AREA AROUND CIRCLE

Example 1 :

Find the area of a ring whose outer radius is 20 cm and inner radius is 15 cm respectively.

Solution :

Let "R" be the radius of the outer circle.

Ler "r" be the radius of the inner circle.

Area of outer circle  =  ΠR²

Area of outer circle  =  Πr²

Radius of outer circle (R)  =  20 cm

Radius of the inner circle (r)  =  15 cm

Area of ring (shaded portion)   

=  Area of outer circle - area of inner circle

=  ΠR² -  Πr²

=  Π (R² - r²)

=  Π (20² - 15²)

=  Π x (400 - 225)

=  Π x 175

=  (22/7) x 175

=  22 x 25 

=  550 cm²

Example 2 :

There is an outside circular path had constructed around a circular garden. If the outer and inner circumferences of the path are 88 cm and 44 cm respectively. Find the width and area of the path.

Solution :

Circumference of outer circle  =  88

2ΠR  =  88

2 x (22/7) x R  =  88

R  =  (88 x 7) /(2 x 22)

R  =  2 x 7

R  =  14 cm

Circumference of inner circle = 44

2Πr  =  44

2 x (22/7) x r  =  44

R  =  (44 x 7) /(2 x 22)

R  =  (2 x 7)/2

R  =  7 cm

Width of path  =  R - r

=  14 - 7

=  7 cm

Area of outer path  =  ΠR²

=  (22/7) x 14²

=  (22/7) x 14 x 14

=  616 cm²

Area of inner path  =  Πr²

=  (22/7) x 7²

=  (22/7) x 7 x 7

=  154 cm²   

Area of path  =  616 - 154

= 462 cm²

Example 3 :

Find the area of the shaded region if radius of two concentric circles are 7cm and 14 cm respectively and ∠𝐴𝑂𝐶 = 40^°.

Solution :

Area of circle shaded = (θ/360) x ΠR² - (θ/360) x Πr²

= (θ/360)Π[R² - r²]

Here R = 14 cm and r = 7 cm

= (40/360)Π[14² - 7²]

= (1/9) x (22/7)[196 - 49]

= (22/9 x 7) x (147)

= (22 x 7/3)

= 154/3

= 51.33 cm²

Example 4 :

A race track is in the form of ring whose outer and inner circumferences are 506 m, 440 m respectively. Find the width of the track.

Solution :

Circumference of large circle = 506 m

2ΠR = 506 m

2 x 22/7 x R = 506

R = 506 x (7/22) x (1/2)

R = 506 x (7/22) x (1/2)

R = 161/2

= 80.5 cm

2Πr = 440 m

r = 440 x (7/22) x (1/2)

r = 440 x (7/22) x (1/2)

r = 70 cm

Radius of large circle = 80.5 cm

Radius of small circle = 70 cm

Area of width in between two circles = ΠR² - Πr²

= Π(R² - r²)

= 3.14(80.5² - 70²)

= 3.14 (6480.25 - 4900)

= 3.14(1580.25)

= 4961.985 cm²

Example 5 :

In the given figure, AB is the diameter where AP = 12 cm and PB = 16 cm. If the value of p is taken 3, what is the perimeter of the shaded region?

Solution :

AB is the diameter of the circle and APB is right triangle.

AB² = AP² + PB²

AB² = 12² + 16²

= 144 + 256

= √400

AB = 20 cm

Radius of circle = 10 cm

= (1/2) Πr² - (1/2) x base x height

= 1/2 [Πr² - 16 x 12]

= 1/2 [3.14 x 10² - 16 x 12]

= (1/2) [314 - 192] 

= (1/2) x 122

= 61 cm²

Example 6 :

In given fig., O is the center of a circle. If the area of the sector OAPB is 5/36 times the area of the circle, what is the value of x? 

Solution :

Area of shaded region = 5/36 x area of circle

(θ/360) x Πr² = 5/36 x Πr²

θ/360 = 5/36

θ = (5/36) x 360

θ = 5 x 10

θ = 50

So, the required angle measure is 50 degree.

Example 7 :

In given figure arcs are drawn by taking vertices , AB and C of an equilateral triangle of side 10 cm, to intersect the side BC , CA and AB at their respective mid-points , DE and F. What is the area of the shaded region? (Use Π = . 3.14 )

Solution :

Area of shaded region = 3 x (θ/360) x Πr²

Length of AB = 10 cm, since F is the midpoint AF = 5 cm

= 3 x (60/360) x Π(5)²

= 39.25 cm²

Related Pages

• Perimeter of sector
  
• practice questions with solution
• Length of arc
• Practice questions on length of arc
  
• Perimeter of square 
• Perimeter of parallelogram
• Perimeter of rectangle
• Perimeter of triangle
• Area of a circle
• Area of Semicircle
• Area of Quadrant
  
• Area of sector
  
• Area of triangle
• Area of equilateral triangle
• Area of scalene triangle
  
• Area of square
  
• Area of rectangle
• Area of parallelogram
• Area of rhombus
• Area of trapezium
• Area of quadrilateral
  
• Area of pathways
• Area of combined shapes
  

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