R S Aggarwal solutions for class 10 Chapter 16 helps students to solve all the questions on the topics Area of Circle, Sector and Segment. Detailed solutions to the R S Aggarwal textbook questions are provided here so that students can compare their answers to the sample responses. This study material contains all the exercise questions provided in the textbook.

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RS Aggarwal Class 10 Maths Solutions for Chapter 16 and clear concepts.

### Access solution to Maths R S Aggarwal Chapter 16 – Area of Circle, Sector and Segment

Get detailed solutions for all the questions listed under the below exercises:

Exercise 16A Solutions : 30 Questions (Short Answers)

Exercise 16B Solutions : 27 Questions (Short Answers)

## Exercise 16A

**Question 1:**

**Solution:**

Given: Circumference of the circle = 39.6 cm.

We know, Circumference of circle = 2πr

Where, r = Radius of the circle

⇨ 2πr = 39.6

r = 39.6/2π

r = 6.3

(put value of π = 22/7)

Area of the circle = πr^{2}

Where, r = radius of the circle

⇨ Area of the circle = π(6.3)^{2}

= 124.74

So, area of circle is 124.74 cm^{2}.

**Question 2:**

**Solution:**

Area of a circle = 98.56 cm^{2}

We know, Area of the circle = πr^{2}

Where, r = radius of the circle

So, πr^{2} = 98.56

Put π = 22/7

r^{2} = 98.56/(22/7) = 31.36

r = 5.6 cm

Circumference of circle = 2πr = 2 ×22/7×5.6 = 35.2 cm

**Question 3:**

**Solution:**

Given: circumference of a circle exceeds its diameter by 45 cm.

⇨ Circumference of circle = 45 + Diameter of circle

2π r = 45 + 2r

2r(π – 1) = 45

2r(22/7 – 1) = 45

15/7 r = 45/2

r = 10.5

Circumference of a circle = 2πr = 2 x 22/7 x 10.5 = 66 cm

**Question 4:**

**Solution:**

Let the square be of side ‘a’ cm and radius of the circle be ‘r’

Area enclosed by the square = 484 cm^{2}

Also, we know that Area of square = Side × Side

Area of the square = a^{2}

⇨ a^{2} = 484

⇨ a = √484

⇨ a = 22

Therefore, side of square is 22 cm.

Perimeter of square = 4 x side = 4 x 22 = 88

From statement, circumference of the circle = Perimeter of square

2πr = 88

r = (88 x 7)/(2×22) = 14

Radius of the circle = 14 cm

Area of circle = πr^{2} = 22/7 x 14 x 14 = 616 cm^{2}.

**Question 5:**

**Solution**:

Area of an equilateral triangle = 121√3 cm^{2} (given)

We know that, Area of an equilateral triangle = √3/4 (a^{2}), where “a” is the side of a triangle.

The area enclosed by the circle is 346.5 cm^{2}

**Question 6:**

**Solution:**

The length of a chain used as the boundary of a semicircular park = 108 m

i.e. circumference of a semicircular park = 108 m

Let the radius of semicircular park = r

Which implies,

Now,

Area of a semicircular park = πr^{2} = 22/7 x 21 x 21 = 1386

Area of the park = 1/2 (area of semicircular park) = 1/2 x 1386 = 693

Therefore, Area of the park is 693 m^{2}.

**Question 7:**

**Solution:**

The sum of the radii of two circles is 7 cm (given)

Difference of circumferences of two circles is 8 cm (given)

Let radii of circles be x and (7-x)

Then,

2πx – 2π(7 – x) = 8

2πx – 14π + 2πx = 8

2πx = 26 …..(1)

Which is circumference of one circle.

Circumference of the another circles = 2π(7 – x) = 14π – 2πx

= 14π – 26

= 14 x 22/7 – 26

= 18

Therefore, Circumference of the circles are 26 cm and 18 cm.

