NCERT Solutions for Class 10 Maths Chapter 12 : Areas Related to Circles

Q: Question 1. What do you understand by a sector with reference to a circle?.

Answer. The sector of the circle refers to that part of the circular region that is enclosed by the two radii and the corresponding arc..

Q: Question 2. What do you understand by a segment with reference to a circle?.

Answer. The segment of the circle refers to that part of the circular region that is enclosed between a chord and the corresponding arc. We can categorize the segment of the circle as a minor segment and a major segment. The major segment covers a larger area corresponding to the minor segment..

Question 1

The radii of two circles are

19 c m

and

9 c m

respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Medium

Solution

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Answer

Let the radius of required circle

= r c m

Radius of 1st circle

r_{1} = 9 c m

Radius of 2nd circle

r_{2} = 19 c m

As per the question

Circumference of the required circle

=

Sum of circumference of two circles

Circumference of small circle

= 2 π r_{1}

= 2 π \times 9

= 18 π

Circumference of small circle

= 2 π r_{2}

= 2 π \times 19

= 38 π

Now,

Circumference of the required circle

=

Sum of circumference of two circles

2 π r = 18 π + 38 π

2 π r = 58 π

r = \frac{5 6 π}{2 π}

r = 28 c m

Hence, radius of new circle is 28 cm.

Answer

Let the radius of required circle

= r c m

Radius of 1st circle

r_{1} = 9 c m

Radius of 2nd circle

r_{2} = 19 c m

As per the question

Circumference of the required circle

=

Sum of circumference of two circles

Circumference of small circle

= 2 π r_{1}

= 2 π \times 9

= 18 π

Circumference of small circle

= 2 π r_{2}

= 2 π \times 19

= 38 π

Now,

Circumference of the required circle

=

Sum of circumference of two circles

2 π r = 18 π + 38 π

2 π r = 58 π

r = \frac{5 6 π}{2 π}

r = 28 c m

Hence, radius of new circle is 28 cm.

Question 2

The radii of two circles are

8

cm and

6

cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.

A

32 c m

B

23 c m

C

15 c m

D

10 c m

Medium

Solution

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Answer

Let the radius of required circle is

r

cm.
So,

π r^{2} = π r_{1}^{2} + π r_{2}^{2}

r^{2} = r_{1}^{2} + r_{2}^{2}

r^{2} = (8)^{2} + (6)^{2}

r^{2} = 100

r = 10 c m

Answer

Let the radius of required circle is

r

cm.
So,

π r^{2} = π r_{1}^{2} + π r_{2}^{2}

r^{2} = r_{1}^{2} + r_{2}^{2}

r^{2} = (8)^{2} + (6)^{2}

r^{2} = 100

r = 10 c m

Question 3

Fif. 12.3 depicts an archery target marked with its five scoring regions 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 each of the five scoring regions.

Medium

Solution

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Answer

SOLUTION:

GIVEN:

Diameter of Gold region= 21 cm

Radius of gold region=

21/2= 10.5 cm

Area of Gold Region = πr²

= π(10.5)² =( 22/7)× 110.25 = 346.5 cm²

Area of Gold Region= 345.5

cm²

Radius of red region =

Radius for gold + red region= 10.5 + 10.5= 21 cm

Area of Red Region = π²(21² - 10.5²)

[Area of a ring= π (R²-r²), where R= radius of outer ring & r= radius of inner ring]

= 22/7 (21² – 10.5²) [ a²-b²= (a+b)(a-b)]

= 22/7 (21 + 10.5)(21 – 10.5)

= (22/7 )x 31.5 x 10.5 = 1039.5 cm²

Area of Red Region =

1039.5 cm²

Radius of blue region =

Radius of blue region = Now radius for gold + red+ blue region= 21+10.5= 31.5 cm

Area of Blue Region = π(31.5² – 21²)

= 22/7 (31.5² - 21²)

= 22/7 (31.5 +21)(31.5 - 21)

