mathamatics xam idea

Now, putting the value of x in equation (i), we have

h = 3 × 20 = 20 3

Hence, the height of the pole is 20 3 m and the distance of the point from first pole is 20 m and

that of the second pole is 60 m.

9. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly

opposite the tower, the angle of elevation of the top of the tower is 60°. From another point

20 m away from this point on the line joining this point to the foot of the tower, the angle of

elevation of the top of the tower is 30° (Fig. 11.42). Find the height of the tower and the width

of the canal. [NCERT]

Sol. Let height of the tower be h metres and width of the canal be x metres, so AB = h m and BC = x m

Now in DABC, we have

tan 60° = h ⇒ 3 = h ⇒ h = 3 x ...(i)
x x

Now, in DADB, we have

tan 30° = AB ⇒ 1 = h x ⇒ 20 + x = 3 h ...(ii)
DB 3 20 +
From (i) and (ii), we have

20 + x = 3 × 3 x ⇒ 20 + x = 3x

⇒ 20 = 3x – x = 2x ⇒ x= 20 = 10 m
2

Now, putting the value of x in equation (i), we have

h = 3 × 10 = 10 3 ⇒ h = 10 3 m Fig. 11.42

Hence, height of the tower is 10 3 m and width of the canal is 10 m.

10. A person standing on the bank of a river observes that the angle of elevation of the top of a tree
standing on the opposite bank is 60°. When he moves 40 metres away from the bank, he finds
the angle of elevation to be 30°. Find the height of the tree and the width of the river.

Sol. Let AB be the tree of height h metres standing on the bank of a river. Let C be the position of
man standing on the opposite bank of the river such that BC = x m. Let D be the new position
of the man. It is given that CD = 40 m and the angles of elevation of the top of the tree at C and
D are 60° and 30°, respectively, i.e.,

∠ACB = 60° and ∠ADB = 30°

In DACB, we have

tan 60° = AB ⇒ tan 60° = h
BC x

⇒ 3 = h ⇒ x= h ...(i)
x 3

In DADB, we have

tan 30° = AB ⇒ 1 = x h
⇒ BD 3 + 40
Fig. 11.43

3 h = x + 40 ...(ii)

Substituting x= h in equation (ii), we get
3

3h = h + 40 ⇒ 3h − h = 40
3 3

Heights and Distances 293

⇒ 3h − h = 40 ⇒ 2h = 40
3 3

⇒ h = 40 × 3 ⇒ h = 20 3 = 20 × 1.732 = 34.64 m
2

Substituting h in equation (i), we get x = 20 3 metres = 20 metres
3

Hence, the height of the tree is 34.64 m and width of the river is 20 m.

11. The angles of elevation and depression of the top and bottom of a lighthouse from the top of
a building, 60 m high, are 30° and 60° respectively. Find

(i) the difference between the heights of the lighthouse and the building.

(ii) distance between the lighthouse and the building.          [CBSE Delhi 2014]

Sol. Let AB be the building and CE be the lighthouse (Fig. 11.44).

In right-angled DABC,

    t an 60° = AB
BC

⇒ 3 = 60    ⇒   3 BC = 60
BC

⇒      BC = 60 × 3 = 60 3 = 20 3
3 3 3

= 20 (1.732) = 34.64 m ( 3 = 1.732)

As      AD = BC (Opposite sides of a rectangle)

\       AD = 20 3 m Fig. 11.44
In right-angled DADE,

tan 30° = DE ⇒ 1 = DE
AD 3 20 3

⇒ 3 DE = 20 3 ⇒ DE = 20 m

\ (i) Difference between the heights of lighthouse and building = EC – DC = DE = 20 m

and (ii) Distance between the lighthouse and the building = BC = 34.64 m.

12. In Fig. 11.45, from the top of a building AB, 60 metres high, the angles of depression of the
top and bottom of a vertical lamp post CD are observed to be 30° and 60°, respectively. Find

(i) the horizontal distance between AB and CD.

(ii) the height of the lamp post.

Sol. Given AB is the building and CD is the vertical lamp post. Then, DE is the horizontal distance
between AB and CD.

Let CD = h metres, then ∠EDB = 30° and ∠ACB = 60°, AE = CD = h
metres and EB = (60 – h) m

In DABC,

  CA = cot 60° or CA = 1
AB 60 3

\   CA = d60 # 1 n = 20 3m
3

\   DE = CA = 20 3 m Fig. 11.45

294 Xam idea Mathematics–X

In DBDE, 1
3
BE = tan 30° =
DE

or BE = 1   or BE = 20 m
20 3 3

⇒ 60 – h = 20,  i.e., h = 40 m

\ (i) Horizontal distance between AB and CD = 20 3 m = 20 × 1.732 = 34.64 m

(ii) The height of lamp post = 40 m

13. A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at

an elevation of 30°. A girl standing on the roof of 20 metre high building, finds the angle of

elevation of the same bird to be 45°. Both the boy and the girl are on opposite sides of the bird.

Find the distance of bird from the girl. [Given 2 = 1.414 ] [CBSE 2019 (30/5/1)]

Sol. Let B be the position of bird. O and P be the positions of boy and girl respectively and PQ be the
building

We have, ∠AOB = 30° and, ∠BPM = 45°
In DAOB, we have

sin 30° = AB ⇒ 1 = AB
100 2 100

⇒ AB = 50 m

\ BM = AB – AM = 50 – 20 = 30 m Fig. 11.46

In DPBM, we have

BM ⇒ 1 = 30   ⇒    BP = 30 2m
sin 45° = BP 2 BP

Hence, distance of bird from girl is 30 2 m.

14. The angles of depression of the top and the bottom of a 8 m tall building from the top of a
multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed
building and the distance between the two buildings.

Sol. In Fig. 11.47, PC denotes the multi-storeyed building and AB denotes the 8 m tall building. We
are interested to determine the height of the multi-storeyed building, i.e., PC and the distance
between the two buildings, i.e., AC.

Let PD = x m

In right DPDB, we have

PD = tan 30° = 1 or x = 1
BD 3 BD 3

or x 3 = BD …(i)

In right DPCA, we have

PC = tan 45° =1 ⇒ x+8 =1 ( AC = BD)
AC BD
Fig. 11.47
⇒ x + 8 = BD ⇒ x + 8 = x 3 (Using (i))

⇒ x( 3 – 1) = 8 ⇒ x= 8 × 3 + 1 = 8( 3 + 1) = 4( 3 + 1)
3− 3 + 1 3−1
1

From (i), BD = [4( 3 + 1)] 3 = 4(3 + 3 )

Heights and Distances 295

So, the height of the multi-storeyed building is {4( 3 + 1) + 8} m = 4(3 + 3 ) m

and the distance between the two buildings is also 4(3 + 3 ) m.

15. The angle of elevation of the top of a tower at a distance of 120 m
from a point A on the ground is 45°. If the angle of elevation of the
top of a flagstaff fixed at the top of the tower, at A is 60°, then find

the height of the flagstaff. (Use 3 = 1.73)

Sol. Let CD be the tower which is at a distance of 120 m from A.

BD = x be the length of flagstaff.

In DACD, tan 45° = CD ⇒ 1 = h
CA 120

⇒ h = 120 m Fig. 11.48

In DABC, tan 60° = BC ⇒ 3 = x+h
AC 120

120 3 = x + 120

x = 120 × 1.73 – 120

x = 120 (1.73 –1) = 120 × 0.73

x = 87.6 m

16. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of
elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high,
find the height of the building.

Sol. Let AB be the building of height h m and CD be the tower of height 50 m. We have,

∠ACB = 30° and ∠DAC = 60°

In DACD, we have

tan 60° = DC ⇒ 3 = 50
AC AC

⇒ AC = 50 = 50 × 3 = 50 3 ...(i)
3 3 3 3

⇒ AC = 50 3 m
3
Fig. 11.49
In DABC, we have

tan 30° = AB
AC

⇒ 1 = h ⇒ AC = 3h
3 AC

⇒ h = =AC3 5=03 3 50 = 16 2 m (From equation (i))
3 3

3

Hence, the height of the building is 16 2 m.
3

17. The shadow of a tower at a time is three times as long as its shadow when the angle of elevation

of the sun is 60°. Find the angle of elevation of the sun at the time of the longer shadow.

[CBSE (F) 2017]

296 Xam idea Mathematics–X

Sol. Let AB be the tower and BC be the length of its shadow when the Sun rays meet the ground at
an angle of 60°. Let q be the angle between the Sun rays and the ground when the length of the
shadow of the tower is BD. Let h be the height of the tower (Fig. 11.50).

Fig. 11.50

Let BC = x \ BD = 3x and CD = 2x

In DABC, we have AB h
BC x
tan 60° = ⇒ 3 = ⇒ h = 3x

In DADB, we have

tan q = AB
BD

⇒ tan q = h ⇒ tan q = 3x (a h = 3x )
3x 3x

⇒ tan q = 1 ⇒ tan q = tan 30° ⇒ q = 30°
3

18. From the top of a hill, the angles of depression of two consecutive kilometre stones due east

are found to be 45° and 30° respectively. Find the height of the hill. [CBSE (F) 2017]

Sol. Let the height of the hill be h m, C and D are two consecutive stones having distance 1000 m
between them and AC = x m.

B

45°

h

45° 30°

A x C 1 km D
1000 m

Fig. 11.51

In DABC, tan 45° = h
x

⇒ x = h ...(i)

In DABD, h 1 h
1000 3 x + 1000
tan 30° = x +   ⇒  =

⇒ h + 1000 = 3 h (From equation (i))

⇒ h ^ 3 –1h = 1000

⇒ h= 1000 or 500^ 3 +1h m
3 –1

Hence, the height of the hill = 500^ 3 + 1h m

Heights and Distances 297

HOTS [Higher Order Thinking Skills]

1. A man standing on the deck of a ship, which is 10 m above the water level, observes the angle
of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°.
Calculate the distance of the hill from the ship and the height of the hill.

Sol. In Fig. 11.52, let C represents the position of the man on the deck of the
ship, A represents the top of hill and D its base.

Now in right-angled triangle CWD,

tan 30° = 10 ⇒ WD = 10
WD tan 30o

⇒   WD = 10 = 10 3 = 17.3 m
1

3

Also, in right-angled triangle ABC, we have

tan 60° = AB or AB    (From fig. BC = WD)
⇒ BC WD
Fig. 11.52
AB
3 = 10 3 ⇒ AB = 10 3 × 3 = 30 m

Now, AD = AB + BD = 30 m + 10 m = 40 m

Therefore, the distance of the hill from the ship is 17.3 m and the height of the hill is 40 m.

2. A spherical balloon of radius r subtends an angle q at the eye of an observer. If the angle of
elevation of its centre is f, find the height of the centre of the balloon. [NCERT Exemplar]

Sol. In Fig. 11.53, O is the centre of balloon, whose radius OP = r and ∠PAQ = q. Also, ∠OAB = f.

Let the height of the centre of the balloon be h. Thus OB = h.

In DOAP, we have

sin θ = r , where OA = d ...(i)
2 d

Also in DOAB,

sin f = h ...(ii)
d

From (i) and (ii), we get

sin φ h h
r
θ = d = or h = r sin f cosec θ Fig. 11.53
2 r 2
sin d

3. From a window (h metres high above the ground) of a house in a street, the angles of elevation

and depression of the top and the foot of another house on the opposite

side of the street are q and f respectively. Show that the height of the

opposite house is h (1 + tan q cot f).

Sol. Let W be the window and AB be the house on the opposite side.

Then, WP is the width of the street (Fig. 11.54).

