Class 10 Maths Extra Questions Chapter 11 Area related to circles

(iv)Angle described by hour hand in 12 hours = 360°

360°
Angle described by hour hand in one hour = = 30°
12

30°  1  
Angle described by hour hand in one minute = = 
60 2

 1 
Thus, hour hand rotates through an angle of   in one minute.
2

VERY SHORT ANSWER QUESTIONS
1. If the diameter of a semi circular protactor is 14 cm, then find its perimeter.
2. If circumference and the area of a circle are numerically equal, find the diameter
of the circle.
3. Find the area of the circle ‘inscribed’ in a square of side a cm.
4. Find the area of a sector of a circle whose radius is r and length of the arc is l.
5. The radius of a wheel is 0.25 m. Find the number of revolutions it will make to
travel a distance of 11 kms.
6. If the area of a circle is 616 cm², then what is its circumference?
7. What is the area of the circle that can be inscribe in a square of side 6 cm?
8. What is the diameter of a circle whose area is equal to the sum of the areas of
two circles of radii 24 cm and 7 cm?
9. 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?
10. What is the angle subtended at the centre of a circle of radius 6 cm by an arc of
length 3 cm?
11. Write the formula for the area of a sector of angle (in degrees) of a circle of
radius r.
12. If the circumference of two circles are in the ratio 2:3, what is the ratio of their
areas?
13. If the difference between the circumference and radius of a circle is 37 cm,
22
then find the circumference of the circle. ( Use  = )
7
170 Mathematics-X

14. If diameter of a circle is increased by 40%, find by how much percentage its
area increases?
15. The minute hand of a clock is 6 cm long. Find the area swept by it between
11:20 am and 11:55 am.
16. The perimeter of a sector of a circle of radius 14 cm is 68 cm. Find the area of
the sector. (CBSE 2020)
17. The circumference of a circle is 39.6 cm. Find its area.
22
(Use  =) (CBSE 2020)
7
18. The length of the minute hand of a clock is 14 cm. Find the area swept by the
22
minute hand in one minute. (Use  = )
7
19. Area of a sector having length of corresponding arc ‘l’ and radius ‘r’ is ______.
20. Circumference of a circle of radius s is ___________ .
21. Area of a circle of radius is _______ ,
22. Length of an arc of a sector of a circle with radius r and angle  is _______ .
23. Area of a sector with radius r and angle with degrees measure  is _______ .
24. Area of segment of a circle = Area of the corresponding sector ________ .

SHORT ANSWER TYPE QUESTIONS (1)
25. Find the area of a quadrant of a circle whose circumference is 22 cm.
22
(Use  =)
7
26. What is the angle subtended at the centre of a circle of radius 10 cm by an arc
of length 5 cm?
27. If a square is inscribed in a circle, what is the ratio of the area of the circle and
the square?
28. Find the area of a circle whose circumference is 44 cm. (CBSE 2020)
29. If the perimeter of a circle is equal to that of square, then find the ratio of their
areas.

Mathematics-X 171

30. What is the ratio of the areas of a circle and an equilateral triangle whose
diameter and a side are respectively equal?
5
31. In fig., O is the centre of a circle. The area of sector OAPB is of the area of
18
the circle. Find x.

O
x
A B

P
32. Find the perimeter of the given fig, where AED is a semicircle and ABCD is a
rectangle. (CBSE 2015)
20 cm
B A
14 cm

E

C D
20 cm
33. In fig. OAPBO is a sector of a circle of radius 10.5 cm. Find the perimeter of
the sector.
P

A B

60°

O
34. In the given fig, APB and CQD are semi circles of diameter 7 cm each, while
ARC and BSD are semicircles of diameter 14 cm each. Find the perimeter of
22
the shaded region. (Use  = ) (Delhi, 2011)
7