= (22/7 )x 52.5 x 10.5 = 1732.5 cm²

Area of Blue Region =1732.5

cm²

Now,

Radius of black region =

radius for gold + red+ blue + black region= 31.5+10.5= 42 cm

Area of Black Region = π(42² – 31.5²)

= 22/7 (42²-31.5² )

= 22/7 (42+31.5)(42-31.5)

= (22/7 )x 73.5 x 10.5 = 2425.5 cm²

Area of Black Region =2425.5

cm²

Now

Radius of white region =

radius for gold + red+ blue + black+ white region= 42+10.5= 52.5 cm

Area of White Region= π(52.5² – 42²)

= 22/7 (52.5²-42² )

= 22/7 (52.5+42)(52.5-42)

= (22/7 )x 94.5 x 10.5 = 3118.5 cm²

Area of white Region =3118.5

cm²

Answer

SOLUTION:

GIVEN:

Diameter of Gold region= 21 cm

Radius of gold region=

21/2= 10.5 cm

Area of Gold Region = πr²

= π(10.5)² =( 22/7)× 110.25 = 346.5 cm²

Area of Gold Region= 345.5

cm²

Radius of red region =

Radius for gold + red region= 10.5 + 10.5= 21 cm

Area of Red Region = π²(21² - 10.5²)

[Area of a ring= π (R²-r²), where R= radius of outer ring & r= radius of inner ring]

= 22/7 (21² – 10.5²) [ a²-b²= (a+b)(a-b)]

= 22/7 (21 + 10.5)(21 – 10.5)

= (22/7 )x 31.5 x 10.5 = 1039.5 cm²

Area of Red Region =

1039.5 cm²

Radius of blue region =

Radius of blue region = Now radius for gold + red+ blue region= 21+10.5= 31.5 cm

Area of Blue Region = π(31.5² – 21²)

= 22/7 (31.5² - 21²)

= 22/7 (31.5 +21)(31.5 - 21)

= (22/7 )x 52.5 x 10.5 = 1732.5 cm²

Area of Blue Region =1732.5

cm²

Now,

Radius of black region =

radius for gold + red+ blue + black region= 31.5+10.5= 42 cm

Area of Black Region = π(42² – 31.5²)

= 22/7 (42²-31.5² )

= 22/7 (42+31.5)(42-31.5)

= (22/7 )x 73.5 x 10.5 = 2425.5 cm²

Area of Black Region =2425.5

cm²

Now

Radius of white region =

radius for gold + red+ blue + black+ white region= 42+10.5= 52.5 cm

Area of White Region= π(52.5² – 42²)

= 22/7 (52.5²-42² )

= 22/7 (52.5+42)(52.5-42)

= (22/7 )x 94.5 x 10.5 = 3118.5 cm²

Area of white Region =3118.5

cm²

Question 4

Find the area of a sector of a circle with radius

6

cm if angle of the sector is

6 0^{o}

.

Hard

Solution

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Answer

Area of sector

= \frac{θ}{3 6 0 ^{o}} \times π r^{2}

where

θ = 6 0^{o}

r = 6

cm

\Rightarrow \frac{6 0 ^{o}}{3 6 0 ^{o}} \times π \times 6 \times 6 = \frac{1}{6} \times \frac{2 2}{7} \times 36 = \frac{2 2}{7} \times 6

= \frac{1 3 2}{7} c m^{2}

.

Answer

Area of sector

= \frac{θ}{3 6 0 ^{o}} \times π r^{2}

where

θ = 6 0^{o}

r = 6

cm

\Rightarrow \frac{6 0 ^{o}}{3 6 0 ^{o}} \times π \times 6 \times 6 = \frac{1}{6} \times \frac{2 2}{7} \times 36 = \frac{2 2}{7} \times 6

= \frac{1 3 2}{7} c m^{2}

.

