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OP Malhotra Class 9 ICSE Maths Solutions Chapter 17 Circle Circumference and Area
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students for the Solutions of OP Malhotra Class 9 ICSE Math Book, Chapter 17, Circle-Circumference and Area. Here students can easily find step by step solutions of all the problems for Circle-Circumference and Area, Exercise 17a and 17b Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 17 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Circle-Circumference and Area Exercise 17(a) Solution :
Question no – (1)
Solution :
(i) Given, Circles diameters are 49 cm
∴ Circumference,
= (πd) = 22/7 × 49
= 154 cm
Thus, the circumference of the circles will be 154 cm.
(ii) Given, Circles diameters are = 14 cm.
∴ Circumference,
= (πd) = 22/7 × 14
= 44 cm
Thus, the circumference of the circles will be 44 cm
(iii) Given, Circles whose diameter is = 9.8 cm.
Circumference,
= (πd)= (22/7 × 9.8) cm
= 30.8 cm
Thus, the circumference of the circle will be 30.8 cm
(iv) Circle whose diameter is 7 cm.
∴ Circumference,
= (πd) = (22/7 × 7) cm
= 22 cm
Thus, the circumference of the circle will be 22 cm.
Question no – (2)
Solution :
(i) Circle whose radii is 7 cm.
∴ Circumference,
= 2 π r = 2 × 22/7 × 7
= 44 cm
Thus, the circumference of the circle will be 44 cm.
(ii) Circle whose radii is 28 cm.
∴ Circumference,
= 2 π r = (2 × 22/7 × 28) cm
= 176 cm
Thus, the circumference of the circle will be 176 cm.
(iii) Circle whose radii is = 3.5 cm.
Now, Circumference
= 2 π r = (2 × 22/7 × 3.5) m
= 22 cm
Thus, the circumference of the circle will be 22 cm.
(iv) Circle whose radii is = 98 m
Now, Circumference,
= 2 π r
= (2 × 22/7 × 98)m
= 616 m
Therefore, the circumference of the circle will be 616 m.
Question no – (3)
Solution :
(i) Radius = 4.5 cm
∴ Circumference,
= (2 × 22/7 × 4.5) cm
= 28.26 cm
Therefore, the circumference of the circle will be 28.26 cm.
(ii) Diameter = 15 cm
∴ Circumference,
= (3.14 × 15)
= 47.1 cm
Hence, circumference of the circle will be 47.1 cm
Question no – (4)
Solution :
(i) Circle whose circumference is = 22 cm.
∴ Length of diameter,
= c/π = 22 × 7/22
= 7 cm
Therefore, the length of the diameter will be 7 cm.
(ii) Circle whose circumference is = 8.8 cm.
Now, Diameter,
= c/π
= 8.8/22/7
= 8.8 × 7/22
= 2.8 cm
Hence, the length of the diameter will be 2.8 cm.
(iii) Circle whose circumference is = 44 cm.
Now, Diameter,
= c/π
= 44 × 7/22
= 14 cm
Therefore, the length of the diameter will be 14 cm.
Question no – (5)
Solution :
(i) Circle whose circumference is = 11 cm.
We know, C = 2 π r
∴ r = C/2π
= 11/2 × 3 . 14
= 1.75
Therefore, the length of the radii will be 1.75 cm
(ii) Circle whose circumference is = 55 cm.
We know, C = 2 π r
∴ Radius (r)
= 55/2 × 3 .14
= 8 .76 cm
Hence, the length of the radii will be 8.76 cm.
(iii) Circle whose circumference is = 13.2 m.
As we know,
= C = 2 π r
= r = C/2 π
∴ Radius (r)
= 13.2/2× 3.14
= 2.1
Hence, the length of the radii will be 2.1 m.
(iv) Circle whose circumference is = 440 m.
As we know that,
= C = 2 π r
= r = C/2 π
∴ Radius (r)
= 440/2 × 3.14
= 70.06
Thus, the length of the radii will be 70.06 m.
Question no – (6)
Solution :
(i) Circumference is = 100 m.
∴ Radius,
= (100/2 × 3.14) cm
= 15.9 m
Thus, the radii of the circle will be 15.9 m.
(ii) Circumference is = 6.4 dm.
∴ Radius (r)
= 6.4/2 × 3 .14
= 1.02 dm
Thus, the radii of the circle will be 1.02 dm.
Question no – (7)
Solution :
As per the given question,
Diameter of Venus planet is = 12,278 km
Circumference = (π) d
= 22/7 × 12.278
= 38588 km
Therefore, the circumference of Venus planet will be 38588 km.
Question no – (8)
Solution :
(i) According to the given question,
Semi-circular plate of radius 3.85 cm.
∴ Perimeter of semi-circular plate,
= π r + 2r
= r (ii + 2)
= 3.85 (3.14 + 2)
= 3.85 π 5.14
= 19.8 cm
Therefore, the perimeter of semi-circular plate will be 19.8 cm.
(ii) Let, (r) be the radius,
∴ 2 π r – 2r = 16.8
= 2r (π – 1) = 16.8
= r 16.8/(22/7 – 1) × 2
= 3.92 cm
Therefore, the radius of the circle will be 3.92 cm.
