# NCERT Exemplar Class 10 Maths Solutions Chapter 11 Areas Related to Circles

NCERT Exemplar class 10 maths solutions chapter 11 is an extension to the learnings of previous classes and chapters, including circles and sectors. In this chapter, we will emphasize the area of Sectors and circles. The topics of NCERT class 10 maths for ‘Areas related to Circles’ can be studied and practiced by these NCERT exemplar class 10 maths chapter 11 solutions prepared by our accomplished mathematics experts. These NCERT exemplar class 10 maths chapter 11 solutions are highly exhaustive and accurate to hone the skills based on areas related to circles-based problems. These NCERT exemplar class 10 maths solutions chapter 11 are prepared on the guidelines of the CBSE syllabus class 10 maths.

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### Question:1

If the sum of the areas of two circles with radii R_{1} and R_{2} is equal to the area of a circle of radius R, then

(A) R_{1} + R_{2 }=R (B) R_{1}^{2} + R_{2}^{2}=R^{2} (C) R_{1} + R_{2 }<R (D)R_{1}^{2} + R_{2}^{2}<R^{2}

### Answer:

(B) R_{1}

^{2}+ R

_{2}

^{2}=R

^{2}

Radius of first circle= R1

Area of first circle =πR

_{1}

^{2}

Radius of second circle =R

_{2}

Area of second circle =πR

_{2}

^{2}

Radius of third circle = R

Area of third circle=πR

^{2}

According to question

πR

_{1}

^{2}+ πR

_{2}

^{2}=πR

^{2}

π(R

_{1}

^{2}+ R

_{2}

^{2})=πR

^{2}

R

_{1}

^{2}+ R

_{2}

^{2}=R

^{2}

Hence option B is correct.

### Question:2

If the sum of the circumferences of two circles with radii R_{1} and R_{2} is equal to the circumference of a circle of radius R, then

(A) R_{1} +R_{2}=R

(B) R_{1} +R_{2}>R

(C) R_{1} +R_{2}<R

(D) Nothing definite can be said about the relation among R_{1}, R_{2} and R.

### Answer:

(A) R_{1}+ R

_{2}=R

Radius of first circle= R

_{1}

circumference of first circle =2πR

_{1}

Radius of second circle =R

_{2}

circumference of second circle =2πR

_{2}

Radius of third circle = R

circumference of third circle=2πR

According to question

2πR

_{1}+ 2πR

_{2}=2πR

2π(R

_{1}+ R

_{2})=2πR

R

_{1}+ R

_{2}=R

Hence option A is correct.

### Question:3

If the circumference of a circle and the perimeter of a square are equal, then

(A) Area of the circle = Area of the square

(B) Area of the circle > Area of the square

(C) Area of the circle < Area of the square

(D) Nothing definite can be said about the relation between the areas of the circle and square.

### Answer:

(B) Area of the circle > Area of the squarecircumference of a circle=

Let the radius of the circle = r

perimeter of a square =

let the side of a square = a

According to question

circumference of a circle = perimeter of a square

And

Hence Area of the circle > Area of the square.

### Question:4

Area of the largest triangle that can be inscribed in a semi-circle of radius r units is

(A) square unit (B) square unit (C) 2 square unit (D) square unit

### Answer:

(A) square unitBase of triangle = diameter of triangle

= 2 x r

=2r {r is radius}

Height of triangle = r

### Question:5

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

Solution

According to question

(Because perimeter of circle = 2πr Perimeter of square =4 side)

(here side of square =a)

(Using area of square = a

^{2})

Hence ratio of their areas is 14: 11

### Question:6

It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be

(A) 10 m (B) 15 m (C) 20 m (D) 24 m

Solution

Diameter of first circle (D) = 16 m

Radius(R) =

Area =

Diameter of second circle (d) = 12 m

Radius(r) =

Area =

Let radius of new park = R

_{1}

Area =

According to question

R = – 10 is not possible because Radius must be positive.

