CBSE Class 10 Maths Chapter-12 Areas Related to Circles – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 12 – Areas Related to Circles prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-12 Areas Related to Circles Important Questions
CBSE Class 10 Maths Important Questions Chapter 12 – Areas Related to Circles
2 Mark Questions
1. 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.
Ans. Let R be the radius of the circle which has area equal to the sum of areas of the two circles, then
According to the question,
R2 = 64 + 36
R2 = 100
R = 10 cm
2. Figure 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 the five scoring regions.
Ans. Gold: Diameter = 21 cm
Radius = 
cm
Area of gold scoring region = 
= 
= 346.5 cm2
Red: Area of red scoring region = 
= 
= 
= 1386 – 346.5 = 1039.5 cm2
Blue: Area of blue scoring region = 
= 
= 
= 3118.5 – 1386 = 1732.5 cm2
Black: Area of black scoring region = 
= 
= 
= 5544 – 3118.5 = 2425.5 cm2
White: Area of white scoring region = 
= 
= 
= 8662.5 – 5544 = 3118.5 cm2
3. 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?
Ans. Diameter of wheel = 80 cm
Radius of wheel 
= 40 cm
Distance covered by wheel in one revolution = 
= 
= 
cm
Distance covered by wheel in 1 hour = 66 km = 66000 m = 6600000 cm
Distance covered by wheel in 10 minutes = 
= 1100000 cm
No. of revolutions = 
= 
= 4375
4. Tick the correct answer in the following and justify your choice: If the perimeter and area of a circle are numerically equal, then the radius of the circle is:
(A) 2 units
(B) 
units
(C) 4 units
(D) 7 units
Ans. (A) Circumference = Area
= 2 units
Unless stated otherwise, take 
5. Find the area of a sector of a circle with radius 6 cm, if angle of the sector is 
Ans. Here, = 6 cm and 
Area of sector = 
= 
= 
cm2
6. Find the area of a quadrant of a circle whose circumference is 22 cm.
Ans. Given, = 22 
cm
We know that for quadrant of circle, 
Area of quadrant =
= 
= 
cm2
7. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Ans. Here, = 14 cm and 
Area swept =
= 
= 
cm2
8. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment, (ii) major segment. 
Ans. i) Here, = 10 cm and
Area of minor sector =
= 
= 78.5 cm2
Area of 
OAB = 
x Base x Height
= 
= 50 cm2
Area of minor segment = Area of minor sector – Area of OAB
= 78.5 – 50 = 28.5 cm2
(ii) For major sector, radius = 10 cm and 
Area of major sector = = 
= 235.5 cm2
9. 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 (see figure). 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 cm.
Ans. (i) Area of quadrant with 5 m rope = = 
= 19.625 m2
(ii) Area of quadrant with 10 m rope = = = 78.5 m2
The increase in grazing area = 78.5 – 19.625 = 58.875 m2
10. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Ans. (i) Diameter = 35 mm
Radius = 
mm
Circumference = = 
= 110 mm ……….(i)
Length of 5 diameters = 35 x 5 = 175 mm ……….(ii)
Total length of the silver wire required = 110 + 175 = 285 mm
(ii) 
mm and 
The area of each sector of the brooch =
= 
= 
mm2
11. An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Ans. Here, = 45 cm and 
Area between two consecutive ribs of the umbrella =
= 
= 
cm2
12. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 
Find the total area cleaned at each sweep of the blades.
Ans. Here, = 25 cm and 
The total area cleaned at each sweep of the blades = 
= 
= 
cm2
13. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 
to a distance of 16.5 km. Find the area of the sea over which the ships are warned.
Ans. Here, = 16.5 km and 
The area of sea over which the ships are warned =
= 
= 189.97 km2
14. Tick the correct answer in the following:
Area of a sector of angle 
(in degrees) of a circle with radius R is:
(A) 
(B) 
(C) 
(D) 
Ans. (D) Given, = R and 
Area of sector = = 
= 
= 
15. Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and 
AOC = 
Ans. Area of shaded region = Area of sector OAC – Area of sector OBD
= 
= 
= 
= 
= cm2
16. Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Ans. Area of shaded region
= Area of square ABCD – (Area of semicircle APD + Area of semicircle BPC)
= 
= 
= 196 – 154 = 42 cm2
17. Find the area of the shaded region in figure, 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.
Ans. Area of shaded region
= Area of circle + Area of equilateral triangle OAB – Area common to the circle and the triangle
= 
= 
= 
= 
= 
cm2
18. 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 figure. Find the area of the remaining portion of the figure.