Question no – (9)
Solution :
Here, circumference (c)
= 2 π r
= 2 × 22/7 × 42
= 264 cm
So, perimeter of square = 264 cm
∴ Side of the square,
= 264/4
= 66 cm
Hence, the side of the square will be 66 cm.
Question no – (10)
Solution :
In the given question,
The radius is = 3.5 cm
Perimeter of quarter,
= 2 π ×1/4 + 2r
= π r/2 + 2r
= 22/7 × 1/2 × 3.5 + 2 × 3.5
= 12.5 cm
Thus, the perimeter of the quarter of the circle will be 12.5 cm.
Question no – (11)
Solution :
(i) Radius (R),
= 440 × 7/2 × 22
= 70 m
∴ Width of truck = 14 m
∴ Radius of order circle,
= (70 + 14)
= 84 m
∴ Diameter,
= (2 × 84) m
= 168 m
Hence, the diameter of the outer circle of the track will be 168 m.
Question no – (12)
Solution :
According to the given question,
A garden roller has a circumference of 3 metres.
revolutions need to moving a distance of 21 metres = ?
Number of revolution will be,
= 21/3
= 7
Therefore, 7 revolutions will be needed.
Question no – (13)
Solution :
Circumference (c) = (π) d
= (22/7 × 7)
= 220 cm
Distance travel in 25 revolution,
= (25 × 220)
= 5600 cm
= 55 m
Now, distance travelled in 1 hr,
= 55/10 × 60 × 60
= 19800 m
= 19.8 km
= 19.8 km
∴ Speed = 19.8 km
Therefore, the bicycle travelling in 19.8 km/h
Question no – (14)
Solution :
Circumference = π d
= 22/7 × 91
= 286 cm
∴ Distance travel in 1 h = 10 km
∴ Distance travelled 1 m,
= 60/60 km
= 10 × 1000/60
= 500/3
∴ No, of revolution in 1 min,
= 500/3 × 100/286
= 58 188/429
Therefore, the number of revolutions made by each wheel per minute will be 58 188/429
Question no – (15)
Solution :
First, Circumference of field = 2 π r
= 2 × 22/7 × 35
= 220 m
Now, Distance travelled = (4 × 220)
= 880 m
∴ Time = 880/5 km/h
= 880 × 60/5000
= 10.56 min
= 10 min 34 sec
Hence, a boy who circles a circular field four times will take 10 minutes and 34 seconds.
Question no – (16)
Solution :
First, Circumference = π d
= (22/7 × 84)
= 264 cm
Now, Distance cover 1 sec
= (4 × 264)
1056 cm
∴ Cover in 1 hr,
= 1056 × 60 × 60/100 × 1000
= 38.016 km
Therefore, Speed is 38.016 km/h
Question no – (17)
Solution :
Here, circumference of two semicircle,
= 2 × 1/12 π d
= π d
= 22/7 × 70
= 220 m
Now, Length of 1 complete round,
= 220 + 100 + 100
= 420 m
Therefore, athlete will run 420 m.
Circle-Circumference and Area Exercise 17(b) Solution :
Question no – (1)
Solution :
(i) Given in the question,
diameter = 7 cm.
∴ Radius = 7/2 cm
Now, Area,
= 22/7 × 7/2 × 7/2
= 38.5 cm2
Thus, the area of the circle 38.5 cm2
(ii) Circle whose radius = 14 cm
Now, Area of the circle,
= 22/7 × 14 × 14
= 616 cm2
Hence, the area of the circle will be 616 cm2
(iii) diameter = 2.8 cm
∴ Radius = 2.8/2 = 1.4 cm
Now, Area of the circle,
= 22/7 × 1.4 × 1.4
= 6.16 cm2
Hence, the area of the circle will be 6.16 cm2
Question no – (2)
Solution :
As per the question,
A horse is tied to a pole with 28 m long string.
∴ Radius = 28 m
Now, Area horse graze,
= 22/7 × 28 × 28
= 2464 m2
∴ The horse can graze a area of 2464 m2
Question no – (3)
Solution :
(i) Let, radius = ‘r’
πr2 = 154
= r2 = 154 × 7/22
= r2 = 49
= r = √49
= r = 7 cm
Hence, the radius of the circular field will be 7 cm.
(ii) Circular field whose area is 1386 cm²
Let, radius of field = r
∴ πr2 = 1386
= r2 = 1386 × 7/22
= r2 = 441
= r = 21 cm (Radius)
Hence, the radius of a circular field will be 21 cm.
Question no – (4)
Solution :
Given, The area of a circle is 24.64 cm²
Let, ‘r’ be the radius
∴ πr2 = 24.64
= r2 = 24.64 × 7/22
= 7.84
= r = 2.8 cm
Now, Circumference
= 2πr
= 2 × 22/7 × 2.8
= 17.6 cm
Therefore, circumference of the circle will be 17.6 cm.