Question 5

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

Easy

Solution

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Answer

w e h a v e C = 22 c m \Rightarrow 2 π r = 22 c m \Rightarrow 2 \times \frac{2 2}{7} \times r = 22 ∴ r = 3.5 c m N o w, a r e a o f q u a d r a n t o f a c i r c l e = \frac{π r ^{2}}{4} = \frac{2 2}{7} \times \frac{( 3 . 5 ) ^{2}}{4} = \frac{2 2 \times 1 2 . 2 5}{2 8} = 9.625 c m^{2} .

Answer

w e h a v e C = 22 c m \Rightarrow 2 π r = 22 c m \Rightarrow 2 \times \frac{2 2}{7} \times r = 22 ∴ r = 3.5 c m N o w, a r e a o f q u a d r a n t o f a c i r c l e = \frac{π r ^{2}}{4} = \frac{2 2}{7} \times \frac{( 3 . 5 ) ^{2}}{4} = \frac{2 2 \times 1 2 . 2 5}{2 8} = 9.625 c m^{2} .

Question 6

The length of the minute hand of a clock is 14cm. find the area swept by the minute hand in 15 min.

Medium

Solution

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Answer

Minute hand completes full circle degree in

60

minutes.

Angle swept by minute in

60

minutes

= 360 °

Angle swept by the minute hand in

15

minutes

= \frac{3 6 0 °}{6 0} \times 15 = 90 °

Therefore,

θ = 90 °

Length of minute hand

= r = 14 c m

Area swept by minute hand in

15

minutes

=

Area of sector

As we know that area of sector is given as-

A = \frac{θ}{3 6 0 °} \times π r^{2}

Therefor,

Area swept by the minute hand in

15

minutes

= \frac{9 0 °}{3 6 0 °} \times (\frac{2 2}{7} \times (14)^{2}) = \frac{1}{4} \times 616 = 154 c m^{2}

Hence the area swept by the minute hand in

15

minutes is

154 c m^{2}

.

Hence the correct answer is

154 c m^{2}

.

Answer

Minute hand completes full circle degree in

60

minutes.

Angle swept by minute in

60

minutes

= 360 °

Angle swept by the minute hand in

15

minutes

= \frac{3 6 0 °}{6 0} \times 15 = 90 °

Therefore,

θ = 90 °

Length of minute hand

= r = 14 c m

Area swept by minute hand in

15

minutes

=

Area of sector

As we know that area of sector is given as-

A = \frac{θ}{3 6 0 °} \times π r^{2}

Therefor,

Area swept by the minute hand in

15

minutes

= \frac{9 0 °}{3 6 0 °} \times (\frac{2 2}{7} \times (14)^{2}) = \frac{1}{4} \times 616 = 154 c m^{2}

Hence the area swept by the minute hand in

15

minutes is

154 c m^{2}

.

Hence the correct answer is

154 c m^{2}

.

Question 7

In a circle of radius

21 c m

, an arc subtends and angle of

6 0^{\circ}

at the centre. Find:
(i) The length of the arc (ii) Area of the sector formed by the arc (iii) Area of the segment formed by the corresponding chord

Hard

Solution

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Answer

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Question 8

A chord of a circle of radius

15 c m

subtends an angle of

6 0^{\circ}

at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use

π = 3.14

and

3 = 1.73

)

Medium

Solution

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Answer

In the mentioned figure,

O

is the centre of circle,

A B

is a chord

A X B

is a major arc,

O A = O B =

radius

=

15

cm
Arc

A X B

subtends an angle

6 0^{o}

at

O .