Hence Radius is 10m

### Question:7

The area of the circle that can be inscribed in a square of side 6 cm is

(A) 36 cm^{2} (B) 18 cm^{2} (C) 12 cm^{2} (D) 9 cm^{2}

### Answer:

(D) 9 cm^{2}

Solution

Diameter of circle (d) = 6 cm

Radius (r)=

Area = (area of circle = πr

^{2})

### Question:8

The area of the square that can be inscribed in a circle of radius 8 cm is

(A) 256 cm^{2} (B) 128 cm^{2} (C) 64 cm^{2} (D) 64 cm^{2}

### Answer:

(B) 128 cm^{2}

Area of square =a

^{2}

Diagonal of square = Diameter of circle

Diagonal of square =8 2 =16cm

Let the side of the square = a cm

Using Pythagoras theorem in ABC

(16)

^{2}=a

^{2}+a

^{2}

2a

^{2}=256

a

^{2}=128

Area of square ABCD= a

^{2}

=128 cm

^{2}

### Question:9

The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36cm and 20 cm is

(A) 56 cm (B) 42 cm (C) 28 cm (D) 16 cm

### Answer:

(C) 28 cmCircumference of circle =

Diameter of first circle (d

_{1}) = 36

Radius (r

_{1}) =

Diameter of second circle (d

_{2}) = 20 cm

Radius (r

_{2})=

Let Radius of 3

^{rd}circle = R cm

According to question

### Question:10

### Answer:

(D) 50 cmarea of circle =

Radius of first circle (r1) = 24

Area =

Radius of second circle (r

_{2}) = 7 cm

Area

Radius of third circle = R

Area of third circle =

According to question

(Because radius is always positive)

Radius of circle = 25 cm

Diameter =2 R=2 25=50cm

### Question:1

### Answer:

FalseUse area of circle =

Side of square = a

Diameter of a circle = a { circle inscribed in square}

Radius =

Area =

= (Because Radius = )

Hence given statement is not true because area of circle inscribed in a square of side a cm is

### Question:2

### Answer:

[true]Perimeter of a square = 4 side

Radius of circle = a

Diameter of circle =2 radius= 2a

Side of square =diameter of circle= 2a

Perimeter of square =4 side=4 2a=8a

Perimeter of square is 8a

Hence given statement is true.

### Question:3

### Answer:

Diameter of circle = d

Side of biggest square = d

Area of the biggest square is = side × side

=d d=d

^{2}

Diagonal of smallest square = d

Let side = a

d

^{2}=a

^{2}+a

^{2}{using Pythagoras theorem}

d

^{2}=2a

^{2}

^{}

Area of smallest square

Here we found that area of outer square is not 4 times the area of inner square.

### Question:4

### Answer:

[False]From the above figure it is clear that the given statement is true Only in case of minor segment. But area of major segment is always greatest

Hence given statement is False.

### Question:5

### Answer:

FalseCircumference of circle =2r

Diameter = d

Radius =

Circumference =2r

=

Here we found that the distance travelled by a circular wheel of diameter d cm in one revolution is πd which is not equal to 2πd.

Hence the given statement is False.

### Question:6

### Answer:

[True]Circumference of circle =

Radius of circular wheel = r m

Circumference of wheel =

Distance covered in One revolution= circumference of wheel =

In covering a distance of s number of revolution required =

Hence the given statement is True

### Question:7

### Answer:

FalseArea of circle =

Circumference of circle =

Case 1:

Let r = 1

Area of circle ==

Circumference of circle = =

Case 2:

Let r = 3

Area of circle = πr2 = π(3)2 = 9π

Circumference of circle = 2πr = 2π(3) = 6π

Conclusion:- In case (1) we found that the area is less than the circumference but in case (2) we found that the area is greater than the circumference.

So, from conclusion we observe that it depend on the value of radius of the circle.

Hence the given statement false.

### Question:8

### Answer:

TrueFormula of length of arc=

Let Radius of first circle = r

Length of arc = ….. (1) { is the angle of first circle}

Radius of second circle = 2r

Length of arc=

= …..(2) { is the angle of second circle}

According to question

No, this statement is True

### Question:9

### Answer:

FalseLet the radius of first circle is r

_{1}and of other is r

_{2}

The length of arcs of both circles is same.

Let the arc length = a.

length of arc (a)=

Area of sector of first circle = (because area of sector = )

Area of sector of second circle =

Here we found that the area of sector is depending on radius of circles.

When the circle is same then radius is also same then the given statement is true.

But in case of different circles then the radius is also different

Hence the given statement is False.

### Question:10

### Answer:

TrueLet the radius of first circle is r

_{1}and of other is r

_{2}

Let the arc length of both circles are same.

