Chapter – 16 : Circumference Of Circle | Chapter Solution Class 9

(i)  AC = 8 m, BC = DE = 5 m, CD = 4 m

AB = 8 m – 5 m = 3 m

∴ AE = \sqrt{\mathrm{AB}^2+\mathrm{BE}^2} \\

= \sqrt{(3)^2+(4)^2} \\

=\sqrt{9+16} \\

= \sqrt{25} = 5 m

∴  Perimeter of a semi-circle with a diameter of 4 m

= \frac{\pi \mathrm{d}}{2}(\because \mathrm{d} \text { is diameter of the circle }) \\

= \frac{22}{7} \times \frac{1}{2} \times 4=\frac{44}{7} cm .

Perimeter of the given figure

= (8+5+5+\frac{44}{7})=(18+\frac{44}{7}) \mathrm{m} . \\

= 24 \frac{2}{7} m

(ii) AB = 14 cm, AD = CD = 14 cm, BC = 14 cm

Radius of the circle (r) = 7 cm

The perimeter of a circle  =\frac{22}{7} \times 7 = 22 cm

∴ Perimeter of the given figure

= 14 cm + 14 cm + 14 cm + 22 cm

= 64 cm

Length of the wire = 2 πr

=2 \times \frac{22}{7}   \\ × 35 m

= 220 m

Circumference of the wheel = 2 πr

= 2 \times \frac{22}{7} × 0.35

= 2.2 m

One revolution = 2.2 m

∴ 450 revolution = 2.2 × 450 m = 990 m

In 1 minute, distance covered = 990 m

In 60 minutes (1 hour), distance covered =990 × 60m

= 59400 m = 59.4 km

∴ Velocity of the wheel = 59.4 km/h

Raduius of Circular field = 280 m

Circumference of the field = 2 πr

= 2 × \frac{22}{7} × 280

= 1760 m

Speed of chaitali = 5.5 km/h

= \frac{55 \times 1000}{10 \times 60 \times 60} m/s

= \frac{55}{36} m/s

Time = \text{distance}\over \text {speed}

= \text{1760}\over {\text {55} \over \text{33}}

= 1152 s or 19.2 m

Perimeter of rectangle = 2 (length + breadth)

= 2 (18 + 15) cm

= 2 × 33 cm = 66 cm

Let ‘ r ‘ be the radius of the copper wire.

Then, circumference = 66

or, 2πr = 66

or, 2 \times \frac{22}{7} \times r =66

or, \frac{2 r}{7}=3

or, 2r = 21

or, r = 10.5 cm

∴ The radius of circular copper wire = 10.5 cm

Let ‘ r ‘ be the radius of the semicircular field.

Then, Perimeter = 108 m

or, \pi \mathrm{r}+2 \mathrm{r}=108

or, r(\pi+2)=108

or, r(\frac{22}{7}+2)=108

or, r(\frac{22+14}{7})=108

or, r \times \frac{36}{7}=108

or, \frac{r}{7}=3

∴ r = 21

∴ diameter of the field = 2 r = 42 cm

Let ‘ r ‘ be the radius of the wheel.

2πr – 2r = 75

or, 2r (π – r) = 75

or, 2 r (\frac{22}{7}-1) = 75

or, 2 r (\frac{22-7}{7}) =75

or, 2 r (\frac{15}{7}) =75

or, 2 r = 35

or, r = (\frac{35}{2}) = 17.5

∴ The length of the radius of the wheel is 17.5 cm

Diameter of circular track = 56 m

Radius = \frac{56}{2} = 28 m

Circumference of the circular track = one revolution of the circular track = 2πr

= 2 × \frac{22}{7} × 28 m = 176 m

∴ Length of the race i.e, 10 revolution

= 176 × 10 = 1760 m

Jakir is one revolution behind

∴ Puja beats Jakir by 176 m

Perimeter of bore well = 44 cm.

or, 2πr = 44

or, Radius of bore well = \frac{440}{2 \pi} \\

= \frac{440 \times 7}{2 \times 22} = 70 cm

The perimeter of borewell with parapet = 616 cm

or, radius of borewell with parapet = \frac{616}{2 \pi} \\

=\frac{616 \times 7}{2 \times 22} = 98 cm

∴ width of the stone parapet = (98-70)cm = 28cm

Circumference of moter wheel = 2πR

=\frac{22}{7} \times 14 cm=44 cm

Circumference of machine wheel = 2πr

=\frac{22}{7} \times 94.5 cm=297 cm

If the motor’s wheel revolves 27 times in a second.

The length covered by Motor wheel = 44 \times 27 =1188 cm.

No. of revolution covered in 1 second by Machine Wheel

=\frac{1188}{297} = 4

No. of revolution covered in 1 hour by Machine Wheel

= 4 \times 3600

=14400

Hours’s hand goes in 12 hours = 1 rev

= 2πr

= 2 × \frac{22}{7} × 8.4 cm

= 52.8 cm.

∴ Hours’s hand goes in 12 hours = 1 rev = 105.6 cm

Minute’s hand goes in one hour = 1 rev

= 2πR

= 2 \times \frac{22}{7} × 14 cm

= 88 cm.

∴ Minute’s hand goes in 24 hours

= 24 × 88

= 2112 cm

Let \mathrm{d}_1: \mathrm{d}_2=3: 2

Then, Perimeter of 1st circle : Perimeter of 2nd circle

= \pi \mathrm{d}_1: \pi \mathrm{d}_2 =\mathrm{d}_1: \mathrm{d}_2 \\

= 3 : 2

In 60 seconds, Rahim travels = 90 m

In 1 seconds, Rahim travels =\frac{90}{60} m \\

In 40 seconds, Rahim travels =\frac{90}{60} × 40=60 m

Let ‘ r ‘ be the radius of the circular field.