Ans. Area of remaining portion of the square
= Area of square – (4 x Area of a quadrant + Area of a circle)
= 
= 
= 
cm2
19. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region).
Ans. Area of design = Area of circular table cover – Area of the equilateral triangle ABC
= 
………(i)
G is the centroid of the equilateral triangle.
radius of the circumscribed circle = 
cm
According to the question, 
= 48 cm
Now, 
= 3072
Required area = 
[From eq. (i)]
= 
= 
cm2
20. In figure ABCD is a square of side 14 cm. With centers 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.
Ans. Area of shaded region = Area of square – 4 x Area of sector
= 
= = 196 – 154 = 42 cm2
21. Figure depicts a racing track whose left and right ends are semicircular.
Ans. (i) Distance around the track along its inner edge
= 
= 
= 
= 
m
(ii) Area of track = 
= 
= 
= 4320 m2
22. In figure, 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.
Ans. Area of shaded region = Area of circle + Area of semicircle ACB – Area of ACB
= 
= 
= 
= 66.5 cm2
23. 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 figure). Find the area of the shaded region. 
Ans. Area of equilateral triangle = 
= 17320.5
cm
Area of shaded region = Area of ABC – 
24. On a square handkerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.
Ans. Area of remaining portion of handkerchief = Area of square ABCD – Area of 9 circular designs
= 
= 1764 – 1386 = 378 cm2
25. In 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:
(i) quadrant OACB
(ii) shaded region
Ans. (i) Area of quadrant OACB = 
= 
= cm2
(ii) Area of shaded region = Area of quadrant OACB – Area of OBD
= 
= 
= 
= 
cm2
26. In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.
Ans. OB = 
= 
= 
OA = 
cm
Area of shaded region = Area of quadrant OPBQ – Area of square OABC
= 
= 
= 200 x 3.14 – 400
= 228 cm2
27. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If AOB = 
find the area of the shaded region.
Ans. Area of shaded region = Area of sector OAB – Area of sector OCD
= 
= 
= 
= 
= 
= 
cm2
28. In figure, 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.
Ans. In right triangle BAC, BC2 = AB2 + AC2[Pythagoras theorem]
BC2 = (14)2 + (14)2 = 2(14)2 BC = 
cm
Radius of the semicircle = 
cm
Required area = Area BPCQB
= Area BCQB – Area BCPB
= Area BCQB – (Area BACPB – Area BAC)
= 
= 
= 154 – (154 – 98) = 98 cm2
29. Calculate the area of the designed region in figure common between the two quadrants of circles of radius 8 cm each.
Ans. In right triangle ADC, AC2 = AD2 + CD2 [Pythagoras theorem]
AC2 = (8)2 + (8)2 = 2(8)2
AC = 
= 
cm
Draw BM
AC.
Then AM = MC = AC
= 
= 
cm
In right triangle AMB,
AB2 = AM2 + BM2 [Pythagoras theorem]
BM2 = 64 – 32 = 32
BM = cm
Area of ABC = x AC x BM
= 
= 32 cm2
Half Area of shaded region = 
= 
= 
cm2
Area of designed region = 2 x = 
cm2
30. Find the circumference of a circle of diameter 14cm.
Ans. 
Circumference 
31. The diameter of a circular pond is 17.5m. It is surrounded by a path of width 3.5m. Find the area of the path.
Ans. Diameter of pond = 17.5cm
R = 8.75m
Width = 3.5m
Outer radius = R = 8.75 + 3.5 = 12.25m
Now,
Area of path 
32. Find the area of a sector of a circle with radius 6cm, if angle of the sector is 60
.
