The availability of step-by-step S Chand Class 9 Maths Solutions ICSE Chapter 17 Circle: Circumference and Area Ex 17(A) can make challenging problems more manageable.

## S Chand Class 9 ICSE Maths Solutions Chapter 17 Circle: Circumference and Area Ex 17(A)

Take π = \(\frac { 22 }{ 7 }\) unless stated otherwise.

Question 1.

Find the circumference of the circles whose diameters are :

(i) 49 cm

(ii) 14 cm

(iii) 9.8 cm

(iv) 7 cm

Solution:

(i) Diameter of the circle (d) = 49 cm

∴ Circumference = πd = \(\frac { 22 }{ 7 }\) × 49 = 154 cm

(ii) Diameter of the circle (d) = 14 cm

∴ Circumference = πd = \(\frac { 22 }{ 7 }\) × 14 = 44 cm

(iii) Diameter of the circle (d) = 9.8 cm

∴ Circumference = πd = \(\frac { 22 }{ 7 }\) × 9.8 = 30.8 cm

(iv) Diameter of the circle (d) = 7 cm

∴ Circumference = πd = \(\frac { 22 }{ 7 }\) × 7 = 22 cm

Question 2.

Find the circumference of the circles whose radii are :

(i) 7 cm

(ii) 28 cm

(iii) 3.5 cm

(iv) 98 m

Solution:

(i) Radius of the circle (r) = 7 cm

∴ Circumference = 2 πr = 2 × \(\frac { 22 }{ 7 }\) × 7 = 44 cm

(ii) Radius of the circle (r) = 28 cm

∴ Circumference = 2 πr = 2 × \(\frac { 22 }{ 7 }\) × 28 = 176 cm

(iii) Radius of the circle (r) = 3.5 cm

∴ Circumference = 2 πr = 2 × \(\frac { 22 }{ 7 }\) × 3.5 = 22 cm

(iv) Radius of the circle (r) = 98 m

∴ Circumference = 2 πr = 2 × \(\frac { 22 }{ 7 }\) × 98 = 616 cm

Question 3.

Find the circumference of the circles, if

(i) Radius = 4.5 cm

(ii) Diameter = 15 cm (Take π = 3.14 in (i) and (ii) above.)

Solution:

(i) Radius (r) = 4.5 cm

∴ Circumference = 2 πr = 2 × 3.14 × 4.5 = 28.26 cm = 28.3 cm

(ii) Diameter (d) = 15 cm

∴ Circumference = πd = 3.14 × 15 × cm = 47.10 = 47.1 cm

Question 4.

Find the length of the diameter of the circles whose circumference are :

(i) 22 cm

(ii) 8.8 cm

(iii) 44 cm

Solution:

(i) Circumference = 22 cm

Let d be the diameter of the circle

∴ d = \(\frac{\text { circumference }}{\pi}\) = \(\frac{\frac{22}{22}}{7}\) = \(\frac{22 \times 7}{22}\) = 7 cm

∴ Diameter = 7 cm

(ii) Circumference = 8.8 cm

∴ Diameter = \(\frac{\text { circumference }}{\pi}\) = \(\frac{\frac{8.8}{22}}{7}\) = \(\frac{8.8 \times 7}{22}\) = 0.4 × 7 = 2.8 cm

(iii) Circumference = 44 cm

∴ Diameter = \(\frac{\text { circumference }}{\pi}\) = \(\frac{\frac{44}{22}}{7}\) = \(\frac{44 \times 7}{22}\) = 14 cm

Question 5.

Find the lengths of the radii of the circles whose circumferences are

(i) 11 cm

(ii) 55 cm

(iii) 13.2 m

(iv) 440 m

Solution:

(i) Circumference = 11 cm

Let radius of the circle = r

Then 2πr = 11

⇒ \(\frac{2 \times 22}{7}\)r = 11 ⇒ r = \(\frac{11 \times 7}{2 \times 22}\) = \(\frac { 7 }{ 4 }\)

∴ Radius = \(\frac { 7 }{ 4 }\) cm = 1.75 cm

(ii) Circumference = 55 cm

Let radius of the circle = r

Then 2πr = 55

⇒ \(\frac{2 \times 22}{7}\)r = 55 ⇒ r = \(\frac{55 \times 7}{2 \times 22}\)

⇒ r = \(\frac { 35 }{ 4 }\) = 8.75

∴ Radius = 8.75 cm

(iii) Circumference = 13.2 cm

Let radius of the circle = r

Then 2πr = 13.2 m

⇒ \(\frac{2 \times 22}{7}\)r = 13.2 ⇒ r = \(\frac{13.2 \times 7}{2 \times 22}\) = \(\frac { 21 }{ 10 }\) = 2.1

∴ Radius = 2.1 cm

(iv) Circumference = 440 m

Let radius of the circle = r

Then 2πr = 440

⇒ \(\frac{2 \times 22}{7}\)r = 440 ⇒ r = \(\frac{440 \times 7}{2 \times 22}\) ⇒ r = 70

∴ Radius = 70 cm

Question 6.

