2  

Total area =A1+A2,

64π+36π=100πcm2  

Let R be the radius of the circle with area A1+ A2 ,

πR2=100π

R2=100

R=10cm

Hence, required radius = 10cm 

 

6. A chord of a circle of radius 10cm subtends a right angle at the centre. Find the area of the corresponding: (Use π=3.14

i. minor sector

Ans: Given, radius of circle, r=10 cm subtending an angle 90 at the centre. AOB=90.

Area of minor sector =θ360πr2

90360×3.14×(10)2=14(3.14)(100)  

3144=78.5cm2

Area of minor sector is 78.5cm2.

ii. major sector 

Ans: Area of circle =πr2=π(10)2

Area of major sector = Area of circle – Area of minor sector 

π(10)290360π(10)2=(3.14)(100)14(3.14)(100)

31478.50=235.5cm2

Area of major sector is 235.5 cm2.

iii. minor segment

Ans:  Area of minor segment = Area of minor sector OAB – Area of ΔOAB

Area of ΔOAB=12(OA)(OB)sinAOB

=12(OA)(OB)(AOB=90)

Area of sector =θ360πr2

14(3.14)(100)50=25(3.14)50

78.5050=28.5cm2

Area of minor segment is 28.5 cm2.

iv. major segment 

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

π(10)225.5=100(3.14)28.5

31418.5=285.5cm2

Area of major segment is 285.5 cm2.

 

7. In Akshita’s house, there is a flower pot. The sum of radii of circular top and bottom of a flowerpot is 140 cm and the difference of their circumference is88cm, find the diameter of the circular top and bottom. 

Ans: Given, sum of radii of circular top and bottom = 140cm 

Let radius of top =r cm 

Radius of bottom=(140r)cm

Circumference of top=2πr cm 

Circumference of bottom=2π(140r)cm

Difference of circumference =[2πr2π(140r)]cm  

By the given conditions in the problem, 

2πr2π(140r)=88

2π[r140+r]=88

2r140=882(227)

2r=140+14154

r=1542

Radius of top is obtained as 77cm 

Diameter of top can be calculated as,

2×77=154cm

Radius of bottom can be obtained as,

 140  r=140  77

=63cm 

Diameter of bottom 2×63=126cm .

 

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

Ans: Given, radius of circle,r=15 cm subtending an angle 60 at centre.

AOB=60.

Area of a sector of a circle is given by, A=θ360×πr2

Area of minor segment = Area of minor sector OAB area of ΔOAB 60360π(15)234(15)2=16(3.14)(225)(1.73)4(225)

117.7597.3125=20.4375 cm2

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

π(15)220.4375=(3.14)(225)20.4375

706.520.4375=686.0625 cm2

Area of minor segment is 20.4375 cm2 and area of major segment is 686.0625 cm2.

 

9. A round table cover has six equal designs as shown in the figure. If the radius of the cover is28cm , find the cost of making the design at the rate of Rs.0.35  per cm2  . (Use 3=1.7

Ans: Given, radius of cover, r=28 cm. Table has six equal designs and the cost of making the design is Rs. 0.35 per cm2.

Area of equilateral triangle is given by,A=34a2 .

From the figure area of one design =60360×π(28)2– Area of equilateral ΔOAB

π6×(28)234(28)2=(28)2(π634)

(28)2(227×61.74) 

Total cost can be calculated as follows,

6×(28)2(22421740)×0.35×44035721×40

21×(28)2(11211740)=21×(28)2100×44035721×40

28×28×8340=7×28×831000

Rs.16.268

The total cost of making the design is Rs.16.268.

 

10. ABCD is a flower bed. If OA = 21m and DC = 14m. Find the area of the bed.

Ans:  Given,

OA = R = 21m

 OC = r = 14m 

Area of the flower bed (i.e., shaded portion) = area of quadrant of a circle of radius R of the quadrant of a circle of radius r

14πR214πr2=π4(R2r2)

14×227[(21)2(14)2]m2

14×227×(21×14)(2114)m2

14×227×35×7 m2

192.5 m2

Hence, area of the flower bed is 192.5 m2.