Area of sector

A O B = \frac{6 0}{3 6 0} \times π \times r^{2}

= \frac{6 0}{3 6 0} \times 3.14 \times (15)^{2}

= 117.75 c m^{2}

Area of minor segment (Area of Shaded region)

=

Area of sector

A O B -

Area of

△

A O B

By trigonometry,

A C = 15 sin 30

O C = 15 cos 30

And,

A B = 2 A C

∴

A B = 2 \times 15 sin 30 = 15

cm

∴

O C = 15 cos 30 = 15 \frac{3}{2} = 15 \times \frac{1 . 7 3}{2} = 12.975

cm

∴

Area of

△ A O B = 0.5 \times 15 \times 12.975 = 97.3125 c m^{2}

∴

Area of minor segment (Area of Shaded region)

= 117.75 - 97.3125 = 20.4375

c m^{2}

Area of major segment

=

Area of circle

-

Area of minor segment

= (3.14 \times 15 \times 15) - 20.4375

= 686.0625 c m^{2}

Answer

In the mentioned figure,

O

is the centre of circle,

A B

is a chord

A X B

is a major arc,

O A = O B =

radius

=

15

cm
Arc

A X B

subtends an angle

6 0^{o}

at

O .

Area of sector

A O B = \frac{6 0}{3 6 0} \times π \times r^{2}

= \frac{6 0}{3 6 0} \times 3.14 \times (15)^{2}

= 117.75 c m^{2}

Area of minor segment (Area of Shaded region)

=

Area of sector

A O B -

Area of

△

A O B

By trigonometry,

A C = 15 sin 30

O C = 15 cos 30

And,

A B = 2 A C

∴

A B = 2 \times 15 sin 30 = 15

cm

∴

O C = 15 cos 30 = 15 \frac{3}{2} = 15 \times \frac{1 . 7 3}{2} = 12.975

cm

∴

Area of

△ A O B = 0.5 \times 15 \times 12.975 = 97.3125 c m^{2}

∴

Area of minor segment (Area of Shaded region)

= 117.75 - 97.3125 = 20.4375

c m^{2}

Area of major segment

=

Area of circle

-

Area of minor segment

= (3.14 \times 15 \times 15) - 20.4375

= 686.0625 c m^{2}

Question 9

A chord of a circle of radius 12 cm subtends an angle of

12 0^{\circ}

at the centre. Find the area of the corresponding segment of the circle.(Use

π = 3.14

and

3 = 1.73

Medium

Solution

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Answer

Given that:-

Radius of circle

(r) = 12 c m

θ = 120 °

To find:-

Area of segment

A P B = ?

Solution:-

Area of sector

O A P B (A_{1}) = \frac{θ}{3 6 0 °} \times π r^{2}

\Rightarrow A_{1} = \frac{1 2 0}{3 6 0} \times 3.14 \times 12^{2} = 150.72 c m^{2}

Let M be the point on

A B

such that

A B ⊥ O M

∴ ∠ O M A = ∠ O M B = 90 °

Now in

△ O M A

and

△ O M B

,

∠ O M A = ∠ O M B [each 90 °]

O A = O B [∵ O A = O B = r]

O M = O M [common]

By R.H.S. congruency,

△ O M A ≅ △ O M B

Now by CPCT,

∠ A O M = ∠ B O M

A M = B M

Therefore,

∠ A O M = ∠ B O M = \frac{1}{2} ∠ A O B = 60 °

A M = B M = \frac{1}{2} A B . . . . . (1)

Now in right angled

△ A O M

sin 60 = \frac{A M}{O A} [∵ sin θ = \frac{Perpndicular}{hypotenuse}]

\Rightarrow sin 60 = \frac{A M}{1 2}

\Rightarrow A M = 63 c m

cos 60 = \frac{O M}{O A} [∵ cos θ = \frac{base}{hypotenuse}]

\Rightarrow cos 60 = \frac{O M}{1 2}

\Rightarrow O M = 6 c m

Now, from

e q^{n} (1)

, we have

A B = 2 A M = 123 c m

Now, area of

△ A O B (A_{2})

will be-

A_{2} = \frac{1}{2} \times A B \times O M = \frac{1}{2} \times 123 \times 6 = 363 c m^{2} = 62.28 c m

∴

Area of segment

A P B = A_{1} - A_{2} = 150.72 - 62.28 = 88.44 c m^{2}

Hence the aea of the corresponding segment of the circle is

88.44 c m^{2}

.