**Question 8:**

**Solution:**

Radius of outer ring = R = 23 cm

Radius of inner ring = r = 12 cm

Area of outer circle = πR^{2}

= 22/7 x 23 x 23

= 1662.5 cm^{2}

Area of inner circle = π r^{2}

= 22/7 x 12 x 12

= 452.2 cm^{2}

Area of Ring = outer area – inner area

= 1662.5 – 452.2

= 1210

Area of the ring is 1210 cm^{2}

**Question 9:**

**Solution:**

Width of the path = 8 m

Inner radius of the circular park = 17 m

Outer radius of the circular park = (17 + 8) = 25 m

Now,

Area of the path = π[(25^{2} – (17^{2})]

= 22/7 x 42 x 8

= 1056

Area of the path is 1056 m^{2}.

**(ii) A park is in the shape of a circle of diameter 7m. It is surrounded by a path of width 0.7m. Find the expenditure of cementing the path at the rate of ₹ 110 per sq.m.**

**Solution:**

Diameter of park = 7 m

so, radius of park = r_{1} = 3.5 m

Width of path = 0.7 m

Bigger radius of park = r_{2} = 0.7 + 3.5 = 4.2 m

Now, Area of path = area of bigger circle -area of smaller circle

= Pi r_{2}^{2} – Pi r_{1}^{2}

= Pi(r_{2}^{2} – r_{1}^{2})

= 22/7 (4.2^{2}– 3.5 ^{2})

= 22/7(17.64 – 12.25)

=16.94 m^{2}

Also,

Cost of expenditure = rate × area

= 110 × 16.94

= 1864.40

Cost of Expenditure of cementing the path is ₹ 1864.40.

**Question 10:**

**Solution:**

Let r and R be the radii of inner circle and outer boundaries respectively.

Then,

Circumference of inner circle = 2πr = 352 and

Circumference of outer circle = 2πR = 396

r = 352/2π and R = 396/2π

Width of the track = (R – r) = 396/2π – 352/2π

= 44/2π

= 44/2 x 7/22

= 7

Width of the track is 7 m.

Again, R + r = 396/2π + 352/2π = 748/2π

Area of the track = π(R^{2} – r^{2})

= π(R – r) (R+r)

= π x 7 x 748/2π

= 2618

Therefore, the area of the track is 2618 m^{2}.

**Question 11:**

**Solution:**

**Question 12:**

**Solution**:

The radius of the circle = 10 cm.

Chord PQ subtends an angle = 60 degrees at the centre.

PQO is an equilateral triangle, O being the centre of the circle.

Area of minor segment = θ/360 x π r^{2} – 1/2 r^{2} sinθ

= 60/360 x 22/7 x 10 x 10 – 1/2 x 10 x 10 x √3/2

= 9.03

Area of circle = π r^{2} = 22/7 x 10 x 10 = 314 (approx.)

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

= 314 – 9.03

= 304.97 cm^{2} (approx.)

**Question 13:**

**Solution:**

**Question 14:**

**Solution:**

Radius of a circle = 7 cm

Area of circle = πr^{2} = 22/7 x 7 x 7 = 154 cm^{2}

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

= 154 – 14

= 140

Therefore, Area of major segment is 140 cm^{2}.

**Question 15:**

**Solution:**

The lengths of the arcs cut off from a circle of radius (r ) 12 cm by a chord 12 cm long.

Which shows it form a equilateral triangle POQ, and ∠POQ = 60 degrees

Area of minor segment = Area of sector – Area of triangle …(i)

Area of triangle = √3/4 (side)^{2}

= √3/4 (12)^{2}

= 62.28 cm^{2}

Area of sector = θ/360 x πr^{2}

= 60/360 x 3.14 x 12 x 12

= 75.36 cm^{2}

Equation (1) implies

Area of minor segment = 75.36 – 62.28 = 13.08 cm^{2}

Therefore, length of major arc is 62.8 cm and of minor arc is 12.56 cm and area of minor segment is 13.08 cm^{2}.

**Question 16:**

**Solution:**

Radius of circle = 5√2 cm

Chord = 10 cm

To find: Areas of both the segments

Now,

Area of minor segment = (area of sector OACBO) – (area of ∆OAB)

= 39.25 – 25 = 14.25 cm^{2}

**Question 17:**

**Solution:**

**Question 18:**

**Solution:**

Radius of circle = r = 30 cm

Area of minor segment = Area of sector – Area of triangle …(1)

Area of major segment = Area of circle – Area of minor segment …(2)

Area of sector = θ/360 x πr^{2}

= 60/360 x 3.14 x 30 x 30

= 471 cm^{2}

Area of triangle = √3/4 (side)^{2} (Since it form a equilateral triangle)

= √3/4 x 30 x 30

= 389.7 cm^{2}

(1) ⇨

Area of minor segment = 471 – 389.7 = 81.3 cm^{2}

(2) ⇨

Area of major segment = π(30^{2}) – 81.3 = 2744.7 cm^{2}

Answer:

Area of major segment is 2744.7 cm^{2} and of minor segment is 81.3 cm^{2}.