 

11. In a circle of radius 21 cm, an arc subtends an angle of 60 at the centre. Find:

i. the length of the arc. 

Ans: Given radius, r= 21 cm and angle θ=60.

Length of arc =θ360×2πr 

60360×2×227×21  = 22 cm 

ii. area of the sector formed by the arc. 

Ans: Area of the sector =θ360×πr2

 60360×227×21×21  = 231 cm2

iii. area of the segment formed by the corresponding chord 

Ans: Area of segment formed by corresponding chord = θ360×πr2 Area of  ΔOAB 

Area of segment =231  Area of ΔOAB   …(1)

In right triangles OMA and OMB

OA = OB [Radii of same circle] 

OM = OM [Common] 

ΔOMAΔOMB [RHS congruency] 

 AM = BM [By CPCT] 

Here M is the midpoint.

  12AB and  AOM =  BOM

12AOB =12×60=30

In right angled ΔOMA,

cos30=OMOA

32=OM21

OM=2132 cm

Also sin30=AMOA

12=AM21

2AM=21cm

AB=21cm

  Area of ΔAOB=12×AB×OM

12×21×2132=44134cm2

Using eq. (1),

Area of segment formed by corresponding chord = (23144134) cm2 .


12. A chord of a circle of radius 15 cmsubtends an angle of 60 at the centre. Find the area of the corresponding segment of the circle.  (Use π=3.14 and 3=1.73)

Ans: Given radius, r = 15 cm and angle θ=60

Area of minor sector can be obtained from  θ360×πr2 

60360×3.14×15×15=117.75cm2

For, Area of ΔAOB

Draw OM, where OMAB

In right triangles OMA and OMB

OA = OB [Radii of same circle] 

OM = OM [Common] 

ΔOMAΔOMB [RHS congruency] 

 AM = BM [By CPCT] 

  12AB and  AOM =  BOM

12AOB =12×60=30

 In right angled ΔOMA,

cos30=OMOA

32=OM15

OM=1532 cm

Also, sin30=AMOA

12=AM15

2AM=15cm

AB=15cm

Area of ΔAOB=12×AB×OM

12×15×1532=22533

225×1.734=97.3125cm2

Area of minor segment = Area of minor sector – Area of  ΔAOB

117.75  97.3125 = 20.4375 cm2 

And, Area of major segment  =πr2 Area of minor segment 

706.5  20.4375 = 686.0625 cm2 

 

13. 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 )

Ans: Given radius, r  = 12cm and angle θ=120.

Area of corresponding sector can be obtained by using the formula θ360×πr2

120360×3.14×12×12= 150.72 cm2

For, Area of ΔAOB

Draw OM, where OMAB

In right triangles OMA and OMB

OA = OB [Radii of same circle] 

OM = OM [Common] 

ΔOMAΔOMB [RHS congruency] 

 AM = BM [By CPCT] 

  12AB and  AOM =  BOM

12AOB =12×120=60

In right angled ΔOMA,

cos60=OMOA

12=OM12

OM=6 cm

Also  sin60=AMOA

32=AM12

2AM=123cm

AB=123cm

Area of ΔAOB=12×AB×OM

12×123×6=363

36×1.73=62.28cm2

Area of corresponding segment = Area of corresponding sector – Area of ΔAOB

150.72  62.28 = 88.44 cm2 

 

14. Find the area of the shaded region  in  figure,  if PQ = 24 cm, PR = 7 cm and Ois  the centre of the circle. 

Ans: Given radius, r  = 12cm and angle θ=120.

Area of corresponding sector can be obtained by using the formula θ360×πr2

120360×3.14×12×12= 150.72 cm2

For, Area of ΔAOB

Draw OM, where OMAB

In right triangles OMA and OMB

OA = OB [Radii of same circle] 

OM = OM [Common] 

ΔOMAΔOMB [RHS congruency] 

 AM = BM [By CPCT] 

  12AB and  AOM =  BOM

12AOB =12×120=60

In right angled ΔOMA,

cos60=OMOA

12=OM12

OM=6 cm

Also  sin60=AMOA

32=AM12

2AM=123cm

AB=123cm

Area of ΔAOB=12×AB×OM

12×123×6=363

36×1.73=62.28cm2

Area of corresponding segment = Area of corresponding sector – Area of ΔAOB

150.72  62.28 = 88.44 cm2 

 

Long Answer Questions (4 Marks)

1. Three horses are tethered with 7m long ropes at the three corners of a triangular field having sides 20m, 34m and  42 m . Find the area of the plot which can be grazed by the horses. Also, find the area of the plot which remains ungrazed. 