Question no – (5)
Solution :
Here, side of square,
= √121
= 11 cm
∴ Perimeter of square (wire)
= 4 × 11
= 44 cm
∴ Radius (r) = 2πr/2π
= 44 × 7/2 × 22
= 7 cm
∴ Area = πr2
= 22/7 × 7 × 7
= 154 cm2
Therefore, the area of the circle will be 154 cm2
Question no – (6)
Solution :
The area of the square is 484 sq m
∴ Side of square (r),
= √484
= 22 m
∴ Perimeter of square,
= 4 × 22
= 88 m
= Circumference
Now, Radius,
= 28 × 7/2 × 22
= 14 m
∴ Area of circle,
= 22/7 × 14 × 14
= 616.54 sq.m
Therefore, the area of the circle will be 616.54 sq.m.
Question no – (7)
Solution :
Let, side ‘s’ and radius ‘r’
Circumference = 2πr
= 2 × 22/7 × 42
= 264 cm
∴ Area = πr2
= 22/7 × 42 × 42
= 5544 cm2
Now, from question,
4s = 264
= s = 264/4 = 66 cm
∴ Area of square,
= 4 × (66)2
= 4356 cm2
∴ Ratio,
= 5644 : 4356
= 14 : 11
Question no – (8)
Solution :
As per the given question,
Radius (r) = 7/2 cm
Side of square = 12.5 cm
∴ Area = (12.5)2 cm2
= 156.25 cm2
∴ Area of circular disc = πr2
= 22/7 × 7/2 × 7/2
= 38.5 cm2
∴ Area of remaining part,
= (156.25 – 38.5)
= 117.75 cm2
Now, Total weight of the remaining port in,
= 117.75 × 0.8
= 94.2 gm
Question no – (9)
Solution :
Let, circumference of 1st circle 2x and 2nd circle 3x
∴ Radius of 1st circle = 2x/2π = x/π
∴ Radius of 2nd circle = 3x/2π
∴ Area of 1st circle = πr2
= π × (x/ π)2
= πx2/ π2
= x2/ π
and, Area of 2nd circle
= πr22
= π × (3x/2π)2
= 9x2/4π
∴ Ratio,
= x2/π : 9x2/4π
= 4 : 9
Therefore, the ratio of their areas will be 4 : 9
Question no – (10)
Solution :
Side of square park = 100 m
∴ Area = (100)2
= 10000 m2
∴ and radius of each quadrant at the corner of the park = 14m
∴ Area of one quadrant
= 1/4 πr2
= 1/4 × 22/7 × 14 × 14
= 154 m2
∴ Area of 4th quadrant
= 4 × 154
= 616 m2
∴ Area of remaining portion of the park,
= (10000 – 616)
= 9384 m2
Thus, the area of the remaining part of the park will be 9384 m2
Question no – (11)
Solution :
Here, Radius of circular,
field = 20 m
width = 5m
∴ Inner radius = (20 – 5)
= 15 m
Now, Area path = πr2 – πr2
= π (R2 – r2)
= 22/7 [(20)2 – (15)2] m2
= 22/7 (20 + 15) (20 – 15)
= 22/7 × 35 × 5
= 550 m2
Therefore, the area of the path will be 550 m2
Question no – (12)
Solution :
Here, circumference of a circular plot = 44 m
∴ Inner Radius,
= 44 × 7/2 × 22
= 7m
∴ Outer radius,
= (7 + 3.5) m
= 10.5 m
Area of older road,
= π (R2 – r2)
= π [(10.5)2 – (7)2]
= π (10.5 + 7) (10.5 – 7)
= 22/7 × 17.5 × 3.5
= 192.5 m2
∴ Total cost,
= (192.5 × 10)
= 1925 Rs
Therefore, the cost of paving the road will be 1925 Rs.
Question no – (13)
Solution :
According to the given question,
Radius = 35/2 = 17.5 dm
width of flower bed = 3.5 dm
∴ Outer radius will be,
= (17.5 + 3.5)
= 21 dm
Now, Area of bed,
= π/2 (R2 – r2)
= π/2 [(21)2 – (17.5)2]
= 22/7 × 2 [(21 + 17.5) (21 – 17.5)]
= 11/7 × 38.5 × 3.5
= 211.75 dm2
Thus, the area of the flower bed will be 211.75 dm2
Question no – (15)
Solution :
Here, inner radius
= 88 × 7/2 × 22
= 14 cm
Let, outer radius = R
∴ Area of shaded portion,
= π (R3 – r2)
= 346.5 = π (R2 – 142)
= π (R2 – 196)
= R2 = 346.5 × 7/22 + 196
= 110.25 + 196
= 306.25
∴ R = √306.25
= 17.5
Hence, the radius of the outer circle will be 17.5 cm.
Question no – (17)
Solution :
Diameter of field
= 650/ π
= 650 × 7/22
Square Plot has its vertices on the circumference as the circle
Diagonal of square = diameter of the circle
= 650 × 7/22
∴ Area = 1/2 [650 × 7/22]2
= 1/2 × [4550/22]2
= 21387 m2
Therefore, the area of the square plot will be 21387 m2
Next Chapter Solution :
👉 Chapter 18 👈