Let AP = h' m PB
WP
In DBPW, tan f =

⇒ h = tan f ⇒ WP = h cot f …(i) Fig. 11.54
WP

298 Xam idea Mathematics–X

Now, in DAWP, tan q = AP = h′
WP WP

⇒ h' = WP tan q  ⇒  h' = h cot f tan q

\ Height of house = h' + h

= h cot f tan q + h = h (1 + tan q cot f)

4. The angle of elevation of a aeroplane from a point A on the ground is 60°. After a flight of

30 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a constant height

of 3600 3 metres then find the speed of the aeroplane. [CBSE 2019 (30/5/1)]

Sol. Let the position of plane be at C, after 30 seconds, it well be C C´
at C' (See figure)

Let AB = x m and ABl = y m

In DABC, we have 3600

tan 60° = BC
AB
B’
xm
⇒ 3 = 3600 3 ⇒ x = 3600 m ym
x
Fig. 11.55
Now,
In D ABlCl, we have

tan 30° = BlCl ⇒ 1 = 3600 3
ABl 3 y

⇒ y = 3600 × 3 = 10800 m

\ Distance covered in 30 seconds = y – x

= (10800 – 3600) m = 7200 m

\ Speed of the plane = 7200
⇒ 30

Speed of the plane = 240 m/sec. = 240 × 18 km/h = 864 km/h
5

5. If the angle of elevation of a cloud from a point h metres above a lake is a and the angle of

depression of its reflection in the lake is b, prove that the height of the cloud is h (tan b + tan a)
tan b – tan a

[NCERT Exemplar]

Sol. Let AB be the surface of the lake and let P be a point of observation (Fig. 11.56) such that
AP = h metres. Let C be the position of the cloud and C' be its reflection in the lake. Then,
CB = C'B. Let PM be perpendicular from P on CB. Then, ∠CPM = a and ∠MPC' = b. Let
CM = x.

Fig. 11.56

Heights and Distances 299

Then, CB = CM + MB = CM + PA = x + h

In DCPM, we have

tan a = CM ⇒ tan a = x ( a PM =AB)
⇒ PM AB ...(i)

AB = x cot a

In DPMC', we have

tan b = C′M ⇒ tan b = x + 2h ( a C'M = C'B + BM = x + h + h)
PM AB

⇒ AB = (x + 2h) cot b ...(ii)

From (i) and (ii), we have

⇒ x cot a = (x + 2h) cot b ⇒ x(cot a – cot b) = 2h cot b

⇒ x  1 − 1  = 2h x  tanβ − tan α  = 2h
 tan tan  tan β  tan α tanβ  tan β
 α β  ⇒  

⇒ x = 2h tan α
tan β − tan α

Hence, the height CB of the cloud is given by 2h tan α

CB = x + h ⇒ CB = tan β − tan α + h

⇒ CB = 2h tan α + h tan β − h tan α = h(tan α + tan β)
tan β − tan α tan β − tan α

6. From an aeroplane vertically above a straight horizontal plane, the angles of depression of two
consecutive kilometre stones on the opposite sides of the aeroplane are found to be a and b .

Show that the height of the aeroplane is tan α . tan β .
tan α + tan β

Sol. In Fig. 11.57, let P be the position of plane, A and B be the positions of two stones one kilometre
apart. Angles of depression of stones A and B are a and b respectively. Let PC = h.

Fig. 11.57

In right-angled triangle ACP, we have

⇒ tan a = PC = h , h = AC tan a
AC AC

⇒ AC = h α ...(i)
tan

Now in right-angled triangle PCB, we have

tan b = PC ⇒ tan b = h
CB CB

300 Xam idea Mathematics–X

⇒ (tan b) CB = h h
⇒ CB = tan β ...(ii)

Adding (i) and (ii), we have

AC + CB = h α + h β = h  tan β + tan α 
tan tan  tan α . tan β 


As it is given that AC + CB = 1

\ 1 = h  tan β + tan α  ⇒ h = tan α . tan β
 tan α . tan β  tan α + tan β
 

7. A man in a boat rowing away from a light house 100 m high takes 2 minutes to change the

angle of elevation of the top of the light house from 60° to 30°. Find the speed of the boat in

metres per minute. (Use 3 = 1.732] ) [CBSE 2019 (30/1/2)]

Sol. Let AB be the light house of height = 100 m.

100

xm ym

Fig. 11.58
Also let the distance DC = x metres and CB = y metres.

In DABC, we have

tan 60° = AB
BC

3 = 100 & y = 100 m ...(i)
y 3

In DABD,

tan 30° = AB
BD

1 = 100 & x + y = 100 3 ...(ii)
3 x+y

From equations (i) and (ii), we get

x+ 100 = 100 3 & 3 x + 100 = 300
3

⇒ x = 2030 m

Time taken to travel from C to D = 2 minutes

As we know that speed = distance
Time

= 200 = 200 = 100 = 57.74 m/ min (approrx.)
3 2× 3 1.732

2

Hence, the speed of boat is 57.74 m/ min.

Heights and Distances 301

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) If the angle of elevation of a tower from a distance of 100 m from its foot is 60°, then the
height of the tower is

(a) 100 3 m (b) 200 3 m (c) 50 3 m (d) 300 3 m

(ii) A tower is 100 m high, what will be the angle of elevation, from a point 100 m away from
its foot, to the top of the tower?

(a) 60° (b) 90° (c) 45° (d) 30°

(iii) The shadow of a pole is equal to its height at 10 : 45 am. The Sun's altitude is

(a) 30° (b) 45° (c) 60° (d) 90°

(iv) In Fig. 11.59, the value of CE is

DC

60°

6 cm

E

A 10 cm B

Fig. 11.59

(a) 12 cm (b) 6 cm (c) 9 cm (d) 6 3 cm

(v) If the angles of elevation of a tower from two points distant a and b where a > b from its

foot and in the same straight line from it are 30° and 60°, then the height of the tower is
a
(a) a + b (b) ab (c) a – b (d) b

2. Fill in the blanks.

(i) Angle of elevation and angle of depression are always ___________ angles.

(ii) If the angle of elevation of Sun ___________ then length of shadow of object decreases.

(iii) The angle for which sine and cosine have equal values is ___________.

(iv) When length of the shadow of a pole is equal to its height the angle of elevation of source

of light is ___________.
(v) The angle of ______________ implies that observer is at a higher altitude than the object.

QQ Very Short Answer Questions: [1 mark each]

3. The ratio of the length of a tree and its shadow is 1 : 1 . What is the Sun's angle of elevation?
3

4. If two towers of height h1 and h2 subtend angle of 60° and 30° respectively at the mid-point of the

line joining their feet, then find h1 : h2. [CBSE Delhi 2015]

5. The height of the tower is 100 m. When the angle of elevation of Sun is 30°, then what is the

length of shadow of the tower?

6. The tops of two poles of height 16 m and 10 m are connected by a wire of length l metres. If the
wire makes an angle of 30° with the horizontal, then find l.

302 Xam idea Mathematics–X

7. An observer, 1.5 m tall, is 28.5 m away from a 30 m high tower. Determine the angle of elevation

of the top of the tower from the eye of the observer. [CBSE Delhi 2017 (C)]

8. The angle of elevation of the top of a tower from a point P on the ground is a. After walking a
distance d meter towards the foot of the tower, angle of elevation is found to be b. Which angle
of elevation is greater?

9. What is the angle of elevation of Sun if the ratio of the length of a rod and its shadow is 1 : 1?

10. A ladder, leaning against a wall, makes an angle of 60° with the horizontal. If the foot of the

ladder is 2.5 m away from the wall, find the length of the ladder. [CBSE (AI) 2016]

QQ Short Answer Questions–I: [2 marks each]

State whether the statements are true or false. Justify (Q. 11 to 12)

11. If the length of the shadow of a tower is increasing then the angle of elevation of the sun is also

increasing. [NCERT Exemplar]

12. The height of a tower is 15 m. The length of its shadow is 5 3 m when Sun’s altitude is 30°.

13. An electrician has to repair an electric fault on a pole of height 4 m. He needs to reach a point
1.3 m below the top of the pole to undertake the repair work. What should be the length of the
ladder that he should use which when inclined at an angle of 60° to the horizontal would enable
him to reach the required position?

14. In Fig. 11.60, what is the angle of elevation of point A from C and angle of depression of point D
from A?

Fig. 11.60

15. In Fig. 11.61, height of a building is h + 2 and point C is h m from the foot of the building. Find
the angle of elevation of the top of the building from a point 2 m from point C.

2m
Fig. 11.61

16. On a straight line passing through the foot of a tower, two points C and D are at distances of

4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the

tower are complementary, then find the height of the tower. [CBSE (AI) 2017]

QQ Short Answer Questions–II: [3 marks each]

17. A tree is broken by wind. The top struck the ground at an angle of 30° and at a distance of 30 m
from the root. Find the height of the whole tree.

18. The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 9.5 m
away from the wall. Find the length of the ladder.

Heights and Distances 303

19. From a window 15 m high above the ground in a street, the angles of elevation and depression
of the top and the foot of another house on the opposite side of the street are 30° and 45°,
respectively. Show that the height of the opposite house is 23.66 m.

20. A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole as
observed from a point A on the ground is 60° and the angle of depression of the point A from the
top of the tower is 45°. Find the height of the tower.

21. The length of a string between a kite and a point on the ground is 90 metres. If the string makes
15
an angle q with the ground level such that tan q = 8 , how high is the kite? Assume that there
is no slack in the string.

22. The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From

another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the

tower. [NCERT Exemplar]

23. A man standing on the deck of a ship, which is 10 m above water level, observes the angle of

elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30°. Find the

distance of the hill from the ship and the height of the hill. [CBSE (AI) 2016]

24. The angles of depression of the top and bottom of a 50 m high building from the top of a tower

are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the
tower and the building. (Use 3 = 1.73).
[CBSE Delhi 2016]

25. From the top of a 120 m high tower, a man observes two cars on the opposite sides of the tower

and in straight line with the base of tower with angles of depression as 60° and 45°. Find the

distance between the two cars. (Take 3 = 1.732) [CBSE Delhi 2017 (C)]

QQ Long Answer Questions: [5 marks each]

26. As observed from the top of a 100 m high light house from the sea-level, the angles of depression

of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light

house, find the distance between the two ships. [Use 3 = 1.732 ] [CBSE 2018 (30/1)]

27. The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of depression

from the top of tower to the foot of hill is 30°. If tower is 50 metre high, find the height of the hill.

[CBSE 2018 (C) (30/1)]

28. A man in a boat rowing away from a light house 100 m high takes 2 minutes to change the angle

of elevation of the top of the light house from 60° to 30°. Find the speed of the boat in metres per

minute. [Use 3 = 1.732 ] [CBSE 2019 (30/1/1)]

29. Two poles of equal heights are standing opposite each other on either side of the road, which is

80 m wide. From a point between them on the road, the angles of elevation of the top of the poles

are 60° and 30° respectively. Find the height of the poles and the distances of the point from the

poles. [CBSE 2019 (30/1/1)]

30. Amit, standing on a horizontal plane, finds a bird flying at a distance of 200 m from him at an

elevation of 30°. Deepak standing on the roof of a 50 m high building, finds the angle of elevation

of the same bird to be 45°. Amit and Deepak are on opposite sides of the bird. Find the distance

of the bird from Deepak. [CBSE 2019 (30/2/1)]

31. From a point P on the ground, the angle of elevation of the top of a tower is 30° and that of the
top of the flag-staff fixed on the top of the tower is 45°. If the length of the flag-staff is 5 m, find
the height of the tower. [Use 3 = 1.732 ]

304 Xam idea Mathematics–X

32. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s
altitude is 30° than when it was 60°. Find the height of the tower. [Given 3 = 1.732 ]
[CBSE 2019 (30/3/1)]

33. A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The

angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat

in m/min. [CBSE 2019 (30/4/2)]

34. There are two poles, one each on either bank of a river just opposite to each other. One pole is
60 m high. From the top of this pole, the angle of depression of the top and foot of the other
pole are 30° and 60° respectively. Find the width of the river and height of the other pole.
[CBSE 2019 (30/4/2)]

35. A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an

elevation of 30°. A girl standing on the roof of a 20 m high building, finds the elevation of the

same bird to be 45°. The boy and the girl are on the opposite sides of the bird. Find the distance

of the bird from the girl. Use 2 = 1.414 ] [CBSE 2019 (30/5/1)]

36. The angles of depression of the top and bottom of a 8 m tall building from the top of a tower are

30° and 45° respectively. Find the height of the tower and the distance between the tower and the

building. [CBSE 2019 (C) (30/1/1)]

37. As observed from the top of a lighthouse, 75 m high from the sea level, the angles of depression

of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the

lighthouse, find the distance between the two ships. [CBSE 2019 (C) (30/1/1)]

38. The angles of depression of the top and bottom of a building 50 metres high as observed from the
top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal
distances between the building and the tower.

39. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same
plane is 60° and the angle of elevation of the top of the second tower from the foot of the first
tower is 30°. Find the distance between the two towers and also the height of the other tower.
[NCERT Exemplar]

40. The lower window of a house is at a height of 2 m above the ground and its upper window is

4 m vertically above the lower window. At certain instant, the angles of elevation of a balloon

from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon

above the ground. [NCERT Exemplar]

41. As observed from the top of a lighthouse, 100 m above sea level, the angle of depression of a ship,
sailing directly towards it, changes from 30° to 45°. Determine the distance travelled by the ship
during the period of observation.

42. The angle of elevation of the top of a tower as observed from a point in a horizontal plane
through the foot of the tower is 32°. When the observer moves towards the tower at a distance of
100 m, he finds the angle of elevation of the top to be 63°. Find the height of the tower and the
distance of the first position from the tower. [Take tan 32° = 0.6248 and tan 63° = 1.9626]

43. The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a
distance of 20 metres towards the foot of the tower to a point B, the angle of elevation increases
to 60°. Find the height of the tower and the distance of the tower from the point A.

Heights and Distances 305

44. From a balloon vertically above a straight road, the angle of depression of two cars at an instant
are found to be 45° and 60°. If the cars are 100 m apart, find the height of the balloon.

45. If the angle of elevation of a cloud from a point h metres above a lake is a and the angle of

depression of its reflection in the lake be b, prove that the distance of the cloud from the point of

observation is 2h sec α α .
tan β − tan

46. The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of

the top of the tower from the foot of the hill is 30°. If the tower is 50 m high, what is the height

of the hill? [CBSE Delhi 2017]

47. The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression

of the reflection of the cloud in the lake is 60°. Find the height of the cloud from the surface of the

lake. [CBSE Delhi 2017]

48. From the top of a tower, the angles of depression of two objects on the same side of the tower are

found to be a and b(a > b). If the distance between the objects is p metres, show that the height

h of the tower is given by h = p tan α tan β metres.
tan α − tan β

49. A bird is sitting on the top of a tree, which is 80 m high. The angle of elevation of the bird, from

a point on the ground is 45°. The bird flies away from the point of observation horizontally and

remains at a constant height. After 2 seconds, the angle of elevation of the bird from the point of

observation becomes 30°. Find the speed of flying of the bird. [CBSE Delhi 2016]

50. The angle of elevation of the top Q of a vertical tower PQ from a point X on the ground is 60°.
From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find

the height of the tower PQ and the distance PX. (Use 3 =1.73) [CBSE (AI) 2016]

51. From a point on the ground, the angle of elevation of the top of a tower is observed to be 60°.

From a point 40 m vertically above the first point of observation, the angle of elevation of the

top of the tower is 30°. Find the height of the tower and its horizontal distance from the point of

observation. [CBSE (AI) 2016]

52. Two points A and B are on the same side of a tower and in the same straight line with its base.
The angles of depression of these points from the top of the tower are 60° and 45° respectively.
If the height of the tower is 15 m, then find the distance between these points.
[CBSE Delhi 2017]

53. An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of
depression from the aeroplane of two points on both banks of a river in opposite directions are

45° and 60° respectively. Find the width of the river. [Use 3 = 1·732] [CBSE (AI) 2017]

54. The angles of depression of two ships from an aeroplane flying at the height of 7500 m are 30°
and 45°. If both the ships are in the same line and on the same side of the aeroplane such that

one ship is exactly behind the other, find the distance between the ships. [Use 3 = 1·73]
[CBSE (F) 2017]

306 Xam idea Mathematics–X

Answers

1. (i) (a) (ii) (c) (iii) (b) (iv) (a) (v) (b)

2. (i) acute (ii) increases (iii) 45° (iv) 45° (v) depression

3. 60° 4. 3 : 1 5. 100 3 m 6. 12 m 7. 45° 8. b is greater

9. 45° 10. 5 m 11. False 12. False 13. 93 m 26. 73.2 m
5 30. 50 2 m

14. ∠C = (90 – f), ∠D = q 15. 45° 16. 8 m

17. 30 3 m 18. 19 m 20. 6.83 m 21. 79.41 m 22. 23.66 m

23. 10 3 m, 40 m 24. 118.28 m, 68.25 m 25. 189.28 m

27. height of hill = 150 m

28. speed of boat = 57.73 m/min 29. 60 m, 20 m, 20 3 m

31. height of tower = 6.83 m 32. 34.64 m

33. Speed of boat is (75 – 25 3 ) m/min or 31.7 m/min

34. width of river = 20 3 m and height of pole = 40 m

35. 42.42 m 36. height of tower = 12 + 4 3 m = distance between tower and building

37. 75 ( 3 – 1) m 38. 25 3 m, 75 m 39. 10 3 m, 10 m

40. 8 m 41. 73.2 m 42. 91.65 m, 146.7 m 43. 17.3 m, 30 m

44. 50(3 + 3 )m 46. 150 m 47. 120 m 49. 29.28 m/s

50. 54.6 m, 94.6 m 51. 60 m; 20 3 m 52. 5(3– 3 )m

53. 473.2 m 54. 5475 m

SELF-ASSESSMENT TEST

Time allowed: 1 hour Max. marks: 40

Section A

1. Choose and write the correct option in the following questions. (4 × 1 = 4)

(i) The ratio of the length of a pole and its shadow is 3 :1. The angle of elevation of the sun
is

(a) 30° (b) 45° (c) 60° (d) 90°

(ii) If the angles of elevation of a tower from two points at distances r and s where r > s from

its foot and in the same straight line from it are 30° and 60°, then the height of the tower is

(a) r + s (b) r – s (c) rs (d) r
s

(iii) A lamp post 5 3 m high casts a shadow 5 m long on the ground. The Sun’s elevation at

this point is

(a) 30° (b) 45° (c) 60° (d) 90°

(iv) The angle of elevation of a ladder leaning against a wall is 60° the foot of the ladder is 12.4
m away from the wall. The length of the ladder is

(a) 14.8 m (b) 6.2 m (c) 12.4 m (d) 24.8 m

Heights and Distances 307

2. Fill in the blanks. (3 × 1 = 3)

(i) An angle measured in anticlockwise direction is considered to be _____________ .

(positive/negative)

(ii) The angle of _____________ of an object is the angle for which the eye has to be raised
through from the horizontal in order to look at the object.

(iii) _____________ is the branch of mathematics which deals with the relations between the
sides and angles of triangle and calculation based on them.

3. Solve the following questions. (3 × 1 = 3)

(i) The tops of two poles of height 16 m and 10 m are connected by a wire. If the wire makes
an angle of 30° with the horizontal, find the length of the wire.

(ii) The angle of elevation of a ladder leaning against a wall is 30°, the foot of the ladder is
12 m away from the wall. What is the length of the ladder?

(iii) If the angle of elevation of top of a tower from a point on the ground which is 20 3 m
away from the foot of the tower is 30°, find the height of the tower.

Section B

QQ Solve the following questions. (3 × 2 = 6)

4. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation

is 30° than when it is 60°. Find the height of the tower. [NCERT Exemplar]

5. An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation

of the top of the tower from the eye of the observer. [NCERT Exemplar]

6. The angle of elevation of the top of a tower from two points distant s and t from its foot are

complementary. Prove that the height of the tower is st . [NCERT Exemplar]

QQ Solve the following questions. (3 × 3 = 9)

7. An observer finds the angle of elevation of the top of a tower from certain point is 30°. If the

observer moves 20 m towards the tower, the angle of elevation of the top increases by 15°. Find

the height of the tower. [NCERT Exemplar, CBSE Delhi 2017]

8. The angles of depression of the top and bottom of a 50 m high building from the top of a tower

are 45° and 60° respectively. Find the height of the tower and the horizontal distance between

the tower and the building. (Use 3 = 1.73). [CBSE Delhi 2016]

9. A man observes a car from the top of a tower, which is moving towards the tower with a uniform

speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time

taken by the car now to reach the tower. [CBSE (AI) 2017]

QQ Solve the following questions. (3 × 5 = 15)

10. From the top of a tower h m high, the angles of depression of two objects, which are in line with
the foot of the tower are a and b(b > a). Find the distance between the two objects.
[NCERT Exemplar]

308 Xam idea Mathematics–X

11. A window of a house is h metres above the ground. From the window, the angles of elevation and
depression of the top and the bottom of another house situated on the opposite side of the lane are
found to be a and b respectively. Prove that the height of the other house is h(1 + tan a cot b) metres.
[NCERT Exemplar]

12. A ladder rests against a vertical wall at an inclination a to the horizontal. Its foot is pulled away

from the wall through a distance p so that its upper end slides a distance q down the wall and

then the ladder makes an angle b to the horizontal. Show that p = cos β − cos α .
q sin α − sin β

[NCERT Exemplar]

Answers

1. (i) (c) (ii) (c) (iii) (c) (iv) (d)

2. (i) positive (ii) elevation (iii) trigonometry

3. (i) 12 m (ii) 8 3 m (iii) 20 m

4. 25 3 m 5. 45° 7. 10^ 3 + 1h m

8. 118.25 m, 68.25 m 9. 16 minutes 23 seconds

10. h (cot a– cot b) metre

zzz

Heights and Distances 309

12 Areas Related
to Circles
BASIC CONCEPTS – A FLOW CHART
BQA

310 Xam idea Mathematics–X

OAPB
OBQA

,

Areas Related to Circles 311

MORE POINTS TO REMEMBER

 Area of Sector: Let AOB be a sector of a circle having centre at O and
radius r such that ∠AOB = q, where q<180°. Obviously, if q increases
the area of sector increases and if q = 180°, then area of sector becomes
equal to area of semicircle. Similarly, if q = 360°, area of sector becomes
equal to area of circle. Thus, we conclude that

When degree measure of angle at the centre is 360° then

area of sector = area of circle = pr2

Fig. 12.1

\ When degree measure of angle at centre is 1° then area of sector = rr2
360°

Thus, when degree measure of angle at centre is q then area of sector = rr2 i
360°
 Angle described by minute hand in 1 minute = 6°

 Angle described by hour hand in 1 hour = 30°

 Area of segment: Let APB be a segment of a circle having centre at
‘O’ and radius r such that ∠AOB = q

Obviously, area of segment APB

= area of sector OAPB – area of DOAB
SSSSSSSSSSSSSSRSTIcsnoinsTii22O==BLOBrrLL WWWWWWWWWWWWWWVW
= rr2 i – 1 # AB # OL & BL = r sin i X Fig. 12.2
360 2 & OL = r cos 2
i
2

= rr2 i – 1 # 2BL # OL
360 2

= rr2 i – 1 # 2r sin i #r cos i
360 2 2 2

= rr2 i – 1 # r2 . 2 sin i . cos i     [ sin q = 2 sin i . cos i ]
360 2 2 2 2 2

= rr2 i – r2 . sin i
360 2

Multiple Choice Questions [1 mark]

Choose and write the correct option in the following questions.