Ans: Given, length of ropes at the three corners, l=33 m and the sides of the triangle as 20 m, 34 m and 42 m.

From the figure let, 

A=θ1, B=θ2 and C=θ3

Area which can be grazed by three horses = Area of sector with central angle θ1 and radius 7 m+ Area of sector with central angle θ2 and radius 7 cm+ Area of sector with central angle θ3 and radius 7 cm 

πr2θ1360+πr2θ2360+πr2θ3360

πr2360(θ1+θ2+θ3)

πr2360×180

( Sum of three angles of a  Δ=180)

227×7×7×180360=77m2 

Sides of plot ABC are given in the problem as,

 a= 20m , b = 34m and c=42m

Semi perimeter,s=20+34+422=48m 

Area of triangular plot = Area of ΔABC

s(sa)(sb)(sc)=48×28×14×6=336m2

Area grazed by the horses = 77m2 

Ungrazed area can be calculated by subtracting area grazed by the horses from the area of the triangular plot.

Ungrazed area =(33677) = 259 m2 

 

2. In the given figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m.  If the corner of each circular flower bed is the point of enter section O of the diagonals of the square lawn. Find the sum of the areas of the lawn and the flower beds. 

Ans: Given, side of square lawn =56 m.

Area of the square lawn ABCD, A=(56)2

 Let OA=OB=x

By Pythagoras theorem, 

x2+x2=562

2x2=56×56

x2=28×56

Again, area of sector OAB =90360×πx2=14πx2

Substituting for x, we get

Area of sector OAB=14×227×28×56m2 

Area of  ΔOAB=1456×56m2

(Here, AOB=90 since square ABCD is divided into 4 right triangles)

Area of flowerbedAB=(12×227×28×561456×56)m2

14×28×56(2272) 

7×56×87

Area of flower bed  AB=448m2

Similarly, area of the other flowerbed = 448m2 

Total area of the lawn and the flowerbeds =56×56+448+448=4032m2

 

3. An elastic belt is placed round the rein of a pulley of radius 5 cm. One point on the belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from O. Find the length of the best that is in contact with the rim of the pulley. Also find the shaded area. 

Ans: Given, radius of rein of pulley, OA=5 cm.

OP=10 cm

Thus, cosθ=OAOP

cosθ=510=12 

θ=60

AOB=120

ArcAB=120×2×π×5360 cm

10π3 cm[l=θ360×2πr]

Length of the belt that is in contact with the rim of the pulley can be calculated by subtracting length of arc AB from Circumference of the rim.

Length of the belt =2π×5cm103πcm

Length of the belt =203πcm

Now, area of sector OAQB=120360×π×52cm=253πcm2 [Since Area=θ360×πr2]

[AP=10025=75=53cm]

Area of quadrilateral OAPB is given by, 2(Area of ΔOAP)=253cm2

Hence, shaded area can be obtained as,

Area of shaded region=253253π

Hence, area of shaded region is 17.12 cm2.


Download Class 10 Maths Chapter 11 Important Questions PDF

Class 10 Maths Chapter 11 Important Questions Areas Related to Circles are given here based on the latest pattern of CBSE for 2024-2025. Students who are preparing for the board exams can practice Areas Related to Circles Important Questions to score full marks for the questions from this chapter. Along with the Important Questions for Class 10 Maths Chapter 11 related to Circles, we have also provided Extra Questions of Chapter 11 Class 10 Maths. Students can refer to the solutions whenever they get stuck while solving a problem. Also, in the end, some practice questions are provided for students to boost their preparation for the exam.

