CBSE Grade 10 Maths Chapter 12 - Areas Related To Circles

Circle 

• A circle is a path of a point which moves in such a way that its distance from a fixed point is a constant. The path of the point is called the locus of the point. 
• The fixed point is called the centre of the circle.
• The constant distance is called the radius of the circle. 
  

In our daily life we see a lot of things which are circular in shape. Example : Ring, bangles, coins, wheel, dosa/chapaati, clock etc.

Know about π

• π is a Greek letter. 
• It is a ratio of the circumference to diameter of a circle. 
• We use the value of π as 22/7 or 3.142.

Area of a Circle

• Area of a circle is πr², where π=22/7 or ≈3.14 (can be used interchangeably for problem-solving purposes)and r is the radius of the circle.
• π is the ratio of the circumference of a circle to its diameter.

Area of a circle = πr²

Circumference of a Circle

• The perimeter of all the plain figures is the outer boundary of the figure. Using a tape or thread we can measure the circumference of a circle.
  
• The perimeter of a circle is the distance covered by going around its boundary once. 
• The perimeter of a circle has a special name: Circumference, which is π times the diameter which is given by the formula 2πr.
  

  Circumference of a Circle = 2πr = πd

• Diameter of a circle is twice its radius i.e. d = 2r 
• Diameter is the longest chord of a circle.

 Semi-Circle

A diameter of a circle divides the circle into two halves called Semi - Circles.

Perimeter and Area of the Semi-circle

• The perimeter of the semi-circle is half of the circumference of the given circle plus the length of diameter as the perimeter is the outer boundary of the figure.

  Circumference of a semi-circle = 𝜋r + 2r = r ( π +2) units

• Area of the semi-circle is just half of the area of the circle.

Area of a semi-circle = ( πr²)/2

Note : 

1. 1 whole circle = 2 semicircles
2. Area of semicircle = half of the area of the whole circle.
3. Perimeter of a semicircle ≠ half of the perimeter of the whole circle. 
   

 Area of a Ring

Area of the ring i.e. the shaded part in the above figure is calculated by subtracting the area of the inner circle from the area of the bigger circle.  

 

Area of the ring = πR² - πr² = π( R² - r²)

 

Where, R = radius of outer circle

r = radius of inner circle

Segment of a Circle

• A circular segment  is a region of a circle which is “cut off” from the rest of the circle by a secant or a chord.

Sector of a Circle

• A circle sector/ sector of a circle is defined as the region of a circle enclosed by an arc and two radii.
• θ is the central angle, l is the length of the arc.

• The smaller area is called the minor sector and the larger area is called the major sector.

Angle of a Sector

• The angle of a sector is that angle which is enclosed between the two radii of the sector.

Length of an arc of a sector

• The length of the arc of a sector can be found by using the expression for the circumference of a circle and the angle of the sector, using the following formula:

   Length of arc of a sector= (θ/360°)×2πr

Where θ is the angle of sector and r is the radius of the circle.

Perimeter of a sector 

Perimeter of a sector = 2r + L

where r is the radius of the circle and L is the length of the arc.

Area of a Sector of a Circle

Area of a sector = (θ/360°)×πr²

where ∠θ is the angle of this sector(minor sector in the following case) and r is its radius.

Note : 

• Area of major sector = Area of the circle - Area of minor sector.
• Area of major sector + Area of minor sector = Area of the Circle. 
• When the central angle θ = 360 °,then length of the arc = 2πr
  

Area of a Triangle

• The Area of a triangle is,

Area=(1/2)×base×height

• If the triangle is an equilateral then

Area=(√3/4)×a²

where “a” is the side length of the triangle

 Segment of a Circle 

• A segment of a circle is a portion of the circular region enclosed between a chord and the corresponding arc.
• A chord divides a circle into two portions called segments.

• The segment which contains the centre of the circle is called the major segment. The other one is called the minor segment.
  

 Area of a Segment of a Circle

Area of segment APB (highlighted in blue)

= (Area of sector OAPB) – (Area of triangle AOB)

    =[(∅/360°)×πr²] – [(1/2)×AB×OM]

[To find the area of triangle AOB, use trigonometric ratios to find OM (height) and AB (base)]

Also, Area of segment APB can be calculated directly if the angle of the sector is known using the following formula.

     =[(θ/360°)×πr²] – [r²×sin θ/2 × cosθ/2] .

Thus , Area of Minor Segment = [(θ/360°)×πr²] – [(1/2)r²sin θ]

Where θ is the angle of the sector and r is the radius of the circle

Areas of different plane figures

• Area of a square (side l) =l²
• Area of a rectangle =l×b, where l and b are the length and breadth of the rectangle
• Area of a parallelogram =b×h, where “b” is the base and “h” is the perpendicular height.
• Area of a trapezium =[(a+b)×h]/2, where a & b are the length of the parallel sides and h is the trapezium height  

• Area of a rhombus =pq/2, where p & q are the diagonals.
  

Areas of Combination of Plane figures

For example: Find the area of the shaded part in the following figure: Given the ABCD is a square of side 28 cm and has four equal circles enclosed within.

Looking at the figure we can visualize that the required shaded area = A(square ABCD) − 4 ×A(Circle).

Also, the diameter of each circle is 14 cm.
=(l2)−4×(πr²)
=(282)−[4×(π×49)]
=784−[4×22/7×49]
=784−616
=168cm²

Areas of Combinations of Plane Figures

As we know how to calculate the area of different shapes, so we can find the area of the figures which are made with the combination of different figures.

Example

Find the area of the colored part if the given triangle is equilateral and its area is 17320.5 cm². Three circles are made by taking the vertex of the triangles as the centre of the circle and the radius of the circle is the half of the length of the side of the triangle. (π = 3.14 and √3 = 1.73205)

Solution

Given-

ABC is an equilateral triangle, so ∠A, ∠B, ∠C = 60°

Hence the three sectors are equal, of angle 60°.

Required-

To find the area of the shaded region.

Area of shaded region =Area of ∆ABC – Area of 3 sectors

Area of ∆ABC = 17320.5 cm²

√3/4 * ( Side)² = 17320.5

( Side)² = (17320.5 * 4 )/ 1.73205 = 4*10⁴

Side = 200 cm

As the radius of the circle is half of the length of the triangle, so

Radius = 100 cm

Area of Sector = (θ/360)*πr² 

= (60/360) * π(100)² 

=(1/6) *3.14 * (100)²

= 15700/3 cm²

Area of 3 Sectors = 3 × 15700/3 cm² cm²

Area of shaded region = Area of ∆ABC – Area of 3 sectors

= 17320.5 - 15700 cm²

= 1620.5 cm²

Example

Find the area of the shaded part, if the side of the square is 8 cm and the circumference of the circle is 44 cm.

Solution

Required region = Area of circle – Area of square

= πr² – (side)²

Circumference of circle = 2πr = 44

2* (22/7) *r = 44

r= 44*(7/22) * (1/2) = 7

Radius of the circle = 7 cm

Area of circle = πr²

(22/7) *7*7=154 cm²

Area of square = (side) ² = (8)² = 64 cm²

Area of shaded region = Area of circle – Area of square

= 154 cm² - 64 cm²

= 90 cm²