Annulus: Formula, Area, Perimeter and Solved Examples

Collegedunia Team logo

Collegedunia Team

Content Curator

Annulus is a ring-shaped geometric figure or, more broadly, a term used to designate a ring-shaped object in mathematics. It is also known as the space between two concentric rings. It is similar to the right-hand throw-ring. A round disc with a circular hole is another example. The ring's outer and inner rings are concentric circles (circles that share a common centre point with different radius).

Keyterms: Annulus, Circle, radius, circumference, centre, plane, set, ring


What is an Annulus?

[Click Here for Sample Questions]

A circle is a curve sketched out by a point moving in a plane so that its distance from a given point is constant. Alternatively, it is the shape formed by all points in a plane that are at a set distance from a given point, the centre. A planar figure bordered by the circumference of two concentric circles of two distinct radii is known as an annulus.

Annulus

Annulus

The two radii R, r in the diagram, which are the radii of the outer ring and the inner 'hole,' respectively, determine the dimensions of an annulus. In the figure shown, the shaded portion, representing a ring, is known as the annulus. Individually, we have 2 circles, with a radii R and r, where R>r.

Also Read:


Annulus Formula

[Click Here for Sample Questions]

A two-dimensional shape's perimeter is defined as the route or border that encircles it and the size of the region encompassed inside any 2D form is its area. By the definition of Annulus, we know that it is a 2D figure, therefore, the area and perimeter of Annulus are:

Perimeter = 2π (R + r)

Area: A = π (R2 – r2)


Area of Annulus

[Click Here for Sample Questions]

The area of an annulus is equal to the difference between the areas of the bigger R-radius circle and the smaller r-radius circle.

Area of the annulus

Area of the annulus

Let A be the area of the outer circle and 'a' be the area of the smaller circle. Thus, the area of Annulus is:

Area = A - a

We know that the area of a circle is 

A= πr2

Thus the difference between the areas of two concentric circles having radius R and r is given by,

A = πR2 – πr2 

A = π (R2 – r2)

Read More: Important Questions on Circle

The length of the longest line segment within an annulus, which is the chord tangent to the inner circle, 2d in the accompanying diagram, determines the annulus' area. 

Longest line in the annulus

Longest line in the annulus

Since this line is parallel to the smaller circle and perpendicular to its radius at that point, d and r are sides of a right-angled triangle with hypotenuse R, and the area of the annulus is determined by the Pythagorean theorem.

A = π (R2 – r2) = πd2


Perimeter of Annulus

[Click Here for Sample Questions]

The annulus's perimeter equals the sum of the radii of the large and small circles multiplied by 2 pi. Given that, a radius of the outer circle is R and that of the inner circle is r, therefore by definition,

Perimeter of Annulus = 2πR + 2πr = 2π (R + r) 


Things to Remember

  • An annulus is a flat ring-like form. Its edges are made up of two circles with the same centre.
  • A spherical shell is a three-dimensional version of an annulus in geometry. It's the space between two concentric circles with different radii that a ball occupies.
  • An annulus is certainly something you've seen in your daily life. The colourful rings on an archery target or the face of a CD are two examples of things that create a set of annuli.
  • The area of an annulus is given by,  A = π (R2 – r2)
  • The perimeter of an annulus is given by,  2π (R + r)

Also Read:


Sample Questions

Ques. If the area of an annulus is 1092 inches and its width is 3 cm, then find the radii of the inner and outer circles? (2 marks)

Ans. Let the inner radius of an annulus be r and its outer radius be R.

