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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 cm^{2}

Thus, the area of the circle 38.5 cm^{2}

**(ii)** Circle whose radius = 14 cm

Now, Area of the circle,

= 22/7 × 14 × 14

= 616 cm^{2}

Hence, the area of the circle will be 616 cm^{2}

**(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 cm^{2}

Hence, the area of the circle will be 6.16 cm^{2}

**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 m^{2}

**∴** The horse can graze a area of 2464 m^{2}

**Question no – (3) **

**Solution : **

**(i)** Let, radius = ‘r’

πr^{2} = 154

= r^{2} = 154 × 7/22

= r^{2} = 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

**∴ **πr^{2} = 1386

= r^{2} = 1386 × 7/22

= r^{2} = 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

**∴ **πr^{2} = 24.64

= r^{2} = 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** = πr^{2}

= 22/7 × 7 × 7

= 154 cm^{2}

Therefore, the area of the circle will be **154 cm**^{2}

**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 cm^{2}

Now, from question,

4s = 264

= s = 264/4 = 66 cm

**∴**** Area of square,**

= 4 × (66)^{2}

= 4356 cm^{2}

**∴**** 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} cm^{2}

= 156.25 cm^{2}

**∴**** Area of circular disc** = πr^{2}

= 22/7 × 7/2 × 7/2

= 38.5 cm^{2}

**∴**** Area of remaining part,**

= (156.25 – 38.5)

= 117.75 cm^{2}

**Now, Total weight of the remaining port in,**

= 117.75 × 0.8

= 94.2 gm

**Question no – (9) **

**Solution :**

Let, circumference of 1^{st} circle 2x and 2^{nd} circle 3x

**∴** Radius of 1^{st} circle = 2x/2π = x/π

**∴** Radius of 2^{nd} circle = 3x/2π

**∴** Area of 1^{st} circle = πr^{2}

= π × (x/ π)^{2}

= πx^{2}/ π^{2}

= x^{2}/ π

and, Area of 2^{nd} circle

= πr_{2}^{2}

= π × (3x/2π)^{2}

= 9x^{2}/4π

**∴**** Ratio,**

= x^{2}/π : 9x^{2}/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 m^{2}

**∴** and radius of each quadrant at the corner of the park = 14m

**∴**** Area of one quadrant**

= 1/4 πr^{2}

= 1/4 × 22/7 × 14 × 14

= 154 m^{2}

**∴**** Area of 4**^{th}** quadrant**

= 4 × 154

= 616 m2

**∴**** Area of remaining portion of the park, **

= (10000 – 616)

= 9384 m^{2}

Thus, the area of the remaining part of the park will be 9384 m^{2}

**Question no – (11) **

**Solution : **

Here, Radius of circular,

field = 20 m

width = 5m

**∴** Inner radius = (20 – 5)

= 15 m

Now, Area path = πr^{2} – πr^{2}

= π (R^{2} – r^{2})

= 22/7 [(20)^{2} – (15)^{2}] m^{2}

= 22/7 (20 + 15) (20 – 15)

= 22/7 × 35 × 5

= 550 m^{2}

Therefore, the area of the path will be 550 m^{2}

**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 (R^{2} – r^{2})

= π [(10.5)^{2} – (7)^{2}]

= π (10.5 + 7) (10.5 – 7)

= 22/7 × 17.5 × 3.5

= 192.5 m^{2}

**∴**** 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 (R^{2} – r^{2})

= π/2 [(21)^{2} – (17.5)^{2}]

= 22/7 × 2 [(21 + 17.5) (21 – 17.5)]

= 11/7 × 38.5 × 3.5

= 211.75 dm^{2}

Thus, the area of the flower bed will be 211.75 dm^{2}

**Question no – (15) **

**Solution : **

Here, inner radius

= 88 × 7/2 × 22

= 14 cm

Let, outer radius = R

**∴**** Area of shaded portion, **

= π (R^{3} – r^{2})

= 346.5 = π (R^{2} – 14^{2})

= π (R^{2} – 196)

= R^{2} = 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 m^{2}

Therefore, the area of the square plot will be 21387 m^{2}

**Next Chapter Solution : **

👉 Chapter 18 👈