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

Let us calculate – 16

1. Let us calculate the perimeter of the following figures.

Solution :

(i)  

Given,

\mathrm{AC} =8 cm, \quad \mathrm{BC}=\mathrm{DE}=5 m, \mathrm{BE}=\mathrm{CD}=4 m \\

\mathrm{AB} =(8-5)=3 m \\

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

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

=\sqrt{9+16} \\

=\sqrt{25}=5 \text { units. }

\therefore Perimeter of semi circle with diameter 4 cm

=\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 \text { (Ans) }

(ii)

\mathrm{AB}=14 cm, \mathrm{AD}=\mathrm{CD}=14 cm, \mathrm{BC}=14 cm

The perimeter of a circle with a radius 7 cm. = \pi r(\because r \text{is radius of the circle} )

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

\therefore Perimeter of the given figure

=(14+14+14+22) \mathrm{cm}=64 cm \text { (Ans.) }

2. Let us calculate how long wire will take to make a circular ring of radius 35 meters.

Solution : Length of the wire = 2 \pi \times radius

=2 \times \frac{22}{7} \times 35 m . \\

=220 m \text { (Ans.) }

3. The radius of the wheel of a train is 0.35 meters. If it makes 450 revolutions per minute, let us calculate the velocity of the train per hour.

Solution: Circumference of the wheel = 2 \times \pi \times radius

=2 \times \frac{22}{7} \times .35 m . \\

=2.2 m \text {. (Ans.) }

\text { one revolution } =2.2 m \\

450 \text { revolution } =2.2 \times 450 m . \\

= 990 m.

In 1 minute, distance covered In 60 minutes = 990 m

In 60 minutes = 1 hour ” ” =990 × 60m

= 59400m \quad= 59.4 Km .

\therefore \text{Velocity of the wheel} =59.4 \mathrm{Km} / \mathrm{hr}. (Ans)

4. The radius of the circular field of the village Amadpur is 280 meters. Chaitali wants to go around, the field by walking at a speed of 5.5 km/hour, let us calculate how long be takes by Chaitali to complete one revolution.

Solution : Raduius of Circular field = 280 m

Circumference of the field = 2 \pi r

=2 \times \frac{22}{7} \times 280 m \\

=1760 m . \quad \text { (Ans.) } \\

\therefore \text { Speed }=5.5 \mathrm{Km} / \mathrm{hour} . \\

=\frac{55 \times 1000}{10 \times 60 \times 60} m / \mathrm{sec} . \\

=\frac{55}{36} m / \mathrm{sec} . \\

\text { In } \frac{55}{36} m, \text { Chaitali takes } 1 \mathrm{sec} . \\

\mathrm{m} \ 1 m, \text { Chaitali takes } \frac{36}{55} \mathrm{sec} \\

\mathrm{m} \ 1760 m, \text { Chaitali takes } \frac{36}{55} \times 1760 \mathrm{sec} .

= 1152 Sec. = 19.2 mins.

5. Tathagata bent a copper wire in the form of a Rectangle of which length is 18 cm. and breadth is 15m. I made a circle by bending this copper wire. Let us calculate the length of the radius of a circular wire.

Solution : Perimeter of the rectangle = 2 (length + breadth )

=2(18+15) \mathrm{cm} . \\

=2 \times 33 cm .=66 cm .

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

Then, circumference = 66

or, 2 \pi r=66

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

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

or, 2 r=21

or, r=\frac{21}{2}=10.5 \quad \therefore r=10.5

\therefore Radius of circular copper wire = 10.5 cm (Ans.)

6. The perimeter of a semicircular field is 108 meters. Let us calculate the diameter of the field.

Solution : 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

\therefore r = 21

\therefore diameter of the field = 2r

= 2 \times 21 cm=42 cm (Ans.)

7. The difference between the circumference and the diameter of a wheel is 75 cm, let us calculate the length of radius of the wheel.

Solution: Let ‘ r ‘ be the radius of the wheel.

Then, \quad 2 \pi r-2 r=75

or, \quad 2 r(\pi-1)=75

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

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

or, 2 \mathrm{r} \times \frac{15}{7}=75

or, \frac{2 r}{7}=5 \quad or, \ 2 r=35

\text { or, } r=\frac{35}{2} \quad \therefore r=17.5

\therefore \quad Length of the radius of the wheel 17.5 cm (Ans.)

8. In a race, Puja and Jakir start to complete from the same point and same time on a circular track of the length of diameter is 56 meters. When Puja finishes the race at the competition by 10 revolutions, Jakir is one revolution behind. Let us calculate how many meters is the length of the race and by how many meters Puja beats Jakir.