172 Mathematics-X

R

P
7 cm 7 cm 7 cm
A B C D
Q

S

SHORT ANSWER TYPE II QUESTIONS
35. Area of a sector of a circle of radius 36 cm is 54cm2 . Find the length of the
corresponding arc of the sector.
36. The length of the minute hand of a clock is 5 cm. Find the area swept by the
minute hand during the time period 6:05 am to 6:40 am.
37. In figure ABDC is a quadrant of a circle of a radius 28 cm and a semi circle
BEC is drawn with BC as diameter find the area of shaded region:
 22 
 Use   
 7 

B E
D

A C
38. In fig, OAPB is a sector of a circle of radius 3.5 cm with the centre at O and
AOB = 120°. Find the length of OAPBO.
P

O

120°
A B
39. Circular footpath of width 2 m is constructed at the rate of ` 20 per square
meter, around a circular park of radius 1500 m. Find the total cost of construction
of the foot path. (Take  = 3.14 )
Mathematics-X 173

40. 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 of cycle.
41. In a circle with centre O and radius 4 cm, and of angle 30°. Find the area of
minor sector and major sector AOB. (Use  = 3.14)
42. Find the area of the largest triangle that can be inscribed in a semi circle of
radius r unit. (NCERT Exemplar)
43. Figure ABCD is a trapezium of area 24.5 cm, In it AD||BC, DAB = 90°, AD
= 10 cm, BC= 4cm. If ABE is a quadrant of a circle. Find the area of the shaded
 22 
region.  Use   
 7 

D E A

B
C
44. From each of the two opposite corners of a square of side 8 cm, a quadrant of a
circle of radius 1.4 cm is cut. Another circle of radius 4.2 cm is also cut from
22
the centre as shown in fig. Find the area of the shaded portion. (Use  = ).
7

45. A sector of 100° cut off from a circle contains area 70.65 cm². Find the radius
of the circle. (Use = 3.14 )
46. In fig. 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
portion.
D C

7 cm

A 14 cm B
174 Mathematics-X

47. A square water tank has its each side equal to 40 m. There are four semi circular
grassy plots all around it. Find the cost of turfing the plot at Rs 1.25 per sq. m.
(Use  = 3.14 )
48. Find the area of the shaded region shown in the fig. (NCERT – Exemplar)
8m

4m
6m

49. Find the area of the minor segment of a circle of radius 21 cm, when the angle
of the corresponding sector is 120°.
50. A piece of wire 11 cm long is bent into the form of an arc of a circle subtending
an angle of 45° at its centre. Find the radius of the circle.
51. Find the area of the shaded region.

16 cm

44 cm
52. In fig. 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 semi circle is added on outside the region. Find the area of the
shaded region.
 22 
 Use   
 7 
A B
9 cm

90° E 15 cm
12 cm
D C
53. The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius
of the circle.

Mathematics-X 175

54. Find the area of the shaded region. (NCERT Exemplar)

4m

3m 3m
12 m

4m

26 m

LONG ANSWER TYPE QUESTIONS
55. Two circles touch externally. The sum of their areas is 130 sq. cm and the
distance between their centres is 14 cm. Find the radii of the circles.
56. 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.
57. Find the number of revolutions made by a circular wheel of area 6.16 m² in
rolling a distance of 572 m.
58. All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if
area of the circle is 2464 cm².
59. With vertices A, B and C of a triangle ABC as centres, arcs are drawn with
radius 6 cm each in fig. If AB= 20 cm, BC= 48 cm and CA= 52 cm, then find
 22 
the area of the shaded region.  Use   
 7 
A

52 cm

20 cm

C
B 48 cm
60. ABCDEF is a regular hexagon. With vertices A, B, C, D, E and F as the centres,
circles of same radius ‘r’ are drawn. Find the area of the shaded portion as
shown in the given figure.

