For students of class 10 facing their first real board exam, it is imperative that they learn all the topics of the syllabus properly. NCERT Solutions for class 10 maths chapter Areas Related to Circles pdf is intended to help students to find the areas of circles which are in correlation with a circle. Here we have provided solutions for class 10 maths chapter areas related to circles. NCERT Solutions for class 10 maths chapter 12 Areas Related to Circles will helps students to score well in their class 10 board exams.
NCERT Solutions Class 10 Maths Chapter 12 Exercises
1. If perimeter of a circle is equal to that of a square, What would be the ratio of their areas.
Here, perimeter of circle =
∴ Ratio of their areas =
2. Find the areas of circle that can be inscribed in a square of side 8 cm.
Diameter of the circle = Side of the square
∴ Diameter = 8 cm
∴ Area of the circle =
3. If sum of the areas of two circles with radii
Since sum of areas of two circles with radii
4. If sum of the circumference of two circles with radii radii
Since sum of circumference of two circles with radii
5. If the circumference of a circle and the perimeter of a square are equal, then write the relation between their radii.
Since sum of circumference of a circle = Perimeter of a square
∴ Area of circle =
Area of square =
6. To build a single circular park equal in area to the sum of areas of two circular parks of diameter 8 m and 6 m in a locality. What would be the radius of new park?
Here, area of single park = Area of the park with (d= 8m) + Area of the park with (d= 6m)
7. Find the area of a square that can be inscribed in a circle of radius 6 cm.
Since square is inscribed in a circle
Now, area of the square =
8. Find the radius of a circle whose circumference is equal to the sum of the circumference of two circles of diameter 24 cm and 16 cm.
From the above, we obtain
9. Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 20 cm and 15 cm.
From the above, we obtain
10. Find the area of the largest triangle that can be inscribed in a semicircle of radius r.
Here, among all altitudes, radius is the largest altitude
Hence, area of the largest triangle inscribed in a semicircle r is
11. In the given figure , a square of diagonal 14 cm is inscribed in a circle. Find the area of the shaded region.
Here, diagonal of the square = 14 cm
∴ Area of the square =
Diameter of the circle = 14 cm
∴ Radius of the circle = 7 cm.
Area of the circle =
Thus, area of the shaded region = 154 – 98
= 56 sq.cm
12. The wheel of a motor cycle is of radius 28cm. How many revolutions per minitue must the wheel make so as to keep a speed of 60 Km/h?
Here, speed of the motor cycle = 60 Km/h
= 1000 m/min
Radius of the wheel (r) = 28 cm =
Circumference of the wheel =
= 1.76 m
Number of revolutions per minute =
13. A cow is tied with a rope of length 21 m at the corner of a rectangular field of dimensions 27 m
Length of the rope = 21 m
∴ Area of the field in which the cow can graze
14. In the given fig, arcs are drawn by taking vertices P,Q and R of an equilateral triangle of side 12 cm to intersect the sides QR, RR and PQ at their respective mid-points X, Y and Z. Find the area of the shaded region.
Side of an equilateral
Since, arcs are drawn by taking vertices P,Q and R to intersect the sides QR, RR and PQ at their mid-points X, Y and Z.
∴ Radii of each sector (r) = 6 cm
Required area of the shaded region
15. In the given fig, arcs have been drawn with radii 7 cm each and with centers A, B and C.
Find the area of the shaded region.
Here, radii of each sector at A, B and C is 7 cm
Required area of shaded region
16. A circular park is surrounded by a road 20 m wide. If the radius of the park is 100 m, find the area of the road.
Here, radius of the circular park (r) = 100 m
Width of the road = 20 m
∴ Radius of the outer circle (R) = 100 + 20 = 120 m
Required area of road
17. A piece of wire 22 cm long is bent into the form of an arc of a circle subtending an angle of
Here, length of wire (arc of required circle) = 22 cm
And central angle(
Let r be the radius of the required circle.
18. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand during the time period 6:05 A.M and 6:40 A.M.
Angle swept by the minute hand in one minute =
∴Angle swept by the minute hand in 35 minute =
Length of the minute hand = 7 cm
Area swept by the minute hand =
19. Find the area of the sector of a circle of radius 7 cm, if the corresponding arc length is 4 cm.
Radius of the circle (r) = 7 cm
Length of the arc = 3.5 cm
Now, area of the corresponding arc
20. A circular pond is of diameter 20 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs.30 per
Radius of the circular pond r =
Width of the path = 2 m
∴ Radius of the outer circle (R) = 10 + 2 = 12 m
Area of the path =
Cost of constructing the path @ Rs.30 per
21. Find the number of revolutions made by a circular tier of area 1.32
Area of a circular wheel = 1.32
Distance traveled in one rotation =
= 4.07 m
Total distance to be travelled = 165
∴Number of rotations =
22. Find the difference of the areas of sectors of angle
Radius of the circle (r) = 14 cm
Angle of the sector
∴ Area of the sector =
Area of major sector
∴ The required distance = 308 cm²
23. Find the area of the shaded region in the give figure.
Here, unshaded region is combination of two semicircles and one rectangle.
∴ Area of unshaded region = Area of rectangle with dimensions (26-3-3-) cm by (12-4-4) cm + 2 area of semicircle with radii
Required area of shaded region
= 312 – 64 – 4
= (248 – 4
24. Find the area of the minor segment of a circle of radius 28 cm, when the angle of the corresponding sector is
Radius of the circle (r) = 28 cm
Central angle (
Area of minor segment
= Area of sector PRQ – Area of
= 616 – 392
25. The diameter of front and rear wheels of a tractor 70 cm and 2 m respectively. Find the number of rotations that rear wheel will make in covering a distance in which the wheel makes 1200 rotations.
Radius of the front wheel (r) = 35 cm
Number of rotations = 1200
∴ Total distance covered by front wheel = 1200 2
= 264000 cm
Radius of the rear wheel (R) =
Distance covered by the rear wheel in one rotation =
26. Number of rotations =
Length of the rope = 7 m
Sides of the triangular field are AB = 16 m, BC = 17 m and AC = 15 m
Let cow, buffalo and horse be tied at vertices A, B and C respectively.
∴ Area of
Hence, area of the field which cannot be grazed by three animals =
27. Find the area of the flower bed ( with semi- circular ends) as shown in the figure.
Here, the flower bed is a combination of two semicircular ends and rectangle.
Radius of semicircular ends = 7 cm
Length and breadth of rectangular field are 35 cm and 8 cm
∴ Area of the shaded region
= 154 + 280
Hence, area of the flower bed is 434
28. On a square cardboard sheet of area 196
Area of the square PQRS = 196
Side of the square =
∴ Diameter of each circular plate = 7 cm
And radius of each circular plate = 3.5 cm
Area of four circular plates =
Area of the square sheet not covered by the circular plate = 196 – 154 = 42
29. The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of two sectors as well as the lengths of those corresponding arcs.
Here, radii of two circles are
Central angles are
∴ Areas of two sectors are
Lengths of corresponding arcs are
Arc lengths of two sectors of two different circles may be equal (14.67 cm), but their areas need not be equal.
30. Floor of a room is of dimension 8 m × 5 m and it is covered with circular tiles of diameters 40 cm each as shown in figure. Find the area of floor that remains uncovered with tiles.
Radii of each circular tile = 20 cm
∴ Number of circular tiles along its length = 20
Number of circular tiles along its breadth = 12.5
Total number of circular tiles =20× 12.5 = 250
Area of each circular tile =3.14×20×20
= 1256 cm²
Area of 250 circular tile = 250 × 1256
= 314000 cm²
Area of the rectangle floor = 800 × 500
= 400000 cm²
Thus, area of the floor that remains uncovered with tiles = 400000 – 314000
= 86000 cm²
= 8.6 cm²