1. If the area of a circle is 154 cm2, then its perimeter is [NCERT Exemplar]

(a) 11 cm (b) 22 cm (c) 44 cm (d) 55 cm

2. 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 [NCERT Exemplar]

(a) RR11 + RR22 = R (b) NR1o+neRo2f >R
(c) + < R (d) these

312 Xam idea Mathematics–X

3. If the circumference of a circle and the perimeter of a square are equal, then [NCERT Exemplar]
(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.

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

(b) 1 r2sq. units (c) 2r2sq. units [NCERT Exemplar]
(a) r2sq. units 2 (d) 2 r2sq. units

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

[NCERT Exemplar]

(a) 22 : 7 (b) 14 : 11 (c) 7 : 22 (d) 11 : 14

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

[NCERT Exemplar]

(a) 10 m (b) 15 m (c) 20 m (d) 24 m

7. The area of the circle that can be inscribed in a square of side 6 cm is [NCERT Exemplar]

(a) 36 π cm2 (b) 18 π cm2 (c) 12 π cm2 (d) 9 π cm2

8. The area of the square that can be inscribed in a circle of radius 8 cm is [NCERT Exemplar]

(a) 256 cm2 (b) 128 cm2 (c) 64 2 cm2 (d) 64 cm2

9. The radius of a circle whose circumference is equal to the sum of the circumferences of the

two circles of diameters 36 cm and 20 cm is [NCERT Exemplar]

(a) 56 cm (b) 42 cm (c) 28 cm (d) 16 cm

10. 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 [NCERT Exemplar]

(a) 31 cm (b) 25 cm (c) 62 cm (d) 50 cm

11. If the area of circle is 1386 cm2, then its circumference is

(a) 66 cm (b) 88 cm (c) 132 cm (d) 264 cm

12. The diameter of a wheel is 1 m. The number of revolutions it will make to travel a distance of
22 km will be

(a) 2,800 (b) 4,000 (c) 5,500 (d) 7,000

13. A circular park has a path of uniform width around it. The difference between the outer and

inner circumferences of the circular park is 132 cm. Its width is

(a) 20 cm (b) 21 cm (c) 22 cm (d) 24 cm

14. The ratio of area of two circles whose ratio of circumference is 3:1 will be

(a) 3 : 1 (b) 1 : 3 (c) 1 : 9 (d) 9 : 1

15. The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are
respectively equal is

(a) r : 2 (b) r : 3 (c) 3 : r (d) 2 : r

Areas Related to Circles 313

16. The area of a circular path of uniform width b surrounding a circular region of radius r is

(a) r (2r + b) r sq. units (b) r (2r + b) b sq. units

(c) r (b + r) r sq. units (d) r (b + r) b sq. units

17. The area of a sector of a circle bounded by an arc of length 5π cm is equal to 20π cm2, then its
radius is

(a) 12 cm (b) 16 cm (c) 8 cm (d) 10 cm

18. The area of a circle whose area and circumference are numerically equal is

(a) 2π sq. units (b) 4π sq. units (c) 6π sq. units (d) 8π sq. units

19. The area of a quadrant of a circle whose circumference is 616 cm will be
(a) 7546 cm2 (b) 7500 cm2 (c) 7456 cm2 (d) 7564 cm2

20. If the sum of the areas of two circles with radii r1 and r2 and is equal to the area of a circle of
radius r, then r12 + r22 is

(a) greater than r2 (b) equal to r2 (c) less than r2 (d) none of these

Answers

1. (c) 2. (a) 3. (b) 4. (a) 5. (b) 6. (a)
7. (d) 8. (b) 9. (c) 10. (d) 11. (c) 12. (d)
13. (b) 14. (d) 15. (b) 16. (b) 17. (c) 18. (b)
19. (a) 20. (b)

Fill in the Blanks [1 mark]

Complete the following statements with appropriate word(s) in the blank space(s).
1. If radius of a circle is 14 cm the area of the circle is _______________ .
2. Length of arc of a sector angle 45° of circle of radius 14 cm is ______________ .
3. 2πr is ______________ of a circle.
4. Perimeter of a semi circle ______________ .
5. ______________ is the region between the arc and two radii.
6. Measure of angle in a semi circle is ______________ .
7. Pie (π) is the ratio between circumference and ______________ of the circle.
8. Angle formed by two radii at the centre is known as ______________ .
9. Concentric circles are circles having same ______________ .
10. Segment is the region enclosed between chord and ______________ .
11. If circumferences of two circles are equal, then their areas are ______________ .
12. The area of the largest circle that can be drawn inside a rectangle of length a cm and breadth

b cm (a > b) is ______________ cm2.
13. The perimeter of a square circumscribing a circle of radius a cm is ______________ cm.
14. Two circles are congruent if their ______________ are equal.

15. Angle described by hour hand in one minute is ______________ .

Answers 2. 7 r cm 3. circumference 4. (πr + d) units 5. sector
2 8. central angle 9. centre
1. 616 cm2 7. diameter 10. arc
6. 90° 14. radii
12. rb2 13. 8a 15.  1 °
11. equal 4  2 

314 Xam idea Mathematics–X

Very Short Answer Questions [1 mark]

1. Find the area of a square inscribed in a circle of diameter p cm. Fig. 12.3
Fig. 12.4
Sol. Diagonal of the square = p cm

\ p2 = side2 + side2

⇒ p2 = 2 side2
or p2
side2 = 2 cm2 = area of the square

2. Find the area of the circle inscribed in a square of side a cm.

Sol. Diameter of the circle = a ra2
4
⇒ Radius = a ⇒ Area = r a a 2 = cm2
2 2
k

3. Find the area of a sector of a circle whose radius is r and length of the arc
is l.

Sol. Area of a sector of a circle with radius r

= i # rr2 = i # 2rr r = 1 lr sq. units ca l = 2rri m
360° 360° 2 2 360°

4. A square inscribed in a circle of diameter d and another square is Fig. 12.5
circumscribing the circle. Show that the area of the outer square is twice
the area of the inner square.

Sol. Side of outer square = d [Fig. 12.6]

\ Its area = d2

Diagonal of inner square = d = Diameter of circle

side2 + side2 = d2 Fig. 12.6

2 side2 = d2  ⇒ side2 = d2
2
d2
\ Side = d  ⇒ Area = 2
2
Area of outer square = 2 × Area of inner square.

5. If circumference and the area of a circle are numerically equal, find the diameter of the circle.

Sol. Given, 2pr = pr2

⇒ 2r = r2 ⇒ r2 – 2r = 0

⇒ r(r–2) = 0 or r = 2

i.e., d = 4 units

6. The radius of a wheel is 0.25 m. Find the number of revolutions it will make to travel a

distance of 11 km.

Sol. Number of revolutions = 11 # 1000 = 11 # 1000 # 7 = 7000.
11
2# 22 # 0.25
7 D

7. If the perimeter of a semi-circular protractor is 36 cm, find its diameter.

Sol. Perimeter of a semicircular protractor = perimeter of a semicircle r

= (2r + pr) cm AB

Given, 2r + pr = 36 C

Fig. 12.7

Areas Related to Circles 315

⇒ r c2+ 22 m = rd 36 n = 36 ⇒ r = 7 cm
7 7

Diameter = 2r = 2 × 7 = 14 cm.

8. If the diameter of a semicircular protractor is 14 cm, then find its perimeter.

Sol. Perimeter of a semicircle = pr + 2r

= 22 × 7 + 2 × 7 = 22 + 14 = 36 cm
7

Short Answer Questions-I [2 marks]

1. If a square is inscribed in a circle, what is the ratio of the areas of the circle and the square?

Sol. Let radius of the circle be r units.

Then, diagonal of the square = 2r
⇒ Side of the square
= 2r = 2r
2

\ Area of the circle = ( rr2 = rr2 = r :2
Area of the square 2 r)2 2r2

Fig. 12.8
2. What is the angle subtended at the centre of a circle of radius 10 cm by an arc of length 5p cm?

Sol. Arc length of a circle of radius r = i # 2rr
360°

⇒ 5p = i × 2p × 10 or i = 5r = 1 ⇒ q= 360° = 90°
360° 360° 20r 4 4

3. Difference between the circumference and radius of a circle is 37 cm. Find the area of circle.

Sol. Given 2p r – r = 37

⇒ r (2p – 1) = 37

⇒   r = 37 = 37 = 37 # 7 =7
2r – 1 22 37
2# 7 –1

So area of circle = pr2 = 22 × 7 × 7 = 154 cm2
7

4. 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. [NCERT]

Sol. Let R be the radius of required circle. Then, we have

pR2 = p(8)2 + p(6)2 pR2 = 100 p
⇒ pR2 = 64p + 36p ⇒

\ R2 = 100r =100 ⇒  R = 10 cm
r

Hence, radius of required circle is 10 cm.

5. 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. [NCERT]

Sol. Let R be the radius of required circle. Then, we have

2pR = 2p(19) + 2p (9)

316 Xam idea Mathematics–X

⇒ 2pR = 2p(19 + 9) ⇒ R= 2r # 28 = 28
2r

Hence, the radius of required circle is 28 cm.

6. Find the area of a circle whose circumference is 22 cm. [NCERT]

Sol. Let r be the radius of the circle then, circumference = 22 cm

⇒ 2pr = 22 ⇒ r = 22 = 22 # 7 = 7
2r 2 # 22 2
22 7 7
\ Area of the circle = pr2 = 7 # 2 # 2 = 38.5 cm2

7. The area of a circular playground is 22176 m2. Find the cost of fencing this ground at the rate
of `50 per m.

Sol. Area of circular playground = 22176 m2

pr2 = 22176

⇒ 22 r2 = 22176   ⇒   r2 = 22176 # 7
⇒ 7 22

r = 84 m

\ Circumference of the playground = 2πr = 2 × 22 ×84 = 44 × 12 = 528 m
7

\ Cost of fencing this ground = ` 528 × 50 = ` 26400.

8. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. [NCERT]

Sol. We know that

Area of a sector = i # rr2
360°

Here, q = 60° and r = 6 cm

\ Area of the sector = 60° # r (6)2
360°

= 6 # 22 = 132 cm2 = 18 6 cm2
7 7 7

9. Find the area of a quadrant of a circle whose circumference is 22 cm. [NCERT]

Sol. Let r be the radius of circle, then circumference = 22 cm

⇒ 2pr = 22   ⇒    r = 22 = 22 # 7 = 7
2 # 22 2
2 # 22
7
2
rr2 22 # c 7 22 # 49
4 7 2 m 7 4 4

Now, area of a quadrant of a circle = = 4 =

= 154 = 77 =9 5 = 9.625 cm2
16 8 8

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

5 minutes. [NCERT]

Sol. Since the minute hand rotates through 6° in one minute, therefore, area swept by the minute
hand in one minute is the area of a sector of angle 6° in a circle of radius 14 cm.

Hence, the area swept in 5 minutes = i # rr2 #5
360°

Areas Related to Circles 317

= 6° # 22 # (14)2 # 5= 1 # 22 # 28 # 5
360° 7 60

= 154 cm2 = 51 1 cm2
3 3

11. 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.

(Use p = 3.14) [NCERT]

Sol. We have, r = 16.5 km and q = 80° i
360°
\ Area of the sea over which the ships are warned = # rr2

= 80° × 3.14 × 16.5 × 16.5
360°

= 2 × 3.14 × 16.5 × 16.5 = 189.97 km2
9

12. The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle.

Sol. Let the radius of the circle be r cm. Then,

Diameter = 2r cm and circumference = 2pr cm

According to question,

\ Circumference = diameter + 16.8

⇒ 2pr = 2r + 16.8

⇒ 2# 22 # r = 2r +16.8 ⇒ 44r = 14r + 16.8 × 7
7 or 30r = 117.6

⇒ 44r – 14r = 117.6

⇒ r = 117.6 = 3.92
30

Hence, radius = 3.92 cm.

13. A race track is in the form of a ring whose inner circumference is 352 m, and the outer
circumference is 396 m. Find the width of the track.