176 Mathematics-X

A B

F C

E D

61. ABCD is a diameter of a circle of radius 6 cm. The lengths AB, BC and
CD are equal. Semicircles are drawn on AB and BD as diameter as shown
in the fig. Find the perimeter and area of the shaded region.
 22 
 Use   
 7 

A D
B C

62. A poor artist on the street makes funny cartoons for children and earns
his living. Once he made a comic face by drawing a circle within a circle,
the radius of the bigger circle being 30 cm and that of smaller being
20 cm as shown in the figure. What is the area of the cap given in this
figure?

Cap

63. In the given figure ABCD is a trapzium with
AB || DC, AB = 18 cm, DC = 32 cm and
distance between AB and DC is 14 cm. If arc
of equal radii 7 cm with centres A B, C and D
have been drawn, then find the area of shaded
 22 
region.  Use   
 7 
Mathematics-X 177

64. Find the area of the shaded region as shown in the given figure.

 22 
 Use   
 7 

20 cm

7 cm

ANSWERS AND HINTS

22
1. r + d =  7  14 = 36 cm
7
2. 2r = r2  diameter = 4 units
3. Side of the square is equal to diameter of the circle,

2 a2 a
r =   (side = a, radius = )
4 2

  l  r 2 lr
4. l=  2 r , Area =  r 2 = = sq. units
360 360 2 r 2

distance 11  1000  7  100
5.  = 7000
circumference 2  22  25
6. r2 = 616  r = 14 cm or 2r = 88 cm
7. Side of the square is equal to the diameter of the circle
 r = 3 cm or r2 = (3)2 = 9 cm2 .

8. R 2  r12  r22  R = 25 and diameter = 50 cm.

22 220
9. 2r = 2   35 = 220 cm , Side of square = 55 cm
7 4
Area of square = 55 × 55 = 3025 cm2

178 Mathematics-X

 
10. l =  2r  3   2  6   = 90°
360 360


11.  r 2
360
2
2 
2  r2 
12. 2r1  2  r1  2 r2 or r1   3   4 : 9
2r2 3 3 r22 r22
22
13. (2r – r) = 37 or r = 7, 2r = 2   7 = 44 cm
7
14. 96%
210 22  6  6
15. = 66 cm2 ( = 210°) (11: 20 to 11: 55 = 35 minutes)
360  7
16. 280 cm2
17. 124.74 cm2
18. 10.27 cm2
1
19. A = lr
2
20. 2r
21. s2


22.  2r
360


23.  r 2
360
24. Area of the corresponding triangle
7
25. 2r = 22, r =
2
r 2 22  7  7
Area of quadrant =  = 9.625 cm2
4 74 22

Mathematics-X 179

 
26. l =  2r  5   2  10    90
360 360

27.

If side of square is 1 unit, by Pythagoras Theorem
Diameter 2 unit.
Area of square = 1 × 1 = 1 sq units.

2 2 2  11
Area of Circle = r      =
2 2 2 7
Required ratio = 11 : 7
28. 154 cm2
2r Perimeter of circle
29. 2r = 4 unit or  (Let side of square = 1 unit)
4 unit Perimeter of square
7
r= unit
11
r 2 22 7 7 14
    or 14 : 11
1 7 11 11 11
3 2
30. Area of equilateral triangle = a
4
2
 a
Area of circle =   
2
Required ratio = 3:
 5
31. r 2 = r 2 
360 18
 = 100°
32. 20 cm + 14 cm + 20 cm + r
22
20 cm + 14 cm + 20 cm + × = 76 cm
7
180 Mathematics-X

 60  2  22  105
33.  2r   11 cm
360 360  7  10
Perimeter = 10.5 + 10.5 + 11 cm = 32 cm
34. Perimeter of shaded region = Perimeters of semi circles,
= ARC + APB + BSD + CQD
=  (r1 + r2 + r3 + r4)

22  7 7  22
=  7 7   21 = 66 cm
7  2 2 7
   36  36
35. 54  =
360
 = 15°
 15 2   36
l=  2r = = 3  cm
360 360

 210 22  5  5 1650 5
36. Area =  r 2    45  cm 2
360 360 7 36 6
( = 210° in 35 minutes)
37. AC = 28 cm, BC = 28 2 cm (by Pythagoras Theorem).