**Question 19:**

**Solution:**

Radius of circle = 10.5 cm

Let x cm be the major arc, then x/5 cm be the length of minor arc.

Circumference of circle = x + x/5 = 6x/5 cm

We know, Circumference of circle = 2π r = 2 x 22/7 x 10.5

This implies,

6x/5 = 2 x 22/7 x 10.5

x = 55 cm

Area of major sector = 1/2 x 55 x 10.5 = 288.75

The area of major sector is 288.75 cm^{2}.

**Question 20:**

**Solution:**

The short and long hands of a clock are 4 cm and 6 cm long respectively.

In an hour the hour hand completes one rotation in 12 hours therefore in 24 hours it will complete 2 rotations.

Similarly, the minute hand completes one rotation therefore in 24 hours the minute hand will complete 24 rotations.

Distance travelled by its tip in 2 days = 4(circumference of the circle with radius 4 cm)

= (4 × 2π × 4) cm = 32π cm

(As, in 2 days, the short hand will complete 4 rounds)

In 2 days, the long hand will complete 48 rounds length moved by its tip = 48(circumference of the circle with radius 6 cm)

= (48 x 2π x 6) cm

= 576 π cm

Now,

Sum of the lengths moved

= (32π + 576π)

= 608π cm

= (608 x 3.14) cm

= 1909.12 cm

**Question 21:**

**Solution:**

Radius of circle, say r

Circumference = 88 cm

2πr = 88

r = 14 cm

Area of quadrant of a circle = 1/4 π r^{2}

= 1/4 x 22/7 x 14 x 14

= 154 cm^{2}

**Question 22:**

**Solution: **

At radius = 16 m, cow can graze the area of plot is πr^{2}

= 22/7 x 16 x 16

= 804.5 m^{2}

At radius = 23 m, cow can graze the area of plot is πr^{2}

= 22/7 x 23 x 23

= 1662.57 m^{2}

Additional ground area = 1662.57 – 804.5 = 858 m^{2}

**Question 23:**

**Solution:**

Length of rectangular field = L = 70 m

Breadth of rectangular field = B = 52 m

So, Area of the field = L x B

= 70×52 = 3640

Area of the field is 3640 m^{2}

And, Area of quadrant of radius 21 m = 1/4 x 22/7 x 21 x 21

= 346.5 m^{2}

Area available for grazing = (Area of the field) – (Area of quadrant of radius 21 m)

= 3640 – 346.5

= 3293.5 m^{2}

Thus, 3293.5 m^{2} area is left ungrazed.

**Question 24:**

**Solution:**

Each angle of equilateral triangle = 60^0

Side of an equilateral triangle = 12 m

Length of the rope = 7 m

Area of an equilateral triangle = √3/4 (side)^{2} = √3/4 x 12 x 12 = 62.35 m^{2}

Area of sector with radius 7 m = 60/360 x 22/7 x 7 x 7 = 25.66 m^{2}

Now,

Area which cannot be grazed by horse = Area of an equilateral triangle – Area of sector with radius 7 m

= 62.35 m^{2} – 25.66 m^{2}

= 36.68 m^{2}

**Question 25:**

**Solution:**

**Question 26: **

**Solution:**

**Question 27:**

**Solution**:

Side of a square = 10 cm

Diameter of the inscribed circle = 10 cm

Radius of the inscribed circle = 5 cm

Diameter of the circumscribed circle = Diagonal of the square

Radius circumscribed circle = 5√2 cm

(i) the area of the inscribed circle

= 22/7 x 5 x 5 = 78.57 cm^{2}

(ii) the area of the circumscribed circle

= 22/7 x 5√2 x 5√2 = 157.14 cm^{2}

**Question 28:**

**Solution:**

Diagonal of square = diameter of circle = 2r cm, where r is radius of circle

Area of circle = πr^{2} cm^{2}

Area of Square = 1/2 x diagonal^{2}

= 1/2 x 4r^{2} = 2r^{2} cm

Now, ratio of the areas of the circle and the square

= πr^{2}/2r^{2}

= π/2

Therefore, the required ratio is π:2.