Let the arc length is a.

length of arc (a)=

Area of sector of first circle =

(because area of sector = )

Area of sector of other circle =

Here we found that both areas are equal in the case of when r

_{1}= r

_{2}

Hence the area of two sectors of two different circles would be equal only in case of both the circles have equal radii and equal corresponding arc length.

Hence it is necessary that their corresponding arcs lengths are equal.

### Question:11

### Answer:

FalseDiameter of circle = b

Radius =

Area =

Here we found that the area of the largest circle is not equal πb

^{2}cm

^{2}.

Hence the given statement is False

### Question:12

Circumferences of two circles are equal. Is it necessary that their areas be equal? Why?

### Answer:

TrueUse circumference of circle =

Let two circles having radius r

_{1}and r

_{2}

Here it is given that their circumferences are equal

We know that area of circle =

Area of circle with radius r

_{1}= πr

_{1}

^{2}

Area of circle with radius r

_{2}= πr

_{2}

^{2}……..(1)

Put r

_{2}= r

_{1}in (1) we get

πr

_{2}

^{2}= πr

_{1}

^{2}

Hence the area of given circles are also equal because two circles with equal radii will also have equal areas.

Hence the given statement is True.

### Question:13

Areas of two circles are equal. Is it necessary that their circumferences are equal? Why?

### Answer:

TrueSolution

Use area of circle

Let two circles having radius r

_{1}and r

_{2}

Here it is given that their areas are equal

We know that circumference of circle =

Circumference of circle with radius r

_{1}= 2πr

_{1}

Circumference of circle with radius r

_{2}= 2πr

_{2}…..(1)

Put r

_{2}= r

_{1}in (1) we get

2πr

_{1}= 2πr

_{2}

Hence the circumference of given circle are also equal because two circles with equal radii will also have equal circumference.

Therefore, the given statement is True.

### Question:14

Is it true to say that area of a square inscribed in a circle of diameter p cm is p^{2}cm^{2}? Why?

### Answer:

FalseIn the figure we see that the diameter of circle is equal to diagonal of square

Hence, diagonal of square = p cm

Let side of square = a cm Using Pythagoras theorem we get

Area of square = side × side

Here we found that area of square is not equal to p

^{2}cm

^{2}.

Hence the given statement is False

### Question:1

### Answer:

Radius = 33 cmSolution

Circumference of circle =

The radii of the two circles are and

Let the circumference of these two are and respectively.

Let the circumference of required circle is C with radius R.

So according to question

Hence the radius of the required circle is 33cm.

### Question:2

In Figure, a square of diagonal 8 cm is inscribed in acircle. Find the area of the shaded region.

### Answer:

[18.24 cm^{2}]

Area of square =(side)

^{2},

Area of circle =

Diagonal of square = Diameter of circle = 8 cm

Using Pythagoras theorem in ABC

Area of square ABCD =a

^{2}

Diameter of circle = 8 cm

Radius (r) =

Area of circle =

Area of shaded region = Area of circle – Area of square

=50.24 - 32

=18.24 cm

^{2}

### Question:3

Find the area of a sector of a circle of radius 28 cm and central angle 45°.

### Answer:

ANswer: 308 cm^{2}

Solution

Area of sector =

The radius of the circle r = 28 cm

Angle (q) = 45°

Area of sector =

### Question:4

### Answer:

500 revolutionsCircumference of circle =

The speed of wheel = 66 km per hour =

= 1100 m/min

Radius = (because 1m = 100cm)

Circumference of wheel =

=2.2 m

The distance covered by wheel in one revolution = 2.2m

The number of resolution per minute to keep a speed of 66 km per hour = revolution

### Question:5

### Answer:

154 m^{2}

Solution

Area of circle =

According to question

In the figure we see that the area graze by cow is in the form of fourth part of a circle

Hence area grazes by cow = here

(Because Area of circle

### Question:6

Find the area of the flower bed (with semi-circular ends) shown in Figure.

### Answer:

458.5 cm^{2}

Solution

If we observe the figure

We found that there is a rectangle and two semi circles in it.