Ans. Diameter of pond = 17.5cm
R = 8.75m
Width = 3.5m
Outer radius = R = 8.75 + 3.5 = 12.25m
Now,
Area of path 
33. Find the area of a quadrant of a circle whose circumference is 22 cm.
Ans. Circumference = 22 cm
2πr = 22
Quadrant of circle will subtend 90° angle at the centre of the circle.
Area of such quadrant of the circle
34. Find the area of the shaded region where ABCD is a square of side 14cm.
Ans. Area of square 
Diameter of each circle 
Radius of each circle 
Area of each circle = 
Area of 4 circles = 
Area of shaded region = Area of square – Area of 4 circles
= 196 cm2 – 154 cm2 = 42 cm2
35. The radius of a radius of a circle is 20cm. Three more concentric circles are drawn inside it in such a manner that it is divided into four parts of equal area. Find the radius of the largest of the three concentric circles.
Ans. Let r be the radius of the largest of the three circles
Area of largest circle 
[area of given circle]
36. OACB is a quadrant of a circle with centre O and radius 7 cm. If OD = 4cm, then find area of shaded region.
Ans. Area of quadrant 
Area of shaded region = Area of quadrant 
37. A pendulum swings through on angle of 
and describes an arc 8.8cm in length. Find the length of pendulum.
Ans. Let r be the length of pendulum
38. The cost of fencing a circular field at the rate of Rs. 24 per metre is Rs. 5280. The field is to be ploughed at the rate of Rs.0.50per m2. Find the cost of ploughing the field.
(Take 
)
Ans. Since for Rs. 24, the length of fencing = 1metre
for Rs.5280, the length fencing 
meters
Perimeter i.e., circumference of the field = 220meters
Let r be the radius of the field
Area of the field 
Rate = Rs.0.50 per m2
Total cost of ploughing the field = Rs. 
39. Find the difference between the area of regular hexagonal plot each of whose side 72m and the area of the circular swimming take in scribed in it. (Take )
Ans. Side of hexagonal plot = 72m
Area of equilateral triangle 
Area of hexagonal plot = 
Area of triangle OAB
Area of circular region = 
Difference = 13468 m2 – 12081 m2 = 1385 m2
40. In the given figure areas have been drawn of radius 21cm each with vertices A,B,C and D of quadrilateral ABCD as centers. Find the area of the shaded region.
Ans. Required area = Area of the circle with radius 21
41. Find the area of a sector of a circle with radius 6cm, if angle of the sector is 
.
Ans. We know that Area of sector = 
Required area 
42. A wheel has diameter 84cm, find how many complete revolutions it must make to complete 792 meters.
Ans. Diameter, 2r = 84cm
Distance covered in one revolution= circumference = 
Thus, for distance covered 264cm, number of revolution = 1
For distance covered 792 metres = 79200cm
Number of revolutions = 
43. The given figure is a sector of a circle of radius 10.5cm. Find the perimeter of the sector. (Take 
)
Ans. We know that circumference i.e, perimeter of a sector of angle Pof a circle with radius R
Required perimeter
44. A car had two wipers which do not overlap. Each wiper has a blade of length 25cm sweeping through an angle of 115. Find the total area cleaned at each sweep of the blades.
Ans. Radius of each wiper = 25cm, Angle = 115
Total area cleaned at each sweep of the blades
= 
45. In the given figure arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaped region.
Ans. Area of shaded region
= 
46. The radii of two circles are 19cm and 9cm respectively. Find the radius of the circle which has its circumference equal to the sum of the circumference of the two circles
Ans. C1 = circumference of the 1st circle = 2
(19) = 38cm
C2 = circumference of the 2nd circle = 2(19) = 18cm
47. A car travels 0.99km distance in which each wheel makes 450 complete revolutions. Find the radius of its wheel
Ans. Distance travelled by a wheel in 450 complete revolutions = 0.99 km = 990m
Distance travelled in one revolution = 
Let r be the radius of the wheel
48. A sector is cut from a circle of diameter 21cm. If the angle of the sector is 150find its area.
Ans. We have,
Diameter = 21cm 
radius = 
cm
Angle of sector = 
Area of the sector 
49. In the given figure AOBCA represent a quadrant of area 9.625
. Calculate the area of the shaded portion.
Ans. Required area = Area of quadrant OAB – Areaof 