**Question 29:**

**Solution:**

Area of circle = π r^{2} (where r be the radius of circle)

= 22/7 x r^{2}

Given: area of a circle = 154 cm^{2}

22/7 x r^{2} = 154

r^{2} = 154 x 7/22

or r = 7

Radius of the circle is 7 cm

Let us consider each side of the equilateral triangle is “a” and height is h, then

r = h/3

or h = 3r = 3 x 7 = 21

Height of the triangle is 21 cm

From right triangle APC:

h^{2} = a^{2} – (a/2)^{2}

h^{2} = 3a^{2}/4

h = √3/2 (a)

21 = √3/2 x a (putting value of h)

a = 14√3 cm

Perimeter of the triangle = Sum of all the sides = 3a = 3 x 14√3

= 42 x 1.73

= 72.66 cm

**Question 30:**

**Solution:**

Radius of the wheel = r = 42 cm (given)

Circumference of wheel = Circumference of circle = 2πr = 2 x 22/7 x 42

= 264

Circumference of wheel is 264 cm.

Again, distance travelled by wheel = 19.8 km = 1980000 cm

(Convert km into cm)

Now, number of revolutions taken by a wheel = 1980000/264 = 7500

## Exercise 16B Page No: 722

**Question 1:**

**Solution:**

Circumference of a circle – Radius = 37cm (given)

We know that, Circumference of a circle = 2 π r (where r = radius)

2 π r – r = 37

2 x 22/7 x r – r = 37

(37/7)r = 37

r = 7 cm

Therefore, circumference of a circle = 2 π r = 2 x 22/7 x 7 = 44 cm

**Question 2:**

**Solution:**

Circumference of a circle = 22 cm

That is, 2 π r = 22

2 x 22/7 x r = 22

r = 7/2 cm

Area of quadrant of circle = ¼ π r^{2} = ¼ x 22/7 x 7/2 x 7/2 = 77/8

Area of quadrant of circle is 77/8 cm^{2}

**Question 3:**

**Solution**:

Area of circle = Area of circle of diameter 10 cm + Area of circle of diameter 24 cm

or Area of circle = Area of circle of radius 5 cm + Area of circle of radius 12 cm

πr^{2} = π(5)^{2} + π(12)^{2}

πr^{2} = 25 π + 144π = 169π

r = 13 So, required diameter is 26 cm.

**Question 4:**

**Solution:**

Area of circle = 2 x circumference of circle

πr^{2} = 2 x 2πr

r = 4

Diameter of circle = 2r = 8 cm

**Question 5:**

**Solution:**

Given, square circumscribes a circle of radius a cm.

Side of the square = 2 x radius of circle = 2a cm

Now, perimeter of the square = (4 x 2a) = 8a cm

Perimeter of the square is 8a cm.

**Question 6:**

**Solution:**

Diameter of circle = 42 cm

Radius = 42/2 = 21 cm

Central angle = 60^{0}

We know that,

Length of the arc = θ/360 (2πr)

= 60/360 x 2 x 22/7 x 21

= 22

Length of the arc is 22 cm

**Question 7:**

**Solution:**

Area of circle = Area of circle of radius 4 cm + Area of circle of radius 3 cm

Area of circle = π(4)^{2} + π(3)^{2}

π r^{2} = 16π + 9π

π r^{2} = 25π

r = 5

Radius of circle = 5 cm

Diameter of circle = 2r = 10 cm

**Question 8:**

**Solution:**

Circumference of circle = 8π

2πr = 8π

r = 4

Area of circle =πr^{2} = π(4)^{2} = 16 π

**Question 9:**

**Solution:**

Diameter of the semicircular protractor = 14 cm

Radius = 14/2 cm = 7 cm

Perimeter of semicircle = πr + d

Perimeter of semicircular protractor = 22/7 x7 + 14 = 22 + 14 = 36 cm

The perimeter of the semicircular protractor is 36 cm.