Find the radii of the circles if circumference is : (π = 3.14)

(i) 100 m

(ii) 6.4 dm

Solution:

(i) Circumference = 100 m

Let radius of the circle = r

∴ 2πr = 100

⇒ 2 × 3.14 × r = 100

⇒ r = \(\frac{100}{2 \times 3.14}\) = \(\frac{100 \times 100}{2 \times 314}\)

= \(\frac { 10000 }{ 628 }\) = 15.92 = 15.9 cm

∴ Radius = 15.9 cm

(ii) Circumference = 6.4 dm

Let radius of the circle = r

Then 2 πr = 6.4

⇒ 2 × 3.14 × r = 6.4

⇒ 6.28r = 6.4 ⇒ r = \(\frac{6.4}{6.28}\)

⇒ r = 1.019 = 1.02 dm

∴ Radius = 1.02 cm

Question 7.

The diameter of Venus planet is 12,278 km. Find its circumference.

Solution:

Diameter of Venus planet = 12,278 km

∴Circumference = πd = \(\frac { 22 }{ 7 }\) × 12278

= 22 × 1754 = 38588 km

Question 8.

(i) Find the perimeter of semi-circular plate of radius 3.85 cm.

(ii) The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle. (Take π = 22/7)

Solution:

(i) Radius of the semicircle = 3.85 cm

∴ Perimeter = πr + 2r

= \(\frac { 22 }{ 7 }\) × 3.85 + 2 × 3.85

= 22 × 0.55 + 7.70

= 12.10 + 7.70 = 19.80 cm

(ii) Let r be the radius of the circle

∴ 2 πr – 2r = 16.8

⇒ 2r (π – 1) = 16.8

⇒ 2r \(\left(\frac{22}{7}-1\right)\) = 16.8 ⇒ 2r \(\left(\frac{15}{7}\right)\) = 16.8

⇒ r = \(\frac{16.8 \times 7}{2 \times 15}\) = \(\frac{168 \times 7}{10 \times 30}\) = \(\frac{56 \times 7}{100}\)

= \(\frac { 392 }{ 100 }\) = 3.92

∴Radius of the circle = 3.92 cm

Question 9.

A wire is in form of a circle of radius 42 cm. It is bent into a square. Determine the side of square.

Solution:

Radius of the circular wire = 42 cm

∴ Circumference = 2 πr = 2 × \(\frac { 22 }{ 7 }\) × 42 cm = 264 cm

On bending it into a square,

The perimeter of square = 264 cm

∴ Side of the square = \(\frac{\text { Perimeter }}{4}\)

= \(\frac { 264 }{ 4 }\) = 66 cm

Question 10.

In the figure, the radius is 3.5 cm. Find the perimeter of the quarter of the circle correct to one decimal place.

Solution:

Radius of the quater

Cirlce =3.5 cm

∴ Perimeter = 2 πr × \(\frac { 1 }{ 4 }\) + 2r

= \(\frac { 1 }{ 2 }\) πr + 2r = \(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 3.5 + 2 × 3.5

= 5.5 + 7.0 = 12.5 cm

Question 11.

(i) The inner circumference of a circular track is 440 m. The track is 14 m wide. Find the diameter of the outer circle of the track. (Take π = 22/7)

(ii) The inner edge of a circular running-track is 440 m long. If the track is 10 m wide, find the length of the outer edge of the track.

Solution:

(i) The inner circumference of the circular track = 440 m

∴ Radius (r) = \(\frac{\text { circumference }}{2 \pi}\)

= \(\frac{440 \times 7}{2 \times 22}\) = 70 m

Width of track = 14 m

∴ Radius of the outer circle (R)

= 70 + 14 = 84 m

and diameter = 2R = 2 × 84 = 168 m

(ii) The inner edge of the circular track = 440 m

∴ Inner radius (r) = \(\frac{\text { circumference }}{2 \pi}\)

= \(\frac{440 \times 7}{22 \times 2}\) = 70 m

Width of track = 10 m

∴ Outer radius (R) = 70 + 10 = 80 m

∴ Length of outer edge (circumference)

= 2πR = 2 × \(\frac { 22 }{ 7 }\) × 80 = \(\frac { 3520 }{ 7 }\) m = 502.86 m

Question 12.