Length and breadth of rectangle is 38cm and 10cm respectively,

Area of rectangle =

Diameter of semi circle = 10cm

Radius of semi circle=

Area of semi circle=

Hence the total Required area = Area of rectangle + 2(Area of semi-circle)

### Question:7

### Answer:

[54.5 cm^{2}]

Given: AC = 6cm and BC = 8cm

In the figure ABC is a right angle triangle.

Hence using Pythagoras theorem

Diameter of circle = AB = 10 cm

Radius =

Area of circle =

Area of

Area of shaded region = Area of circle – Area of DABC

=78.5-24=54.5 cm

^{2}

### Question:8

Find the area of the shaded field shown in Figure.

Here length and breadth of rectangle ABCD is 8m and 4m respectively.

Are of rectangle

Radius of semi circle = 2m

Area of semi circle=

=

Area of shaded field = Area of rectangle ABCD + Area of semi circle

= 32+6.28

= 38.28 m

^{2}

### Question:9

Find the area of the shaded region in Figure.

### Answer:

235.44 m^{2}

There are two semi-circle with diameter (d) 4 cm.

Radius(r) =

Area of semi-circle =

Length and breadth of rectangle ABCD is 16m and 4m respectively

Area of ABCD=16 x 4=64 m

^{2}( Area of rectangle = length× breadth)

Length and breadth of rectangle UVWX is 26m and 12m respectively

Area of UVWX=26 x 12 =312 m

^{2}( Area of rectangle = length× breadth)

Area of shaded region = Area of UVWX – Area of ABCD – 2 × Area of semi-circle

### Question:10

### Answer:

Solution

Here

r=14 cm

Area of segment =

### Question:11

### Answer:

[30.96 cm^{2}]

Solution

Here ABCD is a square of side 12 cm

Area of ABCD= (side)

^{2}=(12)

^{2}=144 cm

^{2}

Area of sector = here

Here PSAP, PQBP, QRCQ, RSDR all sectors are equal

Area of 4 sectors =

Area of shaded region = Area of square – Area of 4 sectors

= 144-113.04

=30.96 cm

^{2}

### Question:12

### Answer:

39.25 cm^{2}

Solution

Angle made by vertices A, B and C = 60° { In equilateral triangle all angles = 60°}

Diameter of circle = 10

Radius =

Area of shaded region = 3 × Area of sector

### Question:13

### Answer:

[ 308 cm^{2}]

Solution

Area of sector with angle

Here

Radius of each circle = 14 cm

There are three sectors

Area of each sector =

Area of shaded region = 3 x (Area of one sector)

### Question:14

### Answer:

15246 m^{2}

Area of circle =

Given that AB = 105m, BC = 21m

Where AB is radius of park and BC is wide of road

AC=AB+BC

AC=105+21=126 m

Area of big circle=

Area of small circle =

Area of road =Area of big circle - Area of small circle

=49896-34650=15246 m

^{2}

### Question:15

### Answer:

[1386cm^{2}]

Area of sector =

Here

Radius = 21 cm

There are four sectors in the figure

Area of sector =

=

Area of shaded region = 4 × Area of one sector

= 4 × 346.5

= 1386 cm

^{2}

### Question:16

### Answer:

Solution

Given

Length of arc = 20 cm

We know that

Length of arc =

### Question:1

### Answer:

[26400]Given:- Area of circular playground = 22176 m2

Rate of fencing = 50 Rs. per meter

Circumference of circle =

We have to find the radius (r) of playground

Area of playground =

Circumference of playground =

Cost of fencing the playground = 528 50

= 26,400 Rs.

Hence the cost of fencing the playground at the rate of 50 per meter is 26400 Rs.

### Question:2

### Answer:

560 revolutions.Solution:

Circumference of circle =

Diameter of front wheel = 80 cm

Radius =

Diameter of rear wheel = 2m = 200 cm

Radius =

We know that distance covered in one revolution =

distance covered by front wheel in 1400 revolution =

Let rear wheel take x revolutions = x × 2π(100)

According to question

Rear wheel will make 560 revolutions.