Then, 2πr – 2r = 60

or, 2r (π – r) = 60

or, 2 r (\frac{22}{7}-1) = 60

or, 2 r (\frac{22-7}{7}) = 60

or, 2 r (\frac{15}{7}) = 60

or, \frac{30 r}{7} = 60

or, 30r = 60 × 7

or, r = 14 m

∴ Length of the diameter = 2 × r = 28 m

Let r₁ and r₂ be the radii of two circles

Perimeter of 1st circle : perimeter of 2nd circle = 2 : 3

or, 2π r₁ : 2π r₂ = 2 : 3

or, r₁ : r₂ = 2 : 3

or, \frac{r_1}{r_2}=\frac{2}{3}

or, r_1=\frac{2}{3} r_2

According to the condition of the problem,

\mathrm{r}_2-\mathrm{r}_1=2 \\

or, r_2-\frac{2}{3} r_2=2 \\

or, \frac{1}{3} \times \mathrm{r}_2=2 \\

∴ r₂ = 6 cm

∴ \quad r_1=\frac{2}{3} \times 6 \\

r₁ = 4 cm

∴ diameter of 1st circle = 2r₁

= 2 × 4

= 8 cm

∴ diameter of 1st circle = 2r₂

= 2 × 6

= 12 cm

Area of square = 196 sq cm

or, side² = 196

or, side² = 14²

or, side = 14 cm

diameter of two circles = 14 cm.

diameter of one circle = \frac{14}{2} = 7 cm

Radius of one circle =\frac{7}{2} cm

∴ Circumference of one circular plate

= 2πr

=2 \times \frac{22}{7} \times \frac{7}{2} = 22 cm

In 60 seconds, Nashifer travels = 80 m

In 1 seconds, Nashifer travels = \frac{80}{60} \\

In 45 seconds, Nashifer travels = \frac{80}{60} \\ × 45 = 60 m.

Let ‘ r ‘ be the radius of the circular field.

Then, 2πr – 2r = 60

or, 2 × r × (\pi-1)=60 = 60

or, 2 × r × (\frac{22}{7}-1) = 60

or, 2 × r × (\frac{22-7}{7}) = 60

or, 2 × r × \times \frac{15}{7} = 60

or, \frac{30 \mathrm{r}}{7} = 60

or, 30r = 60 × 7

or, r = \frac{60 \times 7}{30}

or, r = 14 cm

∴ Length of the diameter = 2r

= 2 × 14 = 28 cm

Width by a path = 7 m 5 dcm

= 7.5 m

In (46 – 44) = 2; Mohim covered = 2πr

= 2 × π × 7.5

= \frac{15}{2} \times \frac{22}{7} \times 44 m \\

= \frac{7260}{7} m \\

Let ‘ r ‘ be the radius of the circle.

Then, 2 \pi \mathrm{r}=\frac{7260}{7}

or, 2 \times \frac{22}{7} \times r=\frac{7260}{7}

or, r = \frac{7260}{2 \times 22} \\

or, r = 165 m

∴ Diameter of the circle = 2r

= 2 × 165 = 330 m

Let the external and internal radii of the circular path be R and r.

Given R – r = 5m

∴ R = 5 + r

Distance covered along external boundary = 2π(5 + r)

Distance covered internal boundary = 2πr

According to the question,

\frac{\text { external circumference }}{\text { Internal circumference }} =\frac{20}{19}

or, \frac{2 \pi(5+r)}{2 \pi r}=\frac{20}{19}

or, 20r = 95 + 19r

or, 20r – 19r = 95

or, r = 95

Internal diameter = 95 × 2 meters = 190 meters

(a) 1 : 12

(b) \frac{x}{100} m

Explanation

By question,

2πr = \frac{\pi x}{100} \\

or, 2r =  \frac{x}{100} \\

or d = \frac{x}{100} \\

Length of the diameter of the circle

= length of the side of a square = 10 cm

∴ (d) is correct option

(a) 5 \sqrt{2} cm

Explanation

Length of the diameter of the circle

= length of diagonal

=\operatorname{side} \sqrt{2} \\

= 5 \sqrt{2} cm

(a) 5 cm

Let r be the radius of the semicircle.

Then, πr + 2r = 36

or,  r (\pi+2) = 36

or, r (\frac{22}{7}+2) = 36

or, r (\frac{22+14}{7}) = 36

or, r \times \frac{36}{7} = 36

or, \frac{r}{7} = 1

or, r = 7 cm

∴ diameter (d) = 2 × r = 14 cm

Required length = \frac{\pi \mathrm{r}}{2} \\

= \frac{22}{7} \times \frac{1}{2} \times 7 \\

= 11 cm

Let a be the side of a square.

Then, radius of small circle = \frac{a}{2}

∴ radius of big circle = \frac{a \sqrt{2}}{2}

∴ Ratio of inscribed circle : ratio of the circumscribed circle =\frac{a}{2}: \frac{a \sqrt{2}}{2} \\

= 1 : √2

In 15 minutes, minute’s hand moves 90°

∴ Required length = \frac{\pi \mathrm{r}}{2} \\

= \frac{22}{7} \times \frac{1}{2} \times 7 \\

= 11 cm

Let ‘ r ‘ be the radius of the circle.

‘ d ‘ be the diameter of the circle.

‘a’ be the side of the square

Then,d = a

The ratio of the side of the square to the perimeter of the circle

a: \pi d \\

= a: \frac{22}{7} d \\

= 7: 22