BC
radius = 14 2 cm =
2
Shaded region = Area of semicircle – Area of segment BCD
1 90 1
= (14 2)2    (28)   28  28
2 360 2
= 392 cm2
240 2  22  35
38. l=
360  7  10
= 14.67
Length of OAPBO = 14.6 + 3.5 + 3.5
= 21.67 cm

Mathematics-X 181

39.  (r22  r12 ) = [(1502) 2 – (1500) 2 ]  20
= 3.14 [(1502)2 – (1500)2] × 20
= ` 377051.2
40. Circumference of cycle = 2r
22
= 2  30 cm
7
= 188.57 cm
18857  140  60
Speed of cycle =
100  1000
= 15.84 km/h

41. Area of Minor sector =  r 2
360
30
=  3.14  4  4 cm 2
360
= 4.19 cm2 (approx.)

Area of major sector =  r 2
360
330
=  3.14  4  4
360
= 46.1 cm2 (approx)
1 C
42. Area of  = base × height
2
1 r
= AB  OC
2 A
r
B
O
1
= 2r  r = r2 square unit
2
43. Let AB = h cm
1
Area of trapezium = ( AD  BC )  AB
2
1
24.5 = (10  4)  h (AB = h)
2
182 Mathematics-X

h = 3.5 cm
90
Area of quadrant ABE =   (3.5) 2 sq.cm
360
= 9.625 sq.cm
Area of shaded region = 24.5 – 9.625
= 14.875 sq.cm
44. Area of shaded portion =
Area of square – Area of circle – (Area of 2 quadrants)
22  42  42 22  14  14  1
= 64  
7  10  10 7  10  10  2
= 64 – 55.44 – 3.08
= 5.48 cm2
2
7065 100 314  r
45. =
100 360  100
7065  360
= r2
100  314
9= r
r = 9 cm.
  DC  
2
   
46. 2 
Area of shaded portion = r  AB  BC   2  
 2 
22  22  7  7 
=  (3.5) 2  98 
7  7  2 
= 38.5 + [98 – 77]
= 38.5 + 21
= 59.5 cm2
47. Four semicircluar means 2 circles ,
Area of 2 circles = 2r 2
= 2 × 3.14 × 20 × 20
= 2512 sq.m
Total cost = 2512 × 1.25
= ` 3140

Mathematics-X 183

r 2
48. Area of shaded region = l  b 
2
2 2
= 8 4  
2
2
= (32 + 2) cm
49. Area of the segment = Area of sector – Area of 
120 22
Area of sector =   21 21 = 462 cm2
360 7
441
Area of  = 3 cm 2
4
 441 
Area of segment =  462  3 cm2

4

=
21
4
 
88  21 3 cm 2


50. l=  2r
360
45 2  22  r
11 = 
360 7
14 = r
r = 14 cm
51. Shaded Area = l × b + r2
= (44 × 16 +  × 8 × 8)
= (704 + 64) cm2
1
52. Shaded Area = 20 × 15 + 28.12  –  12  9
2
= 334.39 cm2
53. 2r = 2r + 16.8

22 168  22  168
2 r  2r = or 2r   1 
7 10  7  10

 15  168 168  7 1176
or, 2r   = or r = = = 3.92 cm
 7 10 10  2  15 300
184 Mathematics-X

54. Area of shaded region = Area of rectangle – [Area of 2 semicircles + Area of
rectangle]

 r 2 
= L  B  2  l  b
 2 

= 26  12  [  2  2  16  4]
= 312 – 4 – 64 = (248 – 4) m2
2 2
55. r12  r22 = 130  r1  r2  130 ...(1)

 r1 + r2 = 14 …(2)
Substitute the value of r1 from (2) in (1) and solve.
2r22 – 28 r2 + 66 = 0

r22 – 14r2 + 33 = 0 (Neglecting – ve)

r2 = 11 cm and r1 = 3 cm
56. Area of shaded region = Area of  – Area of 3 sectors.