### Question:3

### Answer:

Solution

Let their angles of triangle are and rope’s length (radius) = 7 cm

Area of sector with angle A =

Area of sector with angle B=

Area of sector with angle C=

Sum of the areas are

{ Sum of angles of a triangle = }

=77 m

^{2}

Sides of triangular field are 15m, 16m and 17m

Let a =15m, b =16m, c = 17m

Area of triangular field

So area of the field which cannot be grazed by the animals = Area of triangular field – Area of sectors of field

### Question:4

### Answer:

Solution

Area of sector – Area of triangle

Radius = 12 cm

Angle = 60°

Area of sector OAB =

DAOB is isosceles triangles

Let

{ Sum of all interior angles of a triangle is 180°}

x + x +60=180

2x =120

x=60

Here all the three angles are 60° given triangle is an equilateral triangle.

Area of { Area of equilateral triangle }

Area of segment = Area of sector OBCA – Area of DAOB

=

### Question:5

### Answer:

[3064.28 RS ]Given Diameter of circular pond = 17.5 m

Radius (r)=

Width of Path = 2m

R=r+width of path

Area of path =

Area of path = 122.57m

^{2}

Rate of construction = Rs 25 per m

^{2}

Cost of construction

### Question:6

### Answer:

[196 m^{2}]

Given AB = 18 cm

DC = 32 cm

Radius of circle = 7 cm

Area of trapezium = × sum of the parallel sides distance between parallel sides.

= ×(AB+CD) × 14

= ×(18+32) × 14

=(50) × 7

=350 cm

^{2}

= 360° (sum of interior angles of quadrilateral)

Radius = 7 cm

Area of all sectors = here

=

=

=154 cm

^{2}

Area of shaded region = Area of trapezium – Area of all sectors

=350-154=196 cm

^{2}

### Question:7

### Answer:

1.966 cm^{2}

Radius = 3.5

Diameter = 3.5 + 3.5 = 7 cm

Here ABC is equilateral triangle because AB = BC = CA = 7 cm

Area of

In equilateral triangle each angle = 60°

All the sectors are same

Area of all there sectors

Area of enclosed region = Area of DABC – Area of their sectors

=21.217-19.251

= 1.966cm

^{2}

### Question:8

Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.

### Answer:

[8.75 cm^{2}]

Length of the Arc = 3.5 cm

Radius (r)= 5 cm

Area of sector= radius length of arc

### Question:9

### Answer:

[42 cm^{2}]

Area of shaded portion = Area of square – Area of 4 sectors

Radius = 7 cm

Side of square = 14 cm

AB=BC=CD=DA=14 cm

ABCD is a square

Area of square= 14 14 (Area of square = (side)

^{2})

=196 cm

^{2}

We know that each angle of square =

Area of 4 sectors =

Area of shaded portion = Area of square – Area of 4 sectors

=196-154=42cm

^{2}

### Question:10

### Answer:

[168 cm^{2}]

Given area of sheet =784 cm

^{2}

Let the side of the sheet = a

a

^{2}=784

^{2}(Area of square = (side)

^{2})

a=

Diameter of each circular plate =

Radius =

Area of 4 circular plates =

Area of sheet not covered with circular plates = Area of sheet – Area of 4 circular plates.

=784-616=168cm

^{2}

### Question:11

### Answer:

[4.3 m^{2}]

Diameter of tile =50cm=0.5m (1m = 100cm)

Radius =

Number of tiles lengthwise = tiles

Number of tiles widthwise = tiles

Total tiles =

Area of floor not covered by tiles = Area of rectangular floor – Area of 80 tiles

### Question:12

Answer: [800 cm^{2}]

Solution

Given that area of circle =1256cm

^{2}

Diameter of circle = 40 cm

As we know that the diameter of circle is equal

Diagonals of rhombus = Diameters of circle = 40 cm

Each diagonals of rhombus = 40 cm

Area of rhombus = product of digonals

= 40 40

= 800cm

^{2}

Hence required area of rhombus is =800cm

^{2}

### Question:13

### Answer:

[1 : 3 : 5]Solution

d

_{1}:d

_{2}:d

_{3}= 1: 2 : 3 [multiplying by s]