Sol. Let the outer and inner radii of the ring be R m and r m respectively. Then,

2pR = 396 and 2pr = 352

⇒ 2# 22 # R = 396 and 2 # 22 # r =352
7 7

⇒ R = 396 × 7 # 1 and r = 352 × 7 # 1
22 2 22 2

⇒ R = 63 m and r = 56 m

Hence, width of the track = (R – r) = (63 – 56) = 7 m Fig. 12.9

14. A car has 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. [NCERT]

Sol. We have, r = 25 cm and q = 115°.

\ Total area cleaned at each sweep of the blades

= 2 × (area of the sector having radius 25 cm and angle q =115°)

= 2# i # rr2 = 2 # 115° # 22 # 25 # 25
360° 360° 7

= 23 #11# 25 # 25 = 158125 =1254.96 cm2
18 # 7 126

318 Xam idea Mathematics–X

15. In a society there is a park in rectangular shape of dimension 70 m by 90 m. In the centre of

the rectangular park, there is a circular ground of diameter 14 m. The circular ground is used

for children’s play and remaining region is used for other activities of society. After reading

this passage, answer the following questions:

(i) Find the area for children’s play.

(ii) Find the area for other activities of the society.

Sol. 7m
r 90 m
(i) We have,
radius of circular ground = 14 =7 m
2

∴ Area of the ground for children’s play Fig. (a)

= pr2 = p × (7)2
22
= 7 × 49 =154 m2

(ii) We have,

Area of rectangular park = l × b = 70 × 90

= 6300 m2

∴ Area for other activities for the society = Area of the park – area of the circular ground.

= 6300 – 154

= 6146 m2

Short Answer Questions-II [3 marks]

1. In the figure, three section of a circle of radius 7 cm, making angles of 60°, 80° and 40° at the

centre are shaded. Find the area of the shaded region. [CBSE 2019 (30/5/1)]

Sol. We have radius of circle = 7 cm = i ×rr2
Area of sector 360°
60° 22
Area of sector containing 60°angle = 360° × 7 × (7) 2

1 22 80° 60°
6 7
= × ×49

= 11×7 = 77 cm2 40°
3 3
80° 22
Also, area of sector containing 80° angle = 360° × 7 × (7) 2

= 2 ×22×7 = 308 cm2 Fig. 12.10
9 9
40° 22
Again, area of sector containing 40° angle = 360° × 7 × (7) 2

= 1 ×22×7 = 154 cm2
9 9
77 308 154
Area of total shaded region = 3 + 9 + 9

= 231 + 308 + 154
9
693
= 9 = 77 cm2

2. 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? [NCERT]

Sol. The diameter of a wheel = 80 cm

Areas Related to Circles 319

So, radius of the wheel = 40 cm

Now, distance travelled in one complete revolution of wheel = 2p × 40 = 80p

Since, speed of the car is 66 km/h 66 #100000 #10
60
So, distance travelled in 10 minutes = cm

= 11 × 100000 cm = 1100000 cm.

So, number of complete revolutions in 10 minutes = 1100000 = 110000
80r 22
8 # 7

= 110000 # 7 = 70000 = 4375
8 # 22 16

3. An umbrella has 8 ribs which are equally spaced (Fig. 12.11). Assuming umbrella to be a flat
circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. [NCERT]

Sol. We have, r = 45 cm Fig. 12.11

\ Area between two consecutive ribs = 1 # rr2
8

= 1 # 22 # 45 # 45 = 11 # 45 # 45
8 7 4#7

= 2222875 =795.54 cm2

4. 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 (Fig. 12.12). 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 p = 3.14) [NCERT]

Sol. Let the horse be tied at point O and the length of the rope is OH Fig. 12.12
(Fig. 12.13). Thus,

(i) the area of the part of the field in which the horse can graze

= area of the quadrant of a circle (OAHB)

= rr2 = 1 # 3.14 # 5# 5= 78.5 =19.625 m2
4 4 4

(ii) Now r = 10 m and rr2
4
\ Required area =

= 3.14 # (10)2 = 3.14 #100
4 4
Fig. 12.13

= 314 = 78.5 m2
4

320 Xam idea Mathematics–X

Increase in the grazing area

= (78.5 – 19.625) m2

= 58.875 m2

Fig. 12.14

5. In Fig. 12.15, sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the
area of the shaded region.

Sol. Let A1 and A2 be the areas of sectors OAB and OCD respectively. Then, A1 = area of a sector of
angle 30° in a circle of radius 7 cm.

⇒ A1 = ' 30° # 22 # (7)21 cUsing : A= i # rr2 m
360° 7 360°

= ' 1 # 22 # 7 # 71 ⇒ A1 = 77 cm2
12 7 6

A2 = area of a sector of angle 30° in a circle of radius 3.5 cm Fig. 12.15
30° 22 1 22 7 7 77
= ' 360° # 7 # (3.5)21 ⇒ ' 12 # 7 # 2 # 2 1= 24 cm2

\ Area of the shaded region = A1 – A2 = c 77 – 77 mcm2
6 24

= 77 × (4 – 1) cm2 = 77 cm2 = 9.625 cm2
24 8

6. Find the area of the sector of a circle with radius 4 cm and of angle 30°.
Also, find the area of the corresponding major sector. (Use p = 3.14)

Sol. Area of the sector OAPB = i # rr2 O
360°

= 33600°° × 3.14 × 4 × 4 cm2 Fig. 12.16

= 123.56 cm2 = 4.19 cm2 (approx.)
Area of the corresponding major sector = pr2 – area of sector OAPB
= (3.14 × 4 × 4 – 4.19)cm2 = (50.24 – 4.19) cm2
= 46.05 cm2 = 46.1 cm2 (approx.)
7. 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 p = 3.14 and 3 = 1.73) [NCERT]
OR

Find the area of the segment shown in figure if radius of the circle is

21 cm and ∠AOB = 120°. [CBSE 2019 (30/1/2)]

Fig. 12.17

Areas Related to Circles 321

Sol. We have, r = 12 cm and q = 120°

Now, area of the corresponding segment APB of circle

= area of the minor segment

= i ×rr2 – 1 r2 sin i
360° 2
21 cm 21 cm
120° # (12)2 1 # (12)2
= 360° # 3.14 – 2 sin 120°

da sin 120° = sin (180° – 60°)= sin 60° = 3 n
2
Fig. 12.18

= 1 # 3.14 #144 – 1 #144 # 3 =144 <3.314 – 3 F =144 <12.5612– 3 3F
3 2 2 4

= 12 (12.56 – 3 × 1.73) = 12 (12.56 – 5.19) = 12 × 7.37 = 88.44 cm2

OR

Solution is same as above only radius changes from 12 cm to 21 cm.
Ans: 270.85 cm2

8. A round table cover has six equal designs as shown in Fig. 12.19. If the radius of the cover is
28 cm2, find the cost of making the designs at the rate of ™ 0.35 per cm2. (Use 3 = 1.7) [NCERT]

Sol. Area of one design = area of the sector OAPB – area of ∆AOB

= i # rr2 – 1 r2 sin i
360° 2

= 60° # 22 # 28 # 28 – 1 # 28 # 28 # sin 60°
360° 7 2

= 1 # 22 # 28 # 28 – 1 # 28 # 28 # 3
6 7 2 2

= 28 # 28 c 11 – 1.7 m = 28 × 28 × 8.3 = 77.47 cm2 Fig. 12.19
Area of 6 such designs = 77.47 21 4 cm2 84
×6 = 464.82

Cost of making designs = ™(0.35 × 464.82) = ™ 162.69

Hence, cost of making such designs is ™162.69.

9. Find the area of the shaded region in Fig. 12.20, if PQ = 24 cm, PR = 7 cm and O is the centre

of the circle. [NCERT]

Sol. Here, ROQ is the diameter of given circle, therefore ∠RPQ = 90°.

Now, in right angled DPRQ, we have

RQ2 = RP2 + PQ2 (by Pythagoras Theorem)

RQ2 = (7)2 + (24)2 = 49 + 576 = 625

RQ = 625 = 25 cm

Therefore, radius r = 25 cm
2
Now, area of shaded region = area of the semi-circle – area of DRPQ Fig. 12.20

= 1 πr2 − 1 × PQ × RP = 1 × 22 × 25 × 25 − 1 × 24 × 7
2 2 2 7 2 2 2

=  6875 − 84  cm2 =  6875 − 2352  cm2 = 4523 cm2 = 161.54 cm2
 28   28  28

322 Xam idea Mathematics–X

10. Find the area of the shaded region in Fig. 12.21, if radii of the two concentric circles with

centre O are 7 cm and 14 cm respectively and ∠AOC = 40°. [NCERT]

Sol. Area of shaded region = area of sector AOC – area of sector OBD

= 40° × π × (14)2 − 40° × π × (7)2
360° 360°

= 1 × π[(14)2 − (7)2 ] = π × [196 − 49] = 1 × 22 × 147
9 9 9 7

22 × 21 = 154 = 51.33 cm2 Fig. 12.21
9 3
=

11. Find the area of the shaded region in Fig. 12.22, if ABCD is a square of side 14 cm and APD

and BPC are semicircles. [NCERT]

Sol. We have, radius of semicircles = 7 cm

\ Area of shaded region

= area of square ABCD – Area of semi-circles (APD +BPC)

= (14)2 –  1 π × (7)2 + 1 π × (7)2 
 2 2 

= 196 – 22 × (7)2 = 196 – 154 = 42 cm2 Fig. 12.22
7

12. Find the area of the shaded region in Fig. 12.23, 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.
[NCERT, CBSE (F) 2016]

Sol. We have, radius of circular region = 6 cm and each side of DOAB = 12 cm.

\ Area of the circular portion

= area of circle – area of the sector

= πr2 − θ × πr2
360°

= πr2 1 − θ = 22 × (6)2 1 − 60° 
360°  7 360° 
O

= 22 × 36 × 5 = 22 × 30 = 660 cm2
7 6 7 7

Now, area of the equilateral triangle OAB

3 × (side)2 = 3 × (12)2 = 3 3 cm2 A B
4 4 4
= × 144 = 36 Fig. 12.23

\ Area of shaded region = area of circular portion + area of equilateral triangle OAB

=  660 + 36 3  cm 2 = 12 (55 + 21 3) cm2
 7  7

13. 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 2 cm is cut as shown in Fig. 12.24. Find the area of the remaining portion

of the square. [NCERT]

Sol. We have, the side of the square ABCD = 4 cm
\ Area of the square ABCD = (4)2 = 16 cm2

Since, each quadrant of a circle has radius 1 cm.

