In this topic we will discuss and learn about the circumference and area of circle.

**Circumference of circle: ** The distance around the circular region is called its circumference. The ratio of circumference of any circle to its diameter is constant. This constant is denoted by π and is read as pie.

Circumference/Diameter = Pie

i.e., c/d = π or c = πd

We know that diameter is twice the radius, i.e., d = 2r

C = π × 2r

⇒ C = 2πr

Therefore approximate value of π = 22/7 or 3.14.

**Area of circle: **The measure of the region enclosed inside the circle is called its area.

** In case of concentric circles:** The region enclosed between two concentric circles of different radii is called the area of the ring.

**Note: **

Circles having same centre but different radii are called concentric circles.

**Worked-out examples on how do you find the area of a circle and the circumference of circle: **

**1.*** Find the circumference and area of radius 7 cm.*

**Solution:**

Circumference of circle = 2πr

= 2 × 22/7 × 7

= 44 cm

Area of circle = πr²

= 22/7 × 7 × 7 cm²

= 154 cm²

**2.*** A race track is in the form of a ring whose inner circumference is 220 m and outer circumference is 308 m. Find the width of the track.*

**Solution:**

Let r₁ and r₂ be the outer and inner radii of ring.

Then 2πr₁ = 308

2 × 22/7 r₁ = 308

⇒ r₁ = (308 × 7)/(2 × 22)

⇒ r₁ = 49 m

2πr₂ = 220

⇒ 2 × 22/7 × r₂ = 220

⇒ r₂ = (220 × 7)/(2 × 22)

⇒ r₂ = 35 m

Therefore, width of the track = (49 - 35) m = 14 m

**3.*** The area of a circle is 616 cm². Find its circumference.*

**Solution:**

We know area of circle = πr²

⇒ 22/7 × r² = 616

⇒ r² = (616 × 7)/22

⇒ r² = 28 × 7

⇒ r = √(28 × 7)

⇒ r = √(2 × 2 × 7 × 7)

⇒ r = 2 × 7

⇒ r = 14 cm

Therefore, circumference of circle = 2πr

= 2 × 22/7 × 14

= 88 cm

**4.*** Find the area of the circle if its circumference is 132 cm.*

**Solution:**

We know that the circumference of circle = 2πr

Area of circle = πr²

Circumference = 2πr = 132

⇒ 2 × 22/7 × r = 132

⇒ r = (7 × 132)/(2 × 22)

⇒ r = 21 cm

Therefore, area of circle = πr²

= 22/7 × 21 × 21

= 1386 cm²

**5.*** The ratio of areas of two wheels is 25 : 49. Find the ratio of their radii.*

**Solution:**

If A₁ and A₂ are the area of wheels,

A₁/A₂ =25/49

⇒ (πr₁²)/(πr₂²) = 25/49

⇒ (r₁²)/(r₂²) = 25/49

⇒ r₁/r₂ = √(25/49)

⇒ r₁/r₂ = 5/7

Therefore, ratio of their radii is 5 : 7.

**6.*** The diameter of a wheel of a motorcycle is 63 cm. How many revolutions will it make to travel 99 km?*

**Solution:**

The diameter of the wheel of a motorcycle = 63 cm

Therefore, circumference of the wheel of motorcycle = πd

= 22/7 × 63

= 198 cm

Total distance travelled by motorcycle = 99 km

= 99 × 1000

= 99 × 1000 × 100 cm

Therefore, number of revolutions = (99 × 1000 × 100)/198 = 50000

**7.*** The diameter of a wheel of cycle is 21 cm. It moves slowly along a road. How far will it go in 500 revolutions?*

**Solution:**

In revolution, distance that wheel covers = circumference of wheel Diameter of wheel = 21 cm

Therefore, circumference of wheel = πd

= 22/7 × 21

= 66 cm

So, in 1 revolution distance covered = 66 cm

In 500 revolution distance covered = 66 × 500 cm

= 33000 cm

= 33000/100 m

= 330 m

**8.*** The circumference of a circle exceeds the diameter by 20 cm. Find the radius of the circle.*

**Solution:**

Let the radius of circle of = r m.

Then circumference = 2 πr

Since, circumference exceeds diameter by 20

Therefore, according to question;

2 πr = d + 20

⇒ 2 πr = 2r + 20

⇒ 2 × (22/7) × r = 2r + 20

⇒ 44r/7 - 2r = 20

⇒ (44r - 14r)/7 = 20

⇒ 30r/7 = 20

⇒ r = (7 × 20)/30

⇒ r =14/3

So, the radius of circle = 14/3 cm = 42/3 cm

**9.*** A piece of wire in the form of rectangle 40 cm long and 26 cm wide is again bent to form a circle. Find the radius of the circle.*

**Solution:**

Length of wire = Perimeter of rectangle

= 2(l + b)

= 2(40 + 26)

= 2 × 66

= 132 cm

When it is again bent to form a circle, then

Perimeter of circle = Perimeter of rectangle

2 πr = 132 cm

⇒ 2 × 22/7 × r = 132

⇒ r = (132 × 7)/(2 × 22)

⇒ r = 21 cm

Formula is used to solve the different examples on circumference and area of circle with the detailed step-by-step explanation.

**● ****Mensuration**

**Perimeter and Area of Rectangle**

**Area and Perimeter of the Triangle**

**Area and Perimeter of the Parallelogram**

**Circumference and Area of Circle**

**Practice Test on Area and Perimeter of Rectangle**

**Practice Test on Area and Perimeter of Square**

● **Mensuration - Worksheets**

**Worksheet on Area and Perimeter of Rectangles**

**Worksheet on Area and Perimeter of Squares**

**Worksheet on Circumference and Area of Circle**

**Worksheet on Area and Perimeter of Triangle**

**7th Grade Math Problems** **8th Grade Math Practice** **From Circumference and Area of Circle to HOME PAGE**

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