= s : 2s : 3s

Radius of inner circle (r

_{1})=

Radius of middle circle (r

_{2})=

Radius of outer circle (r

_{3})=

Area of region enclosed between second and first circle

Area of region enclosed between third and second circle

Area of first circle

Ratio of area of three regions

### Question:14

### Answer:

Solution

We know that minute hand revolving in 60 min =

In 1 minute it is revolving =

Time difference =(6:40am -6:05am) =35 min

In 6:05 am and 6.40 am there is 35 minutes

In 35 minutes angle between min hand and hour hand =

Length of minute hand (r)=5cm

Area of sector =

Hence required area is

### Question:15

### Answer:

Area of sector=

Angle = 200°

Area of sector = 770 cm

^{2}

Length of the corresponding arc =

### Question:16

### Answer:

Area of sector=Radius of first sector(r

_{1}) = 7 cm

Angle ( ) = 120°

Area of first sector(A

_{1}) =

Radius of second sector(r

_{2}) = 21 cm

Angle ( ) = 40°

Area of sector of second circle (A

_{2})=

Corresponding arc length of first circle =

=

Corresponding arc length of second circle =

=

We observe that the length of arc of both circle are equal.

### Question:17

Find the area of the shaded region given in Figure.

### Answer:

[Area of shaded area=154.88cm^{2}]

Area of square PQRS =(side)

^{2}=(14)

^{2}

=196 cm

^{2}

Area of ABCD (let side a) =(side)

^{2}=(a)

^{2}

Area of 4 semi circle

Area of semi-circle=

Total inner area = Area of ABCD + Area of 4 semi circles

Area of inner region =

Area of shaded area = Area of PQRS – inner region area

### Question:18

### Answer:

[40] revolutionsCircumference of circle =

Area of wheel = 1.54m

^{2}

Distance = 176 m

r = 0.7m

Circumference =

Number of revolution

### Question:19

Answer [32.16cm^{2}]

Solution

By using Pythagoras in ABC

Area of circle =

Area of sector =

Area of

Area of minor segment = Area of sector – Area of DABC

=9.81-6.25

=3.56cm

^{2}

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

=39.28-3.56=35.72cm

^{2}

Required difference = Area of major segment – Area of minor segment

=35.72-3.56=32.16cm

^{2}

### Question:20

Answer: [462 cm^{2}]

Solution

Radius = 21 cm

Angle = 120°

Area of circle =

Area of minor sector with angle 120° OABO =

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

= 1386-462=924cm

^{2}

Required area =924-462=462cm

^{2}

## NCERT Exemplar Solutions Class 10 Maths Chapter 11 Important Topics:

The important topics covered in NCERT exemplar class 10 maths solutions chapter 11 are:

◊ Area of Sector

◊ Area of Circles

◊ NCERT Exemplar class 10 maths solutions chapter 11 discusses the method to find out perimeter and area for any given shape, which can be seen as a combination of circles, sectors, rectangles, triangles.

## NCERT Class 10 Exemplar Solutions for Other Subjects:

NCERT Exemplar Class 10 Maths solutions

NCERT Exemplar Class 10 Science solutions

## NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Chapter |

NCERT exemplar solution for class 10 Maths Chapter 1 Real Numbers |

NCERT exemplar solutions for class 10 Maths Chapter 2 Polynomials |

NCERT exemplar solution for class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables |

NCERT exemplar solution for class 10 Maths Chapter 4 Quadratic Equations |

NCERT exemplar solution for class 10 Maths Chapter 5 Arithmetic Progressions |

NCERT exemplar solution for class 10 Maths Chapter 6 Triangles |

NCERT exemplar solution for class 10 Maths Chapter 7 Coordinate Geometry |

NCERT exemplar solution for class 10 Maths Chapter 8 Introduction to Trigonometry & Its Equations |

NCERT exemplar solution for class 10 Maths Chapter 9 Circles |

NCERT exemplar solution for class 10 Maths Chapter 10 Constructions |

NCERT exemplar solution for class 10 Maths Chapter 11 Areas related to Circles |

NCERT exemplar solution for class 10 Maths Chapter 12 Surface Areas and Volumes |

NCERT exemplar solution for class 10 Maths Chapter 13 Statistics and Probability |

## Features of NCERT exemplar class 10 maths solutions chapter 11:

These class 10 maths NCERT exemplar chapter 11 solutions emphasise the area of sectors and circles.

In this chapter, NCERT exemplar problems are very tricky and will help develop an excellent logical brain. Sometimes the composite figure will be complicated to resolve, but the perfect analysis will resolve it to determine the perimeter and area.

Class 10 students can use these detailed solutions on Area related to Circles based practice problems as reference content.