Areas Related to Circles 323

\ The sum of the areas of four quadrants

= 4 ×  πr2  = πr2 = 22 × (1)2 = 22 cm2
 4  7 7
 

Now, area of the circle of diameter 2 cm = π d2 = π × 4 = π = 22 cm2
4 4 7

\ Area of the remaining portion Fig. 12.24

= area of the square ABCD – sum of the areas of four quadrants

– area of the circle of diameter 2 cm

= 16 − 22 − 22 = 112 − 22 − 22 = 68 = 9.71 cm2
7 7 7 7

14. In Fig. 12.25, ABCD is a square of side 14 cm. 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. [NCERT]

Sol. We have, each side of square ABCD = 14 cm
\ Area of square ABCD = (14)2 cm2 = 196 cm2

Now, radius of each quadrant of circle,r = 14 cm = 7 cm
2

\ T he sum of the area of the four quadrants at the four corners of the

square = 4 × πr2 = πr2 = π(7)2
4

22 Fig. 12.25
= 7 × 7 × 7 = 154 cm2
Now, area of shaded portion

= area of square ABCD – sum of the areas of four quadrants at the four corners of the square

= (196 – 154) cm2 = 42 cm2

15. On a square handkerchief, nine circular designs, each of radius

7 cm are made (see Fig. 12.26). Find the area of the remaining

portion of the handkerchief. [NCERT]

Sol. Total area of circular design = 9 × Area of one circular design

= 9 × p × (7)2

22
= 9 × 7 × 7 × 7 = 1386 cm2

Now, each side of square ABCD = 3 × diameter of circular design Fig. 12.26

= 3 × 14 = 42 cm

\ Area of square ABCD = (42)2 = 1764 cm2

\ Area of the remaining portion of handkerchief

= area of square ABCD – total area of circular design

= (1764 – 1386) cm2 = 378 cm2

324 Xam idea Mathematics–X

16. In Fig. 12.27, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm,

find the area of the (i) quadrant OACB, (ii) shaded region. [NCERT]

OR

In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If

OD = 2 cm, find the area of the shaded region. [CBSE Delhi 2017]

Sol. (i) Area of quadrant OACB = 1 πr2 = 1 × 22 × (3.5)2
4 4 7

= 1 × 22 × 7 × 7 = 77 cm2 = 9 5 cm2
4 7 2 2 8 8

(ii) Now, we find area of DOBD

We have area of DOBD = 1 ×OB × OD = 1 × 3.5 × 2 Fig. 12.27
2 2

= 3.5 cm2 = 7 cm2
2 area of quadrant OACB – area of DOBD
Hence, area of shaded region =

=  77 − 7  cm 2 =  77 − 28  cm2 = 49 cm 2 = 6 1 cm 2
 8 2   8  8 8

17. In Fig. 12.28, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of

the shaded region. (Use p = 3.14). [NCERT, CBSE Delhi 2014]

Sol. We have, = OB = OA2 + AB2 (By Pythagoras Theorem)
Radius of quadrant

= (20)2 + (20)2 = 400 + 400 = 800 = 20 2 cm

Now, area of shaded region

= area of quadrant OPBQ – area of square OABC Fig. 12.28

= 1 p × (20 2 )2 – (20)2 = 1 p × 800 – 400
4 4
= 200 × 3.14 – 400 = 628 – 400 = 228 cm2

18. Calculate the area of the designed region in Fig. 12.29, which is common between the two

quadrants of circles of radius, 8 cm each. [NCERT]

Sol. Here, radius of each quadrant ABPD and BQDC = 8 cm

Sum of areas of quadrants =2× 1 pr2
4

= 1 # 22 # (8) 2 = 11 # 64 = 704 cm2
2 7 7 7

Now, area of the square ABCD = 8 × 8 = 64 cm2

Hence, area of designed region = area of shaded region

= sum of areas of quadrants

– area of the square ABCD Fig. 12.29

= 7074 − 64 = 704 − 448 = 256 = 36.57 cm2
7 7

Areas Related to Circles 325

19. In the given Fig. 12.30, find the area of the shaded region.

Sol. Clearly, diameter of the circle

= diagonal BD of rectangle ABCD

Now, BD = BC2 + CD2 = 62 + 82 cm =10 cm

Let r be the radius of the circle then, r =  120  cm = 5 cm

Fig. 12.30

Hence, area of the shaded region = area of the circle – area of rectangle ABCD

= πr2 – l × b = (3.14 × 5 × 5) – (8× 6)

= (78.50 – 48) cm2 = 30.50 cm2

20. A square park has each side of 100 m. At each corner of the
park, there is a flower bed in the form of a quadrant of radius
14 m as shown in Fig. 12.31. Find the area of the remaining part
of the park.

Sol. We have,

Area of 4 quadrants of a circle of radius 14 m

= 4× 1 (pr2) = 4 × 1 × 22 × 14 × 14 = 616 m2
4 4 7

Area of square park having side 100 m long

= (100 × 100) m2 = 10,000 m2 Fig. 12.31

Hence, area of the remaining part of the park

= area of square – area of 4 quadrants at each corner

= (10,000 – 616) m2 = 9384 m2

21. Find the area of the shaded region in Fig. 12.32, where ABCD is a square of side 14 cm each.

Sol. Area of square ABCD = 14 × 14 cm2 = 196 cm2

14
Diameter of each circle = 2 cm = 7 cm

7
So, radius of each circle = 2 cm

Area of 4 circles = 4pr2 = 4× 22 × 7 × 7 = 154 cm2
7 2 2

\ Area of shaded region = area of square – area of 4 circles Fig. 12.32

= (196 – 154) cm2 = 42 cm2

22. In Fig. 12.33, ABCD is a trapezium of area 24.5 sq. cm. In it, AD || BC, ∠DAB = 90°,

AD = 10 cm and BC = 4 cm. If ABE is a quadrant of a circle, find the area of the shaded region.

(Take p = 22 ) [CBSE (AI) 2014]
7

Sol. Area of trapezium = 24.5 cm2

12 [AD + BC] × AB = 24.5

12 [10 + 4] × AB = 24.5 Fig. 12.33
AB = 3.5 cm ⇒ r = 3.5 cm

326 Xam idea Mathematics–X

Area of quadrant = 1 pr2
4
22
= 0.25 × 7 ×3.5 × 3.5 = 9.625 cm2

The area of shaded region = 24.5 – 9.625 = 14.875 cm2

23. In Fig. 12.34, O is the centre of a circle such that diameter AB =13 cm and AC = 12 cm. BC is

joined. Find the area of the shaded region. (Take p = 3.14) [CBSE (AI) 2016]

Sol. In DABC, ∠ACB = 90° (Angle in the semicircle)
\ BC2 + AC2 = AB2
\ BC2 = AB2 – AC2

= 169 – 144 = 25

\ BC = 5 cm

Area of the shaded region = area of semicircle – area of right DABC

= 1 pr2 – 1 × BC × AC Fig. 12.34
2 2

= 1 (3.14)  13 2 − 1 × 5 × 12
2  2  2

= 66.33 – 30 = 36.33 cm2

24. In Fig. 12.35, are shown two arcs PAQ and PBQ. Arc PAQ is a part of circle with centre O and
radius OP while arc PBQ is a semi-circle drawn on PQ as diameter with centre M.

If OP = PQ = 10 cm show that area of shaded region is 25  3 − π cm2. [CBSE (Delhi) 2016]
Sol. Since OP = PQ = QO  6 

⇒ DPOQ is an equilateral triangle

\ ∠POQ = 60°

Area of segment PAQM

= i rr2 – 3 a2 = 60° r # 102 – 3 #102
360° 4 360° 4

= d 100r – 100 3 n cm2
6 4

Area of semicircle with M as centre = π (5)2 = 25π cm2 Fig. 12.35
2 2

Area of shaded region = 25π −  50π − 25 3  = 25 π − 50 π + 25 3
2  3  2 3

= –25 r + 25 3 = 25a 3 – r kcm2
6 6

Long Answer Questions [5 marks]

1. PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR Fig. 12.36
and RS are equal. Semicircles are drawn on PQ and QS as diameters is
shown in Fig. 12.36. Find the perimeter and area of the shaded region.

Sol. We have,

PS = Diameter of a circle of radius 6 cm = 12 cm

\ PQ = QR = RS = 12 = 4 cm
3
QS = QR + RS = (4 + 4) cm = 8 cm

Areas Related to Circles 327

Hence, required perimeter
= arc of semicircle of radius 6 cm + arc of semi circle of radius 4 cm

+ arc of semi-circle of radius 2 cm

= (p × 6 + p × 4 + p × 2) cm = 12p cm = 12 × 22 = 264 = 37.71 cm.
7 7
Required area = area of semicircle with PS as diameter + area of semicircle with PQ as diameter

– area of semi-circle with QS as diameter

= 1 × 22 × (6)2 + 1 × 22 × (2)2 − 1 × 22 × (4)2
2 7 2 7 2 7

= 1 × 22 (62 + 22 − 42) = 1 × 22 × 24 = 264 cm 2
2 7 2 7 7

= 37.71 cm2

2. Figure 12.37 depicts an archery target marked with its five scoring areas 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. [NCERT]

Sol. The area of Gold region = p(10.5)2 = 22 × 110.25
7
2425.5
= 7 cm2 = 346.5 cm2

The area of Red region = p[(21)2 – (10.5)2]

= p[441 – 110.25]

= 22 × 330.75 = 7276.5 cm2 = 1039.5 cm2
7 7

The area of Blue region = p[(31.5)2 – (21)2] = p[992.25 – 441] Fig. 12.37

= 22 × 551.25 = 12127.5 = 1732.5 cm2
7 7
The area of Black region = p[(42)2 – (31.5)2]

= 22 [1764 – 992.25] = 16978.5 = 2425.5 cm2
7 7

And the area of White region = p[52.5)2 – (42)2]
22
= 7 [2756.25 – 1764]

= 22 × 992.25 = 21829.5 = 3118.5 cm2
7 7

3. Fig. 12.38, depicts a racing track whose left and right ends are semicircular. The distance
between the two inner parallel line segments is 60 m and they are each 106 m long. If the track
is 10 m wide, find:

(i) the distance around the track along its inner edge.

(ii) the area of the track. [NCERT]

Sol. Here, we have Fig. 12.38
OE = O'G = 30 m
AE = CG = 10 m
OA = O'C = (30+10) m = 40 m
AC = EG = FH = BD = 106 m

328 Xam idea Mathematics–X

(i) The distance around the track along its inner edge

= EG + FH + 2 × (circumference of the semicircle of radius OE = 30cm)

= 106 + 106 + 2 1 × 2π × 30  = 212 + 60p
2 

= 212 + 60 × 22 = c212 + 1320 m = c 1484 +1320 m= 2804 = 400 4 m
7 7 7 7 7

(ii) Area of the track = area of the shaded region

= a rea of rectangle AEGC + area of rectangle BFHD + 2 (area of the
semicircle of radius 40 m – area of the semicircle with radius 30 m)

= [21(01162000×++110270626) 0+×+(71002702×=[(1420016)22)0]–++(3202)22102]0×=27243×2(040m)22− 1 × 22 × (30)2 
2 7
=
=

4. 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 Fig. 12.39). Find the area of the shaded region. (Use p = 3.14 and 3 = 1.73205)

[NCERT]

Sol. Let each side of the equilateral triangle be x cm. Then,
Area of equilateral triangle ABC = 17320.5 cm2 (Given)

⇒ 3 x2 = 17320.5 ⇒ 1.73205 x2 = 17320.5
4 4

⇒ x2 = 4 #17320.5 ⇒ x2 = 40000
1.73205

\ x =200 cm Fig. 12.39
Thus, radius of each circle
= 200 cm
2

Now, area of shaded region = area of DABC –3 × area of a sector of angle 60° and radius 100 cm

= 17320.5 – 3 × 60° × π × (100)2
360°

= 17320.5 – 1 × p × 100 × 100 = 17320.5 – 3.14 × 5000
2

= 17320.5 – 15700 = 1620.5 cm2

5. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle

ABC in the middle as shown in Fig. 12.40. Find the area of the design. [NCERT]

Sol. Here, DABC is an equilateral triangle. Let O be the circumcentre of circumcircle.