**Question 10:**

**Solution:**

Perimeter of circle = Area of circle (given)

2πr = πr^{2}

(where r = radius of circle)

r = 2

The radius of a circle is 2 cm

**Question 11:**

**Solution:**

Circumference of Circle = Circumference of circle with radius 19 cm + Circumference of circle with radius 9 cm

2πr = 2π(19) + 2π(9)

2πr = 38π + 18 π

r = 28

Radius of the circle is 28 cm.

**Question 12: **

**Solution:**

Area of Circle = Area of circle with radius 8 cm + Area of circle with radius 6 cm

Πr^{2} = π(8)^{2} + π(6)^{2}

Πr^{2} = 64π + 36 π

r^{2} = 100

or r = 10

Radius of the circle is 10 cm.

**Question 13:**

**Solution:**

Radius = r = 6 cm and θ = 30^{0}

Area of sector = θ/360 (πr^{2})

= 30/360 x 3.14 x (6)^{2}

= 9.42 cm^{2}

**Question 14:**

**Solution:**

Radius = r = 21 cm and θ = 60^{0}

Length of the arc = θ/360 (2πr)

= 60/360 x 2 x 22/7 x 21

= 22 cm

**Question 15:**

**Solution:**

Ratio of circumferences of two circles = 2:3 (given)

Let the two circles be C1 and C2 with radii r1 and r2.

Circumference of circle = 2πr

Circumference of C1 = 2πr1 and

Circumference of C2 = 2πr2

(Circumference of C1) / (Circumference of C2) = (2πr1)/(2πr2)

2/3 = r1/r2

Now,

(area of C1) / (area of C2) = (πr1^{2})/(πr2^{2})

= {(r1)/(r2)}^{2}

= (2/3)^{2}

=4/9

Therefore, the required ratio is 4:9.

**Question 16:**

**Solution:**

Ratio of areas of two circles = 4:9 (given)

Let the two circles be C1 and C2 with radii r_{1} and r_{2}.

Area of circle = πr^{2}

Area of C1 = πr_{1}^{2} and

Area of C2 = πr_{2}^{2}

(Area of C1) / (Area of C2) = (πr1^{2})/(πr2^{2})

4/9 = r_{1}^{2}/r_{2}^{2}

or (r_{1})/(r_{2}) = 2/3

Now,

(Circumference of C1) / (Circumference of C2) = (2πr_{1})/(2πr_{2})

= (r_{1})/(r_{2})

= (2/3)

= 2/3

Therefore, the required ratio is 2:3.

**Question 17:**

**Solution:**

A square is inscribed in a circle (given)

Let r be the radius of circle and ‘x’ be the side of the square.

So, length of the diagonal = 2r

Length of side of square = x = diagonal/√2 = 2r/√2 = √2r

Area of square = (side)^{2} = ( x)^{2} = √2r × √2r = 2r^{2}

Area of circle = πr^{2}

Ratio of areas of circle and square = (area of the circle)/(area of square)

= πr^{2} / 2r^{2}

= π/2

Hence, the ratio of areas of circle and square is π:2.

**Question 18:**

**Solution:**

Circumference of a circle = 8 cm

Central angle = 72^{0}

Now, Circumference of a circle = 2πr

8 = 2πr

r = 14/11 cm

Area of sector = θ/360 x (πr^{2})

= 72/360 x 22/7 x 14/11 x 14/11

= 1.02 cm^{2}

**Question 19:**

**Solution:**

A pendulum swings through an angle of 30^0 and describes an arc 8.8 cm in length.

Length of the pendulum = Radius of sector of the circle

Arc length = 8.8

θ/360 (2πr) = 8.8

30/360 x 2 x 22/7 x r = 8.8

r = 16.8

Therefore, the length of the pendulum is 16.8 cm.

**Question 20:**

**Solution: **

The minute hand of a clock is 15 cm long

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

Angle described by minute hand in 20 minutes = 360/60 x 20 = 120^{0}

So, area swept by it in 20 minutes = Area of the sector having central angle 120^{o} and radius 15 cm

= θ/360 (πr^{2})

= 120/360 x 22/7 x 15 x 15

= 235.5

Therefore, the area swept by minute hand in 20 minutes is 235.5 cm^{2}.