Areas Related to Circles chapter contains many formulas and concepts thus, it is important from the examination perspective, since most of the questions and objective-based questions will come in the exam from this chapter. Students can also refer to Chapter 11 Maths Class 10 Important Questions while preparing for their exam. NCERT Solution is also available at Vedantu to help the students in scoring full marks in the board exam.

Area related to circles includes the area of a circle, segment, sector, angle and length of a circle are provided in this pdf. In this chapter, we will discuss the concepts of the perimeter (also known as circumference) and area of a circle and apply this knowledge in finding the areas of two special parts of a circular region (or a circle) known as sector and segment. You can download Maths NCERT Solutions Class 10 and NCERT Solution for Class 10 Science to help you to revise the complete Syllabus and score more marks in your examinations.


Area of a Circle

The area is the region enclosed by the circumference of the circle. The area covered by one complete cycle of the radius of the circle on a two-dimensional plane is known as the area of that circle.

Area of the circle = πr2 square units.

Where r is the radius of the circle.


Area of Semicircle

To find the area of the semi-circle we need half of the area of the circle.

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The formula for the area of semicircle = πr22 square unit.

Where r is the radius of the circle.


Area of a Ring

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We know the area of circle πr2. Here the ring is having two radii inner and outer radius. Area of the ring i.e. the coloured part shown in the above figure is calculated by subtracting the area of the inner circle from the area of the bigger circle.

Area of the ring = πR2 - πr2 = π(R2 - r2)

Where R = radius of outer circle

r = radius of the inner circle


The Sector of a Circle

The sector of a circle is the region of a circle enclosed by an arc and two radii. The smaller area is called the minor sector and the larger area is called the major sector of the circle.


Areas of Sectors of a Circle

The area formed by an arc and the two radii joining the endpoints of the arc is known as a sector of a circle.

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In the above diagram, OACB is the minor sector. OADB is the major sector.


Minor Sector

The area including arc  ∠AOB with point C is called Minor Sector. So, OACB is the minor sector. The area included in ∠AOB is the angle of the minor sector.

Area of the sector of angle θ=θ360×πr2


Major Sector

The area included in ∠AOB with point D is called the Major Sector. So OADB is the major sector. The angle of the major sector is 360° – ∠AOB.

If the degree measure of the angle at the center is 360, area of the sector = πr2.

Formula to find the area of Major Sector = πr2 -  Area of the Minor Sector

Area of the circle can also be calculated by adding an area of a minor sector and area of a major sector.

Note: Area of Minor Sector + Area of Major Sector = Area of the Circle


To Calculate Length of an Arc of a Sector of Angle θ

An arc is the piece of the circumference of the circle so the length of an arc can be calculated as the θ part of the circumference.

Formula to calculate the length of an arc of a sector of angle θ=θ360×2πr.


Areas of Segments of the Circle

The area made by an arc and a chord of a circle is called the segment of the Circle.

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Minor Segment

The minor segment is defined as the region bounded by the chord and the minor arc.

In the above diagram, AB is a chord of the circle. The area made by chord AB and arc X is known as the minor segment.

Formulas to calculate Area of Minor Segment = Area of Minor Sector – Area of ∆ABO

=θ360×πr212r2sinθ


Major Segment

The major segment is defined as the region bounded by the chord and the major arc.

Formulas to calculate Area of Major Segment = πr2 -  Area of Minor Segment

Note: Area of major segment + Area of minor segment = Area of the circle.


Areas of Combinations of Plane Figures

A plane figure is a geometric figure that has no thickness, it lies entirely on one plane. In this section, we will learn how to calculate the areas of some plane figures which are combinations of more than one plane figure.

To find the area of the composite area of the shapes, simply find the area of each shape and add them together. The order in which we calculate the areas does not matter and the commutative property states that it does not matter which order you add them in.

Following are the steps to find the area of combined figures:

  • Step I: First we divide the given combined figure into its simple geometrical shapes.

  • Step II: Then calculate the area of simple geometrical shapes separately.

  • Step III: Finally, to find the area of the combined figure we need to add or subtract these areas.

Below are some examples of combinations of plane figures:

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Above diagram is a combination of one rectangle and two semi-circles. So, to find an area of the figure, first, find an area of the rectangle and then add the area of two semi-circles.