Radius, r = 32 cm.
Now, area of circle = pr2

= 22 × 32 × 32 = 22528 cm2
Area of DABC 7 7

= 3 × area of DBOC

= 3 × 1 × 32 × 32 × sin 120°
2

Fig. 12.40

Areas Related to Circles 329

(∠BOC = 2∠BAC = 2 × 60° = 120°)

= 3 × 16 × 32 × 3
2

 sin 120° = sin (180° − 60°) = sin 60° = 3
2 

= 3 × 16 × 16 × 3 = 768 3 cm2

\ Area of the design = area of the circle – area of DABC

=  22528 − 768 3 
7 

= (3218.28 – 1330.176) = 1888.1 cm2

6. In Fig. 12.41, 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. [NCERT]

Sol. Here, area of sector OBQC = 39600°° × π × (7)2

= 1 × 22 ×7×7= 77 cm2
4 7 2

and, area of DOBC = 1 ×OC×OB = 1 ×7×7 = 49 cm2
2 2 2

\ Area of the segment BQC = area of sector OBQC – area of DOBC

     = 77 − 49 = 28 = 14 cm2
2 2 2

Similarly, area of the segment APC = 14 cm2

= pr2 = 22 7 7 77 cm2 Fig. 12.41
7 2 2 2
Now, the area of the circle with OD as diameter × × =

Hence, the total area of the shaded region = 14 + 14 + 77  cm 2 =  28 + 77  cm 2
2   2 

= d 56 + 77 ncm2 = 133 cm2 = 66.5 cm2
2 2

7. In Fig. 12.42 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. [NCERT]

Sol. In DABC, we have

BC = ( AC)2 + ( AB)2 (By Pythagoras Theorem)

= (14)2 + (14)2 = 196 + 196 = 392 = 14 2 cm

Now, area of sector ABPC = 90° × π × (14)2
360°

= 1 × 22 × 14 × 14 = 154 cm2 Fig. 12.42
4 7

and, area of D ABC = 1 × AC × AB = 1 × 14 × 14 = 98 cm2
2 2

\ Area of segment BPC = area of sector ABPC – area of DABC

= (154 – 98) cm2 = 56 cm2

330 Xam idea Mathematics–X

Now, we have radius of semi-circle BQC = 14 2 cm = 7 2 cm
2

\ Area of semi-circle = 1 pr2 = 1 × 22 × 7 2×7 2 = 154 cm2
2 2 7

Hence, area of the shaded region = area of the semi-circle BQC – area of the segment BPC

= (154 – 56) cm2 = 98 cm2

8. In Fig 12.43, a circle is inscribed in an equilateral triangle ABC of

side 12 cm. Find the radius of inscribed circle and the area of the

shaded region. (Use p = 3.14 and 3 = 1.73) [CBSE Delhi 2014]

Sol. Construction: Join OA, OB and OC

Draw OZ ⊥ BC, OX ⊥ AB and OY ⊥ AC.

Let the radius of the circle be r cm.

Area of DABC = area of DAOB + area of DBOC + area of DAOC

43 (side)2 = 1 × AB × OX + 1 × BC × OZ + 1 × AC × OY Fig. 12.43
2 2 2 [From Fig. 12.44]

43 (12)2 = 1 × 12× r + 1 × 12× r + 1 × 12× r
2 2 2

43 × 12 × 12 = 3 × 1 × 12 × r
2

r = 2 3 cm

Area of shaded region = area of ∆ABC – area of inscribed circle

=  3 (12)2 − π(2 3)2  cm 2
 4 
 
Fig. 12.44

= 3 × 12 × 12 – 3.14 × 4 × 3
4

= 1.73 × 3 × 12 – 3.14 × 4 × 3

= 62.28 – 37.68 = 24.6 cm2

9. In Fig. 12.45, PSR, RTQ and PAQ are three semicircles of diameters 10 cm, 3 cm and 7 cm

respectively. Find the perimeter of the shaded region. (Use p = 3.14) [CBSE Delhi 2014]

Sol. Radius of semicircle PSR = 1 × 10 cm = 5 cm
2

Radius of semicircle RTQ = 12 × 3 cm = 1.5 cm

Radius of semicircle PAQ = 12 × 7 cm = 3.5 cm

Perimeter of shaded region = c ircumference of semicircle Fig. 12.45
PSR + circumference of
semicircle RTQ + circumference of semicircle PAQ

= 1 × 2π × 5 + 1 × 2π × 1.5 + 1 × 2π × 3.5
 2 2 2

= p[5 + 1.5 + 3.5] = 3.14 × 10 = 31.4 cm

Areas Related to Circles 331

10. An elastic belt is placed around the rim of a pulley of radius 5 cm. (Fig. 12.46). From one point
C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is
at P, 10 cm from the point O. Find the length of the belt that is still in contact with the pulley.

Also find the shaded area. (Use p = 3.14 and 3 =1.73) [CBSE Delhi 2016]

Sol. In DAOP, cos q = 5
10

⇒ cos q = 1 ⇒ q = 60°
2

⇒ Reflex ∠AOB = 240°

\ Length of belt in contact with pully = θ × 2pr
360°
Fig. 12.46
2 × 3.14 × 5 × 240
= 360 = 20.93 cm

Now, AP = tan 60°
OA

PA = 5 3 cm = BP (Tangents from an external point are equal)

Area (DOAP + DOBP) = 2 1 ×5×5 3  = 25 3 = 43.25 cm2
2 

Area of sector OACB = 36i0° rr2 = 25 # 3.14 # 120 = 26.17 cm2
360
Shaded area = 43.25 – 26.17 = 17.08 cm2

11. In Fig. 12.47, a sector OAP of a circle with centre O, containing angle q. AB is
perpendicular to the radius OA and meets OP produced at B. Prove that the

perimeter of shaded region is r tan θ + sec θ + πθ − 1 . [CBSE (AI) 2016]
180°

Sol. Length of arc %AP = 2rr i or rri
In right DAOB 360° 180°

AB = tan q ⇒ AB = r tan q
r
Fig. 12.47
OB = sec q ⇒ OB = r sec q
r

PB = OB – OP = r sec q – r %AP = r tan q + r sec q – r + rri
180°
Perimeter of shaded region = AB + PB +

= r;tan i + sec i – 1 + ri E Hence proved.
180°

12. Find the area of the shaded region in Fig. 12.48, where *APD, *AQB, (BRC , (CSD are semi-

circles of diameter 14 cm, 3.5 cm, 7 cm and 3.5 cm respectively.  Use π = 22  [CBSE (F) 2016]
 7 

Sol. Area of shaded region = 1 × 22 72 +  7 2 − 2 7 2 
2 7   2  4  


= 1 × 22 49 + 49 − 49  = 1 × 22 × 49  9 
2 7 4 8  2 7  8 

= 693 sq cm or 86.625 cm2
8
Fig. 12.48
332 Xam idea Mathematics–X

13. A chord PQ of a circle of radius 10 cm subtends an angle of 60° at the centre of circle. Find the
area of major and minor segments of the circle. [CBSE Delhi 2017]

Sol. Area of minor segment

= area of minor sector having angle 60° at centre – area of equilateral DOPQ

= 22 ×10 ×10 × 60° – 3 ×10 ×10
7 360° 4

= 10 × 10 = 22 × 1 – 3
7 6 4G
10 cm 60°
100 25
= 84 (44 – 21 3) cm2 or 21 (44 – 21 3) cm2 Q

Area of major segment Fig. 12.49

= area of circle – area of minor segment

= 22 ×10 ×10 – 25 (44 – 21 3)F
<7 21

= 2200 – 25 (44 – 21 3) & 6600 – 1100 + 25×21 3
7 21 21

= 5500 + 25×21 3 & 25 ^220 + 21 3 hcm2
21 21

14. In the given Fig. 12.50, ∆ABC is a right-angled triangle in

which ∠A is 90°. Semicircles are drawn on AB, AC and BC
as diameters. Find the area of the shaded region.

[CBSE (AI) 2017]

Sol. a ABC is right angled triangle

\ Hypotenuse BC = 32 + 42 = 5 cm Fig. 12.50
From figure,

Area (R1 + R2) = area of semicircle having diameter BC – area of ∆ABC

= r c d 2 – 1 × base × height = r c 5 2 – 1 ×3×4
2 2 2 2 2 2
m m

= c 25 r – 6m cm2 ...(i)
8

Now, area of shaded region = a rea of semicircle having diameter BA + area of semicircle
having diameter AC – area [R1 + R2]

= r c 3 2 + r (2) 2 – < 285 r – 6F
2 2 2
m

= r c 9 + 4 – 25 m + 6
2 4 4

= r (0) + 6 = 6 cm2
2

15. In Fig. 12.51, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find

the area of the shaded region. [CBSE (AI) 2017]

Sol. In right angle triangle ABC

Diameter BC = 242 + 72 = 25 cm

Areas Related to Circles 333

Area ∆CAB = 1 × base × height A
2
1 B
= 2 × 24 × 7 = 84 cm2 O

Area of shaded region = area of semicircle – a rea of ∆CAB + area of

2 2 quadrant BOD

= r d 25 n– 84 + r d 25 n C
2 2 4 2
D

= d 625r + 625r n – 84 Fig. 12.51
8 16

= c 1875r – 84m = (117.18π – 84) = 283.94 cm2
16

HOTS [Higher Order Thinking Skills]

1. Two circles touch internally. The sum of their areas is 116 pcm2 and distance between their

centres is 6 cm. Find the radii of the circles. [CBSE (F) 2017]

Sol. Let R and r be the radii of the circles [Fig. 12.52].

Then, according to question,

⇒ pR2 + pr2 = 116p

⇒ R2 + r2 = 116 ...(i)

Distance between the centres = 6 cm

⇒ OO' = 6 cm

⇒ R – r = 6 ...(ii) Fig. 12.52

Now, (R + r)2 + (R – r)2 = 2(R2 + r2)

Using the equation (i) and (ii), we get

(R + r)2 + 36 = 2 × 116

⇒ (R + r)2 = (2 × 116 – 36) = 196

⇒ R + r = 14 ...(iii)

Solving (ii) and (iii), we get R = 10 and r = 4

Hence, radii of the given circles are 10 cm and 4 cm respectively.

2. Find the area of the shaded design of Fig. 12.53, where ABCD is a square
of side 10 cm and semicircles are drawn with each side of the square as
diameter (use p = 3.14).

Sol. Let us mark the four unshaded regions as I, II, III and IV (Fig. 12.54).

Area of I + area of III

= area of ABCD – areas of two semicircles of radius 5 cm each Fig. 12.53

= c10×10 – 2× 1 ×r×52 m
2

= (100 – 3.14 × 25) = (100 – 78.5) = 21.5 cm2

Similarly, area of II + area of IV = 21.5 cm2

So, Area of the shaded design

Fig. 12.54

334 Xam idea Mathematics–X

= area of ABCD – area of (I + II + III + IV)
= (100 – 2 × 21.5) cm2

= (100 – 43) cm2 = 57 cm2

3. Two circles touch externally. The sum of their areas is 130 p sq. cm and the distance between
their centres is 14 cm. Find the radii of the circles.

Sol. If two circles touch externally, then the distance between their centres is equal to the sum of their radii.

Let the radii of the two circles be r1 cm and r2 cm respectively [Fig. 12.55].

Let C1 and C2 be the centres of the given circles. Then,

C1C2 = r1 + r2

⇒ 14 = r1 + r2 [ C1C2 = 14 cm given]

⇒ r1 + r2 = 14 …(i)

It is given that the sum of the areas of two circles is equal to 130 p cm2.

\ πr12 + πr22 = 130π …(ii)
⇒ r12 + r22 =130

Now, (r1 + r2)2 = r12 + r22 + 2r1r2

⇒ 142 = 130 + 2r1r2 (Using (i) and (ii))
⇒ 196 – 130 = 2r1r2
⇒
r1r2 = 33 …(iii)
Now,
(r1 – r2)2 = r12 + r22 – 2r1r2 Fig. 12.55

⇒ (r1 – r2)2 = 130 – 2 × 33 (using (ii) and (iii))

⇒ (r1 – r2)2 = 64

⇒ r1 – r2 = 8 …(iv)

Solving (i) and (iv), we get r1 = 11 cm and r2 = 3 cm.

Hence, the radii of the two circles are 11 cm and 3 cm.

4. In Fig. 12.56, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with
AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking
BC as diameter, a semicircle is added on outside the region. Find
the area of the shaded region. [Use p = 3.14] [CBSE (F) 2014]

Sol. Area of shaded region

= area of rectangle – area of triangle + area of semicircle.

In right ∆ADE Fig. 12.56

AD2 = AE2 + DE2

AD = 92 + 122 = 81 + 144 = 225 = 15 cm

Area of ∆AED = 1 × DE × AE = 1 × 12 × 9 = 54 cm2
2 2
In semicircle at BC, diameter = BC = 15 cm.