Solution : Length of diameter of circular track = 56 m

Circumference of the circular track i.e, one revolution of the circular track = \pi \mathrm{d}

=\frac{22}{7} \times 56 m=176 m

\therefore Length of the race i.e, 10 revolution

=176 m \times 10=1760 m \text { (Ans.) }

Jakir is one revolution behind

\therefore \quad Puja beats Jakir by 176 m (Ans.)

9. The perimeter of the borewell of our village is 440 cm. There is an equally wide stone parapet around this borewell. If the perimeter of the borewell with the parapet is 616 cm. Let us write by calculating that how much width of the stone parapet.

Solution : Perimeter of bore well = 44 cm.

\text { 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

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

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

\therefore width of the stone parapet = (98-70)cm = 28cm (Ans.)

10. Niyamat Cacha of the village attaches the motor’s wheel with a machine’s wheel with a belt. The length of the diameter of the motor wheel is 14 cm and the machine wheel is 94.5 cm. If the motor’s wheel revolves 27 times in a second then let us calculate how many times the machine’s wheel will revolve in an hour.

Solution : Circumference of moter wheel =\pi \mathrm{d}

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

Circumference of machine wheel =\pi \mathrm{d}

=\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 cm=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(\text { Ans.) }

11. The length of the hour’s hand and minute’s hand are 8.4 cm and 14 cm respectively of our club clock. Let us calculate how much distance will be covered by each hand in a day.

Solution :

\text { Hours's hand goes in } 12 \text { hours } =2 \times \frac{22}{7} \times 8.4 cm \\

= 52.8 cm.

” ” ” ” ” ” 24 ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” = 5.8 × 2 = 105.6 cm (Ans.)

Minute’s hand goes in one hour = 2 \times \frac{22}{7} \times 14 cm

= 88 cm.

” ” ” ” ” ” ” ” ” ” ” ” ” ” ” ” 24 hours = 24 × 88 = 2112 cm (Ans.)

12. The ratio of the diameter of two circles which were drawn by me and my friend Mihir is

\square : \square it is founded by calculatting that the ratio of perimeter

\square : \square of two circle is

Solution : 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 \text { (Ans.) }

13. The time that Rahim takes to cover up by running a circular field is 40 seconds less when he goes one end to another end diamterically. Velocity of Rahim is 90 meters per minute. Let us calculate the length of the diameter of the field.

Solution : In 60 seconds, Rahim takes 90 m

” ” ” ” ” ” ” 1 ” ” ” ” ” ” ” =\frac{90}{60} m \\

” ” ” ” ” ” 40 ” ” ” ” ” ” ” =\frac{90}{60} \times 40=60 m .

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

Then, \quad 2 \pi r-2 r=60

or, 2 r(\pi-1)=60

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

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

or, 2 \mathrm{r} \times \frac{15}{7}=60

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

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

or, r=\frac{60 \times 7}{30}=14

\therefore r = 14

\therefore \text{Length of the diameter }=2 r=2 \times 14 m=28 m (Ans.)

14. The ratio of the perimeter of two circles is 2: 3 and the difference of the length of radii is 2 cm. Let us calculate the length of the diameter of two circles.

Solution : Let r_1 and r_2 be the radii of two circles

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

or, 2 \pi r_1: 2 \pi r_2=2: 3

or, r_1: r_2=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 \\

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

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

\therefore \mathrm{r}_2=6 \\

\therefore \quad r_1=\frac{2}{3} \times 6 \\

\therefore \mathrm{r}_1=4 \\

\therefore \quad \text { diameter of } 1 \text { st circle }=2 \mathrm{r}_1=2 \times 4 cm=8 cm \\

\therefore " " " " " " 2 \mathrm{nd}=2 \mathrm{r}_2=2 \times 6 cm=12 cm \\(Ans.)

15. The four maximum-sized circular plates are cut out of a brass plate of square size. Having the area 196 sq. cm. ; Let us calculate the circumference of each circular plate.

Solution :

\because Area of square = 196 sq.cm.

\text { or, }(\text { Side })^2=196 \text { sq.cm. } \\

\text { or, Side }=\sqrt{196 cm} \text {. } \\

= 14 cm.

\therefore diameter of two circles = 14 cm.

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

radius of one circle =\frac{7}{2} cm.

\therefore Circumference of one circular plate

\quad=2 \pi \times r \\

=2 \times \frac{22}{7} \times \frac{7}{2}=22 cm (Ans.)

16. The time that Nashifer takes to cover up a circular field from one end to the other end is 45 seconds less when he goes diametrically. If the velocity of Nashifer is 80 m / min. Let us calculate the length of the diameter of this field.