Then width = R – r

3 = R – r

R = 3 + r

We know,

Area of the annulus =  π (R2 – r2) 

or

Area of the annulus = π (R + r) (R – r)

1092 = 22/7 (3 + r + r) (3)

3 + 2r = 1092×722×3

3 + 2r = 115.82

2r = 115.82 – 3

2r = 112.82

r = 56.41

R = 3 + 56.41

= 59.41

So, Inner radius = 56.41 inches

Outer radius = 59.41 inches

Ques. A steel pipe has an outside diameter (OD) of 100mm and an inside diameter (ID) of 80mm, what is the area of the cross-section? (2 marks)

Ans. Convert diameter to radius for both outside and inside circles:

R = 100 mm / 2 = 50 mm

r = 80 mm / 2 = 40 mm

Now calculate area:

Area = π (R2 – r2)

Area = 3.14 × ( 502 − 402 )

Area = 3.14× (2500 − 1600)

Area = 3.14 × 900

Area = 2827 mm2 (to nearest mm2)

Ques. If the inner radius is 4 cm and the outer radius is 5 cm, find the area of the annulus. (2 marks)

Ans. Area = π(R2 - r2)

R = 5cm

r = 4cm

Area = 22/7 (52 - 42 )

Area = 28.26 sq.cm

Ques. If the inner radius is 9 cm and the outer radius is 10 cm, find the area of the annulus. (2 marks)

Ans. Area = π(R2 - r2)

R = 10cm

r = 9cm

Area = 22/7 (102 - 92 )

Area = 59.66 sq.cm

Ques. If the inner radius is 16 cm and the outer radius is 18 cm, find the area of the annulus. (2 marks)

Ans. Area = π(R2 - r2)

R = 18cm

r = 16cm

Area = 22/7 (182 - 162 )

Area = 213.52 sq.cm

Ques. If the inner radius is 23cm and the outer radius is 28 cm, find the area of the annulus. (2 marks)

Ans. Area = π(R2 - r2)

R = 28cm

r = 23cm

Area = 22/7 (282 - 232 )

Area = 800.7sq.cm

Ques. If the inner radius is 50 cm and the outer radius is 56 cm, find the area of the annulus. (2 marks)

Ans. Area = π(R2 - r2)

R = 56cm

r = 50cm

Area = 22/7 (562 - 502 )

Area = 1997.04 sq.cm

Ques. If the inner radius is 44 cm and the outer radius is 50 cm, find the area of the annulus. (2 marks)

Ans. Area = π(R2 - r2)

R = 50cm

r = 44cm

Area = 22/7 (502 - 442 )

Area = 1770.96 sq.cm

Ques. If the inner radius is 102 cm and the outer radius is 110 cm, find the area of the annulus. (2 marks)

Ans. Area = π(R2 - r2)

R = 110cm

r = 102cm

Area = 22/7 (1102 - 1022 )

Area = 5325.44 sq.cm

Ques. If the inner radius is 17 cm and the outer radius is 36 cm, find the area of the annulus. (2 marks)

Ans. Area = π(R2 - r2)

R = 36 cm

r = 17 cm

Area = 22/7 (362 - 172 )

Area = 3161.98 sq.cm

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Check-Out: 

CBSE X Related Questions

1.

Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

(v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

      2.

      The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

      Length (in mm)

      Number of leaves

      118 - 126

      3

      127 - 135 

      5

      136 - 144

      9

      145 - 153

      12

      154 - 162

      5

      163 - 171

      4

      172 - 180

      2

      Find the median length of the leaves. 
      (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

          3.
          The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

              4.
              A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

                  5.

                  The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them

                  Monthly consumption 
                  (in units)

                   Number of consumers

                  65 - 85 

                  4

                  85 - 105

                  5

                  105 - 125

                  13

                  125 - 145

                  20

                  145 - 165

                  14

                  165 - 185

                  8

                  185 - 205

                  4

                      6.
                      Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
                      (i) 2, 4, 8, 16, . . . .
                      (ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
                      (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
                      (iv) – 10, – 6, – 2, 2, . . .
                      (v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
                      (vi) 0.2, 0.22, 0.222, 0.2222, . . . .
                      (vii) 0, – 4, – 8, –12, . . . .
                      (viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
                      (ix) 1, 3, 9, 27, . . . .
                      (x) a, 2a, 3a, 4a, . . . .
                      (xi) a, \(a^2, a^3, a^4,\)  . . . .
                      (xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
                      (xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
                      (xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
                      (xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

                          Comments



                          No Comments To Show