These class 10 maths NCERT exemplar solutions chapter 11 Area related to Circles are sufficient to solve the problems of NCERT class 10 maths, A textbook of Mathematics by Monica Kapoor, RD Sharma class 10 maths, RS Aggarwal class 10 maths.

The students can download the pdf version by using on NCERT exemplar class 10 maths solutions chapter 11 pdf download feature of online tools. This feature specially caters to the students learning in a low internet connectivity environment or those who plan to be offline while studying NCERT exemplar Class 10 maths chapter 11.

## Check Chapter-wise Solutions of Book Questions

Chapter No. | Chapter Name |

Chapter 1 | |

Chapter 2 | |

Chapter 3 | NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables |

Chapter 4 | NCERT solutions for class 10 maths chapter 4 Quadratic Equations |

Chapter 5 | NCERT solutions for class 10 chapter 5 Arithmetic Progressions |

Chapter 6 | |

Chapter 7 | NCERT solutions for class 10 maths chapter 7 Coordinate Geometry |

Chapter 8 | NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry |

Chapter 9 | NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry |

Chapter 10 | |

Chapter 11 | |

Chapter 12 | NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles |

Chapter 13 | NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes |

Chapter 14 | |

Chapter 15 |

## Frequently Asked Question (FAQs) - NCERT Exemplar Class 10 Maths Solutions Chapter 11 Areas Related to Circles

**Question: **For a given perimeter, which two-dimensional shape has the maximum area?

**Answer: **

The circle will have the maximum area among all the two-dimensional shapes for given circumference.

**Question: **What will be the largest square area drawn in a circle of radius R?

**Answer: **

For the largest square area the diameter must be equal to diagonal of the square.

Hence, the area will be 2R2

**Question: **Is the chapter Area related to circles important for Board examinations?

**Answer: **

The chapter Area related to circles is vital for Board examinations as it holds around 2-3% weightage of the whole paper.

**Question: **What type of questions are expected from Area related to Circles?

**Answer: **

Generally, the paper consists of either a Long Short Answer question or multiple short answer questions from this chapter. NCERT Exemplar class 10 maths solutions chapter 11 can help the students develop high-order thinking skills and ace the Area related to Circles related problems.

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## Questions related to CBSE Class 10th

### what is the timmg of result declaration of class 10 cbse

Hello,

Central Board of Secondary Education may release the CBSE Class 10 result 2021 by July 20, 2021. The officials have not declared any timings exactly when the results will be out. The results will out at any point of the day. Keep an eye on the official website for the latest information. The board publishes CBSE 10th result 2021 online on cbse.nic.in, cbse.gov.in and cbseresults.nic.in. To check the CBSE Class 10th result 2021, students need their roll number, date of birth and other details.

### How to Download MP Board /CBSC 10 th class books PDF 2021-2022

Hy,

If you want the soft copies of the books I suggest you to join some telegram groups which provides books, PDFs , question papers etc . It will be beneficial for you .

Secondly you can download NCERT book app from Google play store for Android where you will get every books of NCERT from class 1 to class 12 for every stream.

Thirdly visit the websites given below

https://ncert.nic.in

https://www.ncertbooks.guru

Hopes this helps.

Wish you luck!

### After class 10 result ,how will subjects distribute to students, on the basis of result!?

It depends on the school in which you want admission.

Some schools have the rule that

i. 75%+ can take science.

ii.60% - 75% can take commerce.

iii.Below 60% will take humanities.

The choice of choosing stream totally depends on students.

And some schools also check your marks in the subject you want to take admission.

Example- If you want to take admission in maths stream they will check your maths and science marks.

If you want to take admission in humanities they will check your social science marks.

Some schools also conduct an entrance exam followed by an interview.

Refer to this link if you are confused about which stream to choose after Class 10th.

https://school.careers360.com/articles/which-stream-select-after-class-x

Feel free to ask if you have any more queries.

All the best!

### Board cbse class 10 Is cancelled and promote all student please sir because of corona do not study

Hello aspirant,

No CBSE and ICSE board didn't not cancelled the 10th and 12th exam.

As they had declared that 10th and 12th class exam will not be cancelled and students will have to give it in offline mode in this pandemic also.

Hope this helps you

All the best for your future

### CBSE class 10th malayalam can someone please tell me a good one

Adamas World school is the best CBSE school in Kolkata. To get admission please visit https://www.adamasworldschool.org/