Solution: In 60 seconds, Nashifer takes 80 m

” ” ” ” ” ” 1 ” ” ” ” ” “ =\frac{80}{60} \\

” ” ” ” ” ” 45 ” ” ” ” ” ” =\frac{80}{60} \times 45 \\

= 60 m.

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

Then, 2 \pi r-2 r=60

or, 2 r(\pi-1)=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, 30 \mathrm{r}=60 \times 7

or, r=\frac{60 \times 7}{30} \quad \therefore r=14

\therefore Length of the diameter = 2r

= 2 × 14 = 28cm (Ans.)

17. Mohim takes 46 seconds and 44 seconds respectively to go around along the outer and the inner edges of a circular path with 7 meter 5 cm. width by a cycle. Let us calculate the diameter of the circle along the inner edge of the path.

Solution : width by a path = 7m. 5dcm.

= 7.5 m.

In (46 – 44) = 2 \mathrm{sec} ; Mohim covered 2 \times \pi \times 7.5 m=15 \pi m.

=\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}

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

\therefore \quad r^{\prime}=165 m \\

\therefore \quad \text { diameter of the circle }=2 r \\

=2 \times 165 m=330 m \text { (Ans.) }

18. The ratio of time taken by a cyclist to go around the outer and inner circumference of a circular path is 20: 19; I the path is 15 meter wide, let us calculate the length of diameter of the inner circle.

Solution :

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

Given R – r = 5m

\therefore R = (5 + r)m

Distance covered along external boundary = 2 \pi(5+r) m

and long internal boundary =2 \pi \mathrm{rm} \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.(Ans)

19. (M.C.Q):

(i) The ratio of the velocity of the hour’s hand and minute’s hand at a clock is

(a) 1 : 12

(b) 12 : 1

(c) 1 : 24

(d) 24 : 1

Solution : (a) is the correct option

(ii) Soma takes \pi x \over 100 minutes to go one complete round of a circular path. Soma will take time to go around the park diametrically.

(a) \frac{x}{200} m

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

(c) \frac{\pi}{100} m

(d) \frac{\pi}{200} m

Solution: By question,

\text { or, } 2 \pi r=\frac{\pi x}{100} \\

\text { or, } 2 r=\frac{x}{100} \\

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

\therefore (b) is correct option

(iii) A circle is inscribed by a square. The length of the side of the square is 10 cm. The length of diameter of circle is

(a) 10 cm

(b) 5 cm

(c) 20 cm

(d) 10 \sqrt{2} cm.

Solution: Length of the diameter of the circle

= length of the side of a square = 10 cm

\therefore (d) is correct option(Ans.)

(iv) A circle circumscribe a square. The length of the side of the square is 5 cm. The length of the diameter of the circle is.

(a) 5 \sqrt{2} cm

(b) 10 \sqrt{2} cm.

(c) 5 cm

(d) 10 cm

Solution : Length of diameter of circle

= length of diagonal

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

= 5 \sqrt{2} cm

\therefore \quad (a) is correct option

(v) A circular ring is 5 cm. wide. The difference of outer and inner radius is

(a) 5 cm.

(b) 2.5 cm

(c) 10 cm

(d) none of these

Solution : (a) is correct option

20. Short answer type questions:

(i) The Perimeter of the semicircle is 36 cm. What is the length of the diameter?

Solution : Let r be the radius of semicircle.

Then, \quad \pi \mathrm{r}+2 \mathrm{r}=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

\therefore \quad \mathrm{r}=7

\therefore \text{Length of diameter}  =2 \mathrm{r}=2 \times 7=14 cm (Ans.)

(ii) The length of a minute’s hand is 7 cm. How much length will a minute’s hand to to rotate 90^{\circ}?

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

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

= 11 cm. (Ans.)

(iii) What is the ratio of radii of inscribed and circumscribed circles of a square?

Solution :

Let a be the side of a square.

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

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

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

= 1: \sqrt{2}(\mathrm{Ans})

(iv) The minute’s hand of a clock is 77 cm. How long does the minute’s hand move in 15 minutes?

Solution : In 15 minutes, minute’s hand moves 90\therefore \text { Required length } =\frac{\pi \mathrm{r}}{2} \\

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

= 11cm (Ans.)

(v) What is the ratio of the side of the square and the perimeter of the circle when the length of the diameter of the circle is equal to the length of the square?

Solution : Let ‘ r ‘ be the radius of the circle.

d ” ” ” ” ” ” ” ” diameter ” ” “

a ” ” ” ” ” ” ” ” side ” ” “

Then, \quad 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 \text { (Ans.) }