**Question 21:**

**Solution: **

Area of the sector = 17.6 cm^{2} (given)

We know, Area of the sector = θ/360 (πr^{2}) square units

This implies,

θ/360 (πr^{2}) = 17.6

56/360 x 2/7 x r^{2} = 17.6

r^{2} = 36

or r = 6

Radius of the circle is 6 cm.

**Question 22:**

**Solution:**

Radius = 10.5 cm

Area of the sector of a circle = 69.3 cm^{2}

Area of the sector = θ/360 (πr^{2})

This implies,

θ/360 (πr^{2}) = 69.3

θ/360 x 22/7 x 10.5 x 10.5 = 69.3

θ = 72

Therefore, central angle of the sector is 72 degrees.

**Question 23:**

**Solution**:

Perimeter of a sector of circle = 31 cm

Radius = 6.5 cm

Arc length = 31 – (6.5 + 6.5) = 18 cm

Now, Area of sector = 1/2 x Arc length x radius

= 1/2 x 18 x 6.5

= 58.5 cm^{2}

**Question 24:**

**Solution:**

Radius of a circle = 17.5 cm

Length of arc of circle = 44 cm

Now,

Area of Sector = 1/2 x Arc length x radius

= 1/2 x 44 x 17.5

= 385

Area of Sector is 385 cm^{2}

**Question 25:**

**Solution:**

Length of the rectangular cardboard = 14 cm and

Breadth of the rectangular cardboard = 7 cm

Area of cardboard = Area of rectangle = length × breadth = 14×7 = 98 cm^{2}

Let the two circles with equal radii and maximum area have a radius r each.

2r = 7 or r = 7/2 cm

Again,

Area of two circular cut outs = 2 × πr^{2} = 2 x 22/7 x (7/2)^{2} = 77 cm^{2}

Now,

The area of remaining cardboard = Area of cardboard – Area of two circular cut outs

= 98 – 77

= 21

Therefore, area of remaining cardboard is 21 cm^{2}.

**Question 26:**

**Solution: **

Side of square = 4 cm

Radius of circle = 1 cm

Area of square = (side)^{2} = 4 x 4 = 16 cm^{2}

Area of four quadrants of circle = 4 (1/4 x 3.14 x 1 x 1) = 3.14 cm^{2}

Area of circle with diameter 2 cm = πr^{2} = 3.14 x 1 x 1 = 3.14 cm^{2}

(diameter = radius/2)

Now,

Area of the shaded region = Area of square – (Area of four quadrants of circle + Area of circle with diameter 2 cm)

= 16 – (3.14 + 3.14)

= 9.72

Therefore, the Area of the shaded region is 9.72 cm^{2}.

**Question 27:**

**Solution**:

Length of rectangular sheet of paper = 40 cm

Breadth of rectangular sheet of paper = 28 cm

Radius of the semicircular cut out = 14 cm

Area of rectangular sheet of paper = Area of rectangle = length × breadth = 40 × 28 = 1120 cm^{2}

Area of semicircular cut out = 1/2 πr^{2}

= 1/2 x 22/7 x 14 x 14

= 308 cm^{2}

Now,

Area of remaining sheet of paper = Area of rectangular sheet of paper – Area of semicircular cut out

= 1120 – 308

= 812

Therefore, area of remaining sheet of paper is 812 cm^{2}.

## R S Aggarwal Solutions for Class 10 Maths Chapter 16 Area of Circle, Sector and Segment Topics

In this chapter students will study important concepts on quadratic equations as listed below:

- Area of Circle, Sector and Segment introduction
- Circle and its parts
- Problems on the sector and segment parts
- Areas of combinations of plane figures
- Real-life Problems

### Key Features of R S Aggarwal Solutions for Class 10 Maths Chapter 16 Area of Circle, Sector and Segment

1. R S Aggarwal Solutions assist students in their preparations, provide detailed answers to all questions.

2. Easy and simple language is used.

3. Step by step problem-solving approach is used to help students clear their concepts on Area of Circle, Sector and Segment.

4. Easy for quick revision.

5. This study material prepared is based on the latest CBSE syllabus by Maths subject experts.