A garden roller has a circumference of 3 metres. How many revolutions does it make in moving a distance of 21 metres?

Solution:

Circumference of roller = 3 m

Distance travelled = 21 m

∴ Number of revolutions will be

= \(\frac{\text { Distance }}{\text { circumference }}\) = \(\frac { 21 }{ 3 }\) = 7

Question 13.

A bicycle wheel, diameter 70 cm, is making 25 revolutions in 10 sec. At what speed in km/hr is the bicycle travelling?

Solution:

Diameter of the cycle wheel (d) = 70 cm

∴ Circumference = πd = \(\frac { 22 }{ 7 }\) × 70 = 220 cm

Distance travelled in 25 revolutions

=220 × 25 = 5500 cm = 55 m

∴ Distance of 55 m travelled in = 10 sec.

∴ The distance travelled in 1 hour

= \(\frac { 55 }{ 10 }\) × 60 × 60 m = 19800 m = 19.8 Km

∴ Speed = 19.8 km/hour

Question 14.

A small cart is being driven at 10 km per hour. If each wheel is 91 cm in diameter, find the number of revolutions made by each wheel per minute.

Solution:

Diameter of wheel (d) = 91 cm

∴ Circumference = πd = \(\frac { 22 }{ 7 }\) × 91 = 286 cm

Speed = 10 km/hr.

Distance travelled in 1 hour = 10 km

∴ Distance travelled in 1 minute = \(\frac { 10 }{ 60 }\) Km

= \(\frac{10 \times 1000}{60}\) m = \(\frac { 500 }{ 3 }\) m

∴ Number of revolutions in one minute

= \(\frac { 500 }{ 3 }\) × \(\frac { 100 }{ 286 }\) (circumference = 286cm)

= \(\frac{500 \times 50}{3 \times 143}\) = \(\frac { 25000 }{ 429 }\) = 58\(\frac { 118 }{ 429 }\)

Question 15.

How long will a boy take to go four times round a circular field whose radius is 35 m, walking at 5 km per hour ?

Solution:

Radius of circular field (r) = 35 m

∴ Circumference = 2πr = 2 × \(\frac { 22 }{ 7 }\) × 35 = 220 m

Distance travelled in 4 rounds = 220 × 4 m = 880 m

Speed of walking = 5km/hour

∴ Time taken to travelled 880 m = \(\frac { 880 }{ 5000 }\) hr.

= \(\frac{880 \times 60}{5000}\) = \(\frac { 264 }{ 25 }\) = 10.56 minute

=10 minute 34 second (approx)

Question 16.

A wheel of a cart is making 4 revolutions per second. If the diameter of the wheel is 84 cm, find its speed.

Solution:

Diameter of the wheel of the cart (d) = 84 cm

∴ Circumference = πd = \(\frac { 22 }{ 7 }\) × 84 = 264 cm

Distance covered in 1 sec = 4 revolutions

= 4 × 264 = 1056 cm

∴ Distance covered in 1 hour

= 1056 × 60 × 60 cm = \(\frac{1056 \times 60 \times 60}{100 \times 1000}\) km

= \(\frac { 38016 }{ 1000 }\) = 38.016 km

∴ Speed of the cart = 38.016 km/hr

Question 17.

The figure shows a running track in which AF = CD = 100 m and FED and ABC are semicircles whose diameters FD and AC are equal. How much will an athlete run in taking one complete round of the track?

Solution:

Diameter of semicircular part = 70 m

∴ Circumference of two semicircular part

= 2 × \(\frac { 1 }{ 2 }\)πd = πd = \(\frac { 22 }{ 7 }\) × 70 = 220 m

∴ Length of one complete round

= 220 + 100 + 100 m = 420 m

Question 18.

A circular field has a perimeter of 660 m. A plot in the shape of a square having its vertices on the circumference is marked in the field, calculate the area of the square plot.

\(\left(\pi=\frac{22}{7}\right) .\)

Solution:

Perimeter (circumference) of the circular field = 660 m

∴Length of diagonal of square ABCD

= diameter of the circle = 2r

= 2 × 105 = 210 m