3 3
area  =  14  14 = 196 = 49 3
4 4

60 22
Area of 3 Sectors = 3    7  7 = 77 sq. cm
360 7

 required Area = (49 3  77) cm2

616
57. r2 = or r 2  1.96 or r = 1.4 m
100

22 14 616
2r = 2   = = 8.8 m
7 10 100

572
Number of revolutions = = 65
8.8
58. r2 = 2464 cm2

Mathematics-X 185

r = 28 cm or d = 28 + 28 = 56 cm
1 1
Area of rhombus = d1d 2 or d 22 (d1  d 2 )
2 2
1
=  56  56 = 1568 cm2
2
59. By converse of Pythagoras theorem ABC is right .
 Area of shaded region = Area of  – Area of 3 sectors.

1 r 2
=  48  20  (1  2  3 )
2 360
22  6  6
= 480  (180)
7  360
= 480 – 56.57
= 423.43 cm2
60. 2r2 (Area is equal to 2 circles.)
2r1 2r2 2r3
61. Perimeter =  
2 2 2
 22 6 22 4 22 2 
= 2    2  2 
 7 2 7 2 7 2 
22
= 2  3  2  1 = 37.71 cm
7
 r12 r22 r32 
Area =       
 2 2 2
2
= 31.71 cm
62. Radius of bigger circle O = 30 cm
Radius of Smaller circle O = 20 cm
Difference of their radii = (30 – 20) = 10 cm
AB is tangent to small circle
Radius = OC i.e. OD  AB
O
 OCA = 90° = OCB
In  OCA by Phythagoras Theorem

186 Mathematics-X

AC = 20 2 cm
 AB = 2 × 20 2 cm
= 40 2 cm
CD = Radius of bigger circle-OC O
= 30–10 = 20 cm
1
Area of cap = AB × CD
2
1
=  40 2  20 cm 2
2
= 400 2 cm 2
1
63. Area of trapezium =  h (a  b)
2
1
=  14  (18  32) = 350 cm2
2
r 2
Area of four sectors =  (A  B  C  D )
360
77
=  360 = 154 cm2
360
area of shaded region = 350 – 154 = 196 cm2
 r 2 r 2 r 2 
64. Area of shaded region =  1  2  3 
2 2 2
 17 17 10 10 7  7 
=    
 2 2 2 
= 688.28 cm2

Mathematics-X 187

PRACTICE-TEST
AREAS RELATED TO CIRCLES
Time : 1 Hr. M.M.: 20

SECTION-A
1. If the circumference of two circles are equal, then what is the ratio between
their areas? 1
2. If the diameter of a protractor is 21 cm, then find its perimeter. 1
3. Area of a circle of radius P is ___________ . 1
4. Choose the correct answer.
If the perimeter and the area of a circle are numerically equal then the radius of
the circle is 1
(a) 2 units (b)  units (c) 4 units (d) 7 units

SECTION-B
5. The length of minute hand of a clock is 14 cm. Find the area swept by the
minute hand in 8 minutes. 2
6. Find the area of a circle whose circumference is 22 cm. 2
7. Find the area of a quadrant of a circle whose circumference is 44 cm. 2

SECTION-C
8. A horse is tied to a pole with 28 cm long string. Find the
area where the horse can graze. 3
9. In fig. two concentric circles with centre O, have radii O
21 cm and 42 cm. If AOB = 60° find the area of the
22 60°
shaded region. (Use  = ) 3
7 A B

SECTION-D
10. A chord AB of a circle of radius 10 cm makes a right angle at the centre of the
circle. Find the area of the minor and major segments. (Use  = 3.14) 4

188 Mathematics-X