Answer

Given that:-

Radius of circle

(r) = 12 c m

θ = 120 °

To find:-

Area of segment

A P B = ?

Solution:-

Area of sector

O A P B (A_{1}) = \frac{θ}{3 6 0 °} \times π r^{2}

\Rightarrow A_{1} = \frac{1 2 0}{3 6 0} \times 3.14 \times 12^{2} = 150.72 c m^{2}

Let M be the point on

A B

such that

A B ⊥ O M

∴ ∠ O M A = ∠ O M B = 90 °

Now in

△ O M A

and

△ O M B

,

∠ O M A = ∠ O M B [each 90 °]

O A = O B [∵ O A = O B = r]

O M = O M [common]

By R.H.S. congruency,

△ O M A ≅ △ O M B

Now by CPCT,

∠ A O M = ∠ B O M

A M = B M

Therefore,

∠ A O M = ∠ B O M = \frac{1}{2} ∠ A O B = 60 °

A M = B M = \frac{1}{2} A B . . . . . (1)

Now in right angled

△ A O M

sin 60 = \frac{A M}{O A} [∵ sin θ = \frac{Perpndicular}{hypotenuse}]

\Rightarrow sin 60 = \frac{A M}{1 2}

\Rightarrow A M = 63 c m

cos 60 = \frac{O M}{O A} [∵ cos θ = \frac{base}{hypotenuse}]

\Rightarrow cos 60 = \frac{O M}{1 2}

\Rightarrow O M = 6 c m

Now, from

e q^{n} (1)

, we have

A B = 2 A M = 123 c m

Now, area of

△ A O B (A_{2})

will be-

A_{2} = \frac{1}{2} \times A B \times O M = \frac{1}{2} \times 123 \times 6 = 363 c m^{2} = 62.28 c m

∴

Area of segment

A P B = A_{1} - A_{2} = 150.72 - 62.28 = 88.44 c m^{2}

Hence the aea of the corresponding segment of the circle is

88.44 c m^{2}

.

Question 10

A horse is tied to a peg at one corner of a square shaped grass field of side

15 m

by means of a

5 m

long rope. Find
(i) the area of that part of the field in which the horse can graze.
(ii) The increase in the grazing area if the rope were

10 m

long instead of

5 m

. (Use

π = 3.14

)

Medium

Solution

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Answer

Side of square

= 15 m

Length of rope

= 5 m =

radius

The area available for horse to graze is nothing but "Area of Quadrant of a circle'

∴

Area of Quadrant

= \frac{π \times r ^{2}}{4}

= \frac{3 . 1 4 \times 5 \times 5}{4}

= 19.625 m^{2}

If the length of rope is increased to

10 m

then the new radius ,

= 10 m

∴

Area of new quadrant

= \frac{3 . 1 4 \times 1 0 \times 1 0}{4}

= 78.5 m^{2}

∴

Increase in grazing area

= 78.5 - 19.625 = 58.875 m^{2}

Answer

Side of square

= 15 m

Length of rope

= 5 m =

radius

The area available for horse to graze is nothing but "Area of Quadrant of a circle'

∴

Area of Quadrant

= \frac{π \times r ^{2}}{4}

= \frac{3 . 1 4 \times 5 \times 5}{4}

= 19.625 m^{2}

If the length of rope is increased to

10 m

then the new radius ,

= 10 m

∴

Area of new quadrant

= \frac{3 . 1 4 \times 1 0 \times 1 0}{4}

= 78.5 m^{2}

∴

Increase in grazing area

= 78.5 - 19.625 = 58.875 m^{2}

NCERT Solutions for Class 10 Maths Chapter 12 : Areas Related to Circles

Access NCERT Solutions for Class 10 Maths Chapter 12 : Areas Related to Circles

NCERT Solutions for Class 10 Maths Chapter 12 : Areas Related to Circles

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