All circles are similar to one another. So, the ratio of the circumference to that of diameter is a constant, that is Circumference/ Diameter = constant [say π( pi ) ].

**Circumference of a Circle**

All circles are similar to one another.
So, the ratio of the circumference to that of diameter is a constant, that is *Circumference/ Diameter = constant* [say π( pi ) **]**

Its approate value is 3.14. Therefor
C/d = π.

The diameter is twice the radius (2r),
so the above equation can be written as C/2r = π.

Therefore, the circumference of circle,
C = 2
πr units.

Obviously now, we see that Circumference,
*C* = π* d* and *d
*=*
*2*r *.

Thus for any circle with a given ‘*r’*
or ‘*d*’, we can find ‘*C’* and vice-versa.

**DO YOU KNOW**

1. Calculation of decimal
digits of π is an interesting
task asmong mathematans.

2. The constan π was used in the construction of popular pyraminds of Egypt.

3. Computer helped
mathematicians to calculate the value of π with
more than 12 trillion decimal digits.

**Think**

A
circle has the shortest perimeter of all closed figures with the same area.
Justify with an example.

** **

__Example 2.1____ __

Calculate the circumference of the bangle
shown in Fig. 2.5 (Take π = 3.14 ).

**Solution**

Given, *d* =
6 *cm* , *d* = 2*r* =
6 *cm* , *r* = 3*cm*

circumference of a circle =
2πr* units*

= 2*π* × 3

= 18.84

The circumference is 18.84 *cm*.

** **

__Example 2.2____ __

What is the circumference of the circular
disc of radius 14 *cm*? (use π = 22/7 ).

**Solution**

Radius of circular disc (*r*) =
14 *cm*

Circumference of the disc =
2 *π*r* units*

= 2 ×
22/7 ×14

= 88
*cm*

** **

__Example 2.3 __

If the circumference of the
circle is 132** ***m*. Then calculate the radius and** **diameter (Take π=22/7 ).

**Solution**

Circumference of the circle, *C*
=
2πr *units*

The circumference of the given circle
= 132 *m*

** **

__Example 2.4 __

What is the distance travelled
by the tip of the seconds hand
of a clock in 1 minute, if the length of the hand is 56* mm* (use π = 22/7 ).

**Solution**

Here the distance travelled by the tip
of the seconds hand of a clock in 1 minute is the circumference of the circle and
the length of the seconds hand is the radius of the circle. So, *r* = 56* mm*

Circumference of the circle, *C*
=
2πr* units*

= 2× (22/7) ×56

= 2×22×8

= 352* mm*

Therefore, distance travelled by the
tip of the seconds hand of a clock in 1 minute is 352* mm.*

** **

__Example 2.5 __

The radius of a tractor wheel
is 77** ***cm*. Calculate the distance
covered by it in 35 rotations? (use π = 22/7 ).

**Solution**

The distance covered in one rotation
= the circumference of the circle

= 2 πr* units*

= 2× (22/7) × 77

= 2×22×11

= 484 *cm*

The distance covered in one rotation
=
484 *cm*

The distance covered in 35 rotations
=
484 ×
35

=16940 *cm*

** **

__Example 2.6 __

A farmer wants to fence his
circular poultry farm with barbed wire whose** **radius is 420 *m*. The cost of fencing is ₹12 per metre. He has
₹30,000 with him. How much more
amount will be needed to fence his farm? (Here *π*
=
22/7)

**Solution**

The radius of the poultry farm is = 420 *m*

The length of the barbed wire for fencing
the poultry farm is equal to the circumference

of the circle.

We know that the circumference of the
circle
= 2 πr* units*

= 2 × (22/7) × 420

= 2×22×60

The length of the barbed wire to fence
the poultry farm = 2640 *m*

The cost of fencing the poultry farm
at the rate of ₹12 per metre = 2640 ×12

= ₹31,680

Given that he has ₹30,000 with him.

The excess amount required =
₹31,680 −
₹30,000 =
₹1,680.

** **

__Example 2.7 __

Find the perimeter of the
given shape (Fig.2.9) (Take π
= 22/7 ).

**Solution**

In this shape, we have to calculate the
circumference of semicircle on each side of the rectangle. The outer boundary of
this figure is made up of semicircles of two different sizes. Diameters of each
of the** **semicircles
are 7 *cm* and 14 *cm*. We know that the circumference of the circle =
πd *units*.

Circumference of the semicircular part
=
½ π *d units*

Hence, the circumference of the semicircle
having diameter 7 *cm* is,

= (1/2) × (22/7) × 7 = 11 *cm*

Circumference of the pair of semicircular
parts (II and IV) = 2 × 11 = 22 *cm*

Similarly, circumference of the semicircle
having diameter 14 *cm* is,

= (1/2) × (22/7) × 14 = 22 *cm*

Circumference of the pair of semicircular
parts (I and III) = 2 × 22 = 44 *cm* .

Perimeter of the given shape =
22 +
44 =
66 *cm* .

** **

__Example 2.8 __

Kannan divides a circular
disc of radius 14* cm *into four equal parts. What is the perimeter of
a quadrant shaped disc? (use *π* = 22/7)

**Solution**

To find the perimeter of the quadrant
disc, we need to find the circumference of quadrant shape.

Given that radius (*r*) = 14 *cm*.

We know that the circumference of circle
=
2 π*r* *units*

*.*

So, the circumference of the quadrant arc = 1/4 × 2πr

= πr / 2

= 22/7 × 14/2

= 22 *cm*

Given, the radius of the circle =
14 *cm*

Thus, perimeter of required quadrant
shaped disc = 14 + 14 + 22

= 50 *cm* .

** **

**Think**

(i) Is the circumference
of the semicircular arc and semicircular shaped disc same? Discuss.

(ii) The traffic lights
are circular. Why?

(iii) When you throw a
stone on still water in pond, ripples are circular. Why?

Tags : Measurements | Term 2 Chapter 2 | 7th Maths , 7th Maths : Term 2 Unit 2 : Measurements

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7th Maths : Term 2 Unit 2 : Measurements : Circumference of a Circle | Measurements | Term 2 Chapter 2 | 7th Maths

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