Radius of semicircle = 1 × 15 = 7.5 cm
2

Area of semicircle = πr2 = 3.14 × 7.5 × 7.5 = 88.31 cm2 Fig. 12.57
2 2
Areas Related to Circles 335
Area of rectangle = AB × BC = 20 × 15 = 300 cm2

Area of shaded region = 300 + 88.31 – 54 = 334.31 cm2

5. In the given Fig. 12.58, the side of square is 28 cm and

radius of each circle is half of the length of the side of the

square where O and O′ are centres of the circles. Find the O

area of shaded region. [CBSE Delhi 2017] B
C
Sol. Area of shaded region

= area of square + area of 2 major sectors having

angle 270° at centre

= side × side + 2 × rr2 i O′
360°
Fig. 12.58
= <28×28 + 2× 22 ×14 ×14 × 270° F
7 360°

= 28 × 28 d1+ 11 × 3 n
7 4

= 28 × 28 c1+ 33 m = 1708 cm2 A 21 cm
28

6. In the given Fig. 12.59, ABCD is a rectangle of dimensions

21 cm × 14 cm. A semicircle is drawn with BC as diameter.

Find the area and the perimeter of the shaded region in
[CBSE (AI) 2017] 14 cm
the figure.

Sol. Area of shaded region

= area of rectangle – area of semicircle

= (l × b) – 1 × π × r2 D
2 Fig. 12.59
1
= (21 × 14) – 2 ×π×7×7

= 294 – 7127 ×=227217×cm7 2× 7
= 294 –

Now, perimeter of shaded region = 2l + b + circumference of semicircle i.e.; πr

= 2 × 21 + 14 + 22 ×7
7
= 56 + 22 = 78 cm

PROFICIENCY EXERCISE

QQ Objective Type Questions: [1 mark each]

1. Choose and write the correct option in each of the following questions.

(i) The area of a circle is 49p cm2. Its circumference is

(a) 7p cm (b) 14p cm (c) 21p cm (d) 28p cm

(ii) The circumference of two circles are in the ratio 2 : 3. The ratio of their areas is

(a) 2 : 3 (b) 4 : 9 (c) 9 : 4 (d) None of these

(iii) A square ABCD is inscribed in a circle of radius r. The area of the square is

(a) 3 r2 (b) 2 r2 (c) 4 r2 (d) None of these

(iv) If the perimeter of a semicircular protractor is 36 cm; its diameter is

(a) 14 cm (b) 16 cm (c) 18 cm (d) 12 cm

(v) The area of a sector of a circle with radius 14 cm and central angle 45° is

(a) 76 cm2 (b) 77 cm2 (c) 66 cm2 (d) 55 cm2

336 Xam idea Mathematics–X

2. Fill in the blanks.

(i) Circumference of a sector of angle q of a circle with radius r is _____________ .
(ii) The sector of a circle of angle 90° is also known as _____________ of circle.
(iii) The area of a circle whose area and circumference are numerically equal, is _____________.
(iv) The length of an arc of a sector of angle q is _____________ .
(v) If the perimeter of a circle is equal to that of a square, then ratio of their areas is _____________.

QQ Very Short Answer Questions: [1 mark each]

3. The difference between the area and square of radius of a circle is 105 cm. What will be the
circumference of the circle?

4. What is 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? [NCERT Exemplar]

5. The area of a circle is 220 cm2. What will be the area of a square inscribed in it?

6. What is the area of the circle that can be inscribed in a square of side 6 cm? [NCERT Exemplar]

7. If the area of circle is 616 cm2, then what is its circumference?

8. The circumference of a circle is 50 cm. What will be the side of a square that can be inscribed in
the circle?

9. If the area of a circle increases from 9π to 16π, then what will be the ratio of the circumference
of the first circle to the second circle?

10. A wire can be bent in the form of a circle of radius 35 cm. If it is bent in the form of a square,
then what will be its area?

QQ Short Answer Questions-I: [2 marks each]

11. What is the area of a square inscribed in a circle of diameter x cm?

12. What is the distance travelled by a circular wheel of diameter d cm in one revolution?

13. What is the ratio of areas of two circles whose circumferences are in the ratio 3 : 4?

14. If the area of a sector of a circle is 5 th of the area of that circle, then find the central angle of
18
the sector.

15. If a circle is inscribed in a square, what is the ratio of the area of the circle and the square?

16. Find the radius of semicircle if its perimeter is 18 cm.

17. What is the length of an arc in terms of p that subtends an angle of 72° at the centre of a circle
of radius 10 cm?

18. In a circle of radius 8 cm, an arc subtends an angle of 108° at the centre. What is the area of the
sector in terms of p?

19. Find the perimeter of a square circumscribing a circle of radius a cm.

20. What is the angle subtended at the centre of a circle of radius 5 cm by an arc length 4p cm?

21. Find the area of a quadrant of a circle whose circumference is 616 cm.

22. Find the radius of a semicircular protractor if its perimeter is 36 cm.

23. Find the circumference of a circle if the area of a quadrant of the circle is 154 cm2.

Areas Related to Circles 337

QQ Short Answer Questions-II: [3 marks each]

24. Find the area of the shaded region in Fig. 12.60, where arcs drawn with centres A, B, C and D

intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA respectively of a

square ABCD of side 12 cm. [Use p = 3.14] [CBSE 2018, (30/1)]

×

××

×

Fig. 12.60

25. The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of

distances travelled by their tips in 48 hours. [CBSE 2018, (C) (30/1)]

26. The side of a square is 10 cm. Find the area between inscribed and circumscribed circles of the

square. [CBSE 2018, (C) (30/1)]

27. Find the area of the shaded region in Fig. 12.61, if ABCD is a rectangle with sides 8 cm and 6 cm

and O is the centre of circle. (Take p = 3.14) [CBSE 2019, (30/1/1)]

D8 C
6O B

A

Fig. 12.61

28. Find the area of the segment shown in Fig. 12.62, if radius of the circle is 21 cm and

∠AOB = 120° dUse r = 22 n [CBSE 2019, (30/1/2)]
7

AB
21 cm 120° 21 cm
O

Fig. 12.62

29. In Fig. 12.63, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of

the shaded region. (Use p = 3.14) [CBSE 2019, (30/2/1)]

B

QP

CA

15 cm

O
Fig. 12.63

338 Xam idea Mathematics–X

30. In Fig. 12.64, ABCD is a square with side 2 2 cm and inscribed in a circle. Find the area of the

shaded region. (Use p = 3.14) [CBSE 2019, (30/2/1)]

D

AC

B

Fig. 12.64

31. A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping

through an angle 120°. Find the total area cleaned at each sweep of the blades. d Take r = 22 n
7

[CBSE 2019, (30/3/1)]

32. A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the
22
corresponding minor segment of the circle. (Use r = 7 and 3 = 1.73) [CBSE 2019, (30/3/3)]

33. In Fig. 12.65, two concentric circles with centre O, have radii 21 cm and 42 cm. If ∠AOB = 60°,

find the area of the shaded region. [CBSE 2019, (30/4/2)]

O

60°

AB
Fig. 12.65

34. In Fig. 12.66, find the area of the shaded region, where ABCD is a square of side 14 cm in which
four semi-circles of same radii are drawn as shown. (Take p = 3.14) [CBSE 2019, (C) (30/1/2)]

A 14 cm B

3 cm 3 cm 14 cm

DC

Fig. 12.66

35. In Fig. 12.67, AB and CD are two perpendicular diameters of a circle with
22
centre O. If OA = 7 cm, find the area of the shaded region. cUse r= 7 m.

Fig. 12.67

Areas Related to Circles 339

36. Find the area of the shaded region in Fig. 12.68, if AC = 24 cm, BC = 10 cm and O is the centre
of the circle. (Use p = 3.14)

Fig. 12.68

37. The area of an equilateral triangle is 49 3 cm2. Taking each vertex as centre, circles are described

with radius equal to half the length of the side of the triangle. Find the area of the part of the
22
triangle not included in the circles. (Take 3 = 1.73, p = 7 )

38. In Fig. 12.69, the boundary of shaded region consists of four semicircular arcs, two smallest

being equal. If diameter of the largest is 14 cm and that of the smallest is 3.5 cm, calculate the
22
area of the shaded region. (Use p = 7 )

Fig. 12.69

39. The inner circumference of a circular track is 132 m. The track is 2.5 m wide everywhere.
Calculate the cost of putting up a fence along the outer circle at the rate of ™3.50 per metre.

40. A race track is in the form of a ring whose inner and outer circumferences are 44 cm and 66 cm
respectively. Find the width of the track.

41. A circular park is surrounded by a road 28 m wide. Find the area of the road if the circumference
of the park is 880 m.

42. Find the area of the flower bed (with semicircular ends.) in Fig. 12.70.

Fig. 12.70
43. Find the area of the shaded region in Fig. 12.71.

Fig. 12.71

340 Xam idea Mathematics–X

44. Find the area of the shaded field shown in Fig. 12.72.

Fig. 12.72

45. In Fig. 12.73, arcs are drawn by taking vertices A, B, and C of an equilateral

triangle of side 10 cm to intersect the sides BC, CA and AB at their Fig. 12.73
respective mid-points D, E and F. Find the area of the shaded region.
(Use p = 3.14)

46. In Fig. 12.74, ABCP is a quadrant of a circle of radius 20 cm. With AC as diameter, a semi-circle

is drawn. Find the area of the shaded portion.

Fig. 12.74

47. In Fig. 12.75, ABC is a triangle right angled at A. Find the area of the shaded region if
AB = 6 cm, BC = 10 cm and I is the centre of incircle of DABC.

Fig. 12.75

48. Prove that the area of a circular path of uniform width h surrounding a circular region of radius
is ph (2r + h).

49. In Fig. 12.76, ABCD is a rectangle with AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as
diameter, three semicircles are drawn. Find the area of the shaded region.

Fig. 12.76

Areas Related to Circles 341

50. In Fig. 12.77, three semicircles A, B and C are drawn having diameters 3 Fig. 12.77
cm each and a circle D is drawn with diameter 4.5 cm. Calculate (i) the area
of the shaded region. (ii) the cost of painting the shaded region at the rate
of 25 paise per cm2.

51. A boy is cycling such that the wheels of the cycle are making 140 revolutions
per minute. If the diameter of the wheel is 60 cm, calculate the speed per
hour which the boy is cycling.

QQ Long Answer Questions: [5 marks each]

52. Three circles each of radius 7 cm are drawn in such a way that each of them touches the other
two. Find the area enclosed between the circles.

53. The area of a circular playground is 88704 m2. Find the cost of fencing this ground at the rate
of ™65 per metre.

54. The diameter of front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the
number of revolutions that rear wheel will make in covering a distance in which the front wheel
makes 1400 revolutions.

55. Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central
angle of 60° (Use p = 3.14)

56. Find the difference of the area of a sector of angle 90° and its corresponding major sector of a
circle of radius 9.8 cm.

57. Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm
subtending an angle of 90° at the centre.

58. On a square cardborad sheet of area 784 cm2, four congruent circular plates of maximum size
are placed such that each circular plate touches the other two plates and each side of the square
sheet is tangent to two circular plates. Find the area of the square sheet not covered by the
circular plates.

59. Area of the sector of central angle 200° of a circle is 770 cm2. Find the length of the corresponding
arc of this sector.

60. Find the area of the shaded region given in Fig. 12.78.

Fig. 12.78

61. All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is

2464 cm2. (Hint: radius of circle = 1 diagonal)
2

62. Find the number of revolutions made by a circular wheel of area 6.16 m2
in rolling a distance of 572 m.

63. An archery target has three regions formed by three concentric circles in
Fig. 12.79. If the diameters of the concentric circles are in the ratio 1 : 3 : 5,
then find the ratio of the areas of three regions.

Fig. 12.79

342 Xam idea Mathematics–X