Area of the given figure = Area of the rectangle + 2 x Area of semi-circles.

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Above diagram is the form of a ring.

Area of the  required region = 34(πr12πr22)

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Above diagram is a combination of circle and square. To find the area of a given figure, first, calculate the area of the circle then subtract the area of the square.

Area of the required region = πr2 - a2

Important formula to remember:

  1. Area of circle = πr2

  2. Length of an arc of a sector of a circle with radius r and angle with degree measure θ is θ360×2πr.

  3. Area of a sector of a circle with radius r and angle with degrees measure θ is θ360×πr2

  4. Area of the segment of a circle = Area of the corresponding sector – Area of the corresponding triangle.

 

Conclusion:

Refer to this Important Questions For Class 10 Maths Chapter 11 for boards exam preparation. All the important concepts are available in the given PDF. The formula for the area of a circle also helps us calculate the area of circle sectors and segments as well. Download important questions of ch 11 Maths Class 10 from Vedantu site or app for more information.


Conclusion

Vedantu's offering of "Important Questions for CBSE Class 10 Maths Chapter 11 - Areas Related to Circles" is a valuable resource for students seeking to excel in their mathematics studies. The comprehensive collection of questions and solutions provides a deep understanding of the intricate concepts related to circles and their areas. By using this platform, students can gain confidence in tackling complex problems, enhance their problem-solving skills, and prepare effectively for their CBSE Class 10 examinations. Vedantu's commitment to quality education is evident in the well-structured and engaging content, making it an indispensable tool for students aiming to achieve academic success in mathematics.


Related Study Materials for Class 10 Maths Chapter 11 Areas Related to Circles


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FAQs on Important Questions for CBSE Class 10 Maths Chapter 11 - Areas Related to Circles

1. How many sums are there in class 10 Maths Chapter 11 Areas Related to Circles?

Chapter 11 of Maths for class 10 teaches about the application of the chapter ‘Circles’. There are a total of 35 sums in all the exercises combined together from the chapter. All of these problems are unique in their own way and offer excellent practice and knowledge to students. At times, however, these can get very confusing. The Important questions by Vedantu for this chapter can help students here because of their simple language structure.

2. What students will study in chapter 11 Areas Related to Circles of class 10 Maths?

In this chapter, students learn the different applications of circles in terms of their areas. Concepts like the area of circles, their circumference, and the various parts of it, segment, sector, angle, and length, everything is discussed in this chapter. Important Questions for chapter 11 is specifically designed to focus on the important areas by providing students with questions so that they can practice with them and score good results for questions from this chapter by Vedantu.

3. What are the benefits of using important questions for chapter 10 of class 11 Maths?

Important Questions provided by Vedantu are good for students to learn ‘Areas Related to Circles’. They can access the questions free of cost by visiting the link-Important Questions for class 10 maths, and they can be assured of their quality content as each question is kept up to date and follows the latest curriculum. Moreover, the easily understandable structure provides them with a good practice experience for good marks. Since these questions are from the perspective of the Board exam paper, these are excellent sources for studying.

4. State the important formulae in chapter 11 of class 10 Maths.

Chapter 11 is named 'Areas Related to Circles' and this chapter is practically an application of the previous chapter in class 10 named 'Circles'. So, this chapter has some important formulae to solve various problems: 

  • Area of circle: πr2
  • Area of a semicircle: 12πr2

  • Area of quadrant: πr24 

  • Area of sector: θ360×πr2

  • Area of Ring: πR2πr2

  • The perimeter of the sector of Circle: 2 Radius + θ360×2πr

  • Length of arc = $\dfrac{2\pi r\theta}{360^\circ}

To know more students can download the vedantu app.

5. How to understand chapter 11 of class 10 Maths?

Circles are round figures without any sides or edges. Their applications are taught to students in chapter 11 of class 10 Maths where students learn how to solve different questions related to the areas of circles, the length and circumference, and a whole lot of other things. Since this chapter has a whole lot of things to learn, students can use Important Questions prepared for this chapter as it offers them an examination patterned structure, which is very beneficial.