Chapter 27 - Additional Examples

Additional Example 1
In the figure (a) below, A, B, C and D are points on the circle. 

Compute the angles of the quadrilateral ABCD, and the angles between it's diagonals
Solution:
1. Consider ΔBPC. We get ∠BPC = [180-(30+50)] =[180-80] = 100^o. (∵ sum of interior angles of a triangle is 180^o) This is marked in fig(b)
2. ∠BPC and ∠BPA form a linear pair. So ∠BPA = 180 - ∠BPC = 180 -100 = 80^o.

3. ∠BPC and ∠APD are opposite angles, and are hence equal. So we get ∠APD = ∠BPC = 100^o.

4 Similarly, ∠BPA and ∠DPC are opposite angles, and are hence equal. So we get ∠DPC = ∠BPA = 80^o.

5. Consider arc BC in fig(c). It subtends ∠BAC (= 35^o) on the alternate arc. 

• The same arc subtends ∠BDC on the alternate arc. So ∠BDC = ∠BAC = 35^o.

6. Consider ΔPCD. We get ∠PCD = [180-(80+35)] =[180-115] = 65^o. (∵ sum of interior angles of a triangle is 180^o)

7. Consider arc CD in fig(c). It subtends ∠CBD (= 30^o) on the alternate arc. 

• The same arc subtends ∠CAD on the alternate arc. So ∠CAD = ∠CBD = 30^o.

8. Consider ΔPAD. We get ∠ADP = [180-(100+30)] =[180-130] = 50^o. (∵ sum of interior angles of a triangle is 180^o)

■ Thus we get all the angles of the quadrilateral ABCD. Since it is a cyclic quadrilateral, we can do a check:

(i) Sum of opposite angles ∠BAD and ∠BCD = 35 + 30 + 50 + 65 = 180^o.          

(ii) Sum of opposite angles ∠ABC and ∠ADC = 65 + 30 + 50 + 35 = 180^o.          

Additional Example 2

In the figure (a) below, AB is a diameter of the circle. 

CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Calculate ∠AEB

Solution:

1.Draw OC and OD as shown in fig(b). Consider the ΔOCD

• OC = OD (∵ radii of the same circle)

• Given that chord CD is equal to the radius.

• So ΔOCD is an equilateral triangle. All it's interior angles are equal to 60^o.

2. Draw AD as shown in fig(c)

3. Consider the arc CD. It has a central angle ∠COD = 60^o.

4. This same arc CD subtends ∠CAD on the opposite arc. 

• So ∠CAD = ¹⁄₂×60 = 30^o.  

5. Now consider ∠ADB. AB is a diameter and D is a point on the semi-circle. So ADB = 90^o.

6. Consider ΔAED. The ∠ADB that we considered above is an exterior angle of ΔAED

• Exterior angle of a triangle = sum of remote interior angles

• So ∠ADB = ∠EAD + ∠AED ⇒ 90 = 30 + ∠AED ⇒ ∠AED = 90 - 30 = 60^o.

7. But ∠AED and ∠AEB are the same angles

• Thus the required ∠AEB = 60^o

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Chapter 21.9 - Area of Sectors - Solved examples

In the previous section we completed the discussion on Area of sectors. In this section we will see some solved examples.

Solved example 21.21

Calculate the area of the green coloured portion in the fig.21.39 below

Fig.21.39

Solution:

1. Area of a sector with radius 3 cm and central angle 120^o

=  (^πr2⁄₃₆₀) × x = (^π×32⁄₃₆₀) × 120 = (^9π⁄₃) = 3π cm²

2. Area of a sector with radius 2 cm and central angle 120^o = (^π×22⁄₃₆₀) × 120 = (^4π⁄₃) cm²

3. Area of the green portion = (1) – (2) = 3π - ^4π⁄₃ = ^5π⁄₃ cm²

Solved example 21.22

Centred at each vertex of a regular hexagon, a part of a circle is drawn and a fig.b is cut out as shown below. What is the area of the shape in fig.b?

Fig.21.40

Solution:
1. We want the area of the shape in fig.b. Fig.c shows the details of how the shape in fig.b is obtained.
2. First, there is a regular hexagon of side 2 cm. At each corner of it, circles are drawn. So there are 6 circles. All of them have a radius of 1 cm
3. Any two adjacent sides of the hexagon will form a sector inside the circle. So the radius of the sectors will be 1 cm
4. The central angle of the sectors will be equal to the interior angle of a regular hexagon
5. The measure of each interior angle of a regular polygon is (^[180(n-2)]⁄_n). Where n is the number of sides.
6. We have a regular hexagon. It has 6 sides. So each interior angle is equal to:

(^[180(6-2)]⁄₆) = (^180×4⁄₆) = 120^o.
7. Now we can find the area of each sector:
(^πr2⁄₃₆₀) × x = (^π×12⁄₃₆₀) × 120 = (^π⁄₃) cm².
8. Total area of 6 sectors = 6 × (^π⁄₃) = 2π cm².
This much area is removed from the total area of the regular hexagon.
9. So we have to find the area of the regular hexagon. We did this in solved example 21.11
Total area of the regular hexagon = 6 × Area of one equilateral triangle of side 's'
Where 's' is the side of the regular hexagon)
10. We can find the area of any equilateral triangle, if we know the side. It is given by:
Area = (^√3⁄₄)s². Details here.
11. So total area of our regular hexagon = 6 × (^√3⁄₄)s² =  6 × (^√3⁄₄) × 2² = 6√3 cm².
12. So the area of the shape in fig.b = (11) - (8) = (6√3 - 2π) cm²

Solved example 21.23

In fig.21.41(a), two circles are drawn, each passing through the centre of the other. 

Fig.21.41

Calculate the area of the region which is common to both the circles.

Solution:

1. We are given fig.a. Let us add some more details to it. In the fig.b, names are given to important points:

• The centre of the left circle is A. The centre of the right circle is B

• The top point of intersection is C. The bottom point of intersection is D

2. Consider the circle with centre A on the left. The point B lies on this circle. But it is given that AB = 2 cm. Thus we get the radius of the left circle as 2 cm

3. Consider the circle with centre B on the right. The point A lies on this circle. But it is given that AB = 2 cm. Thus we get the radius of the right circle also as 2 cm

4. Points C and D lies on the circle with centre A. So we get AC = AD = 2 cm

5. Points C and D lies on the circle with centre B also. So we get BC = BD = 2 cm

6. So ΔABC and ΔABD are two equilateral triangles. These details are shown in fig.c

7. Consider the sector ABC in the left circle. It's boundaries are:

The minor arc BC, and the two radial lines AC and AB

8. The equilateral triangle ABC can be separated out from this sector. The remaining portion will be a curved region.

9. We know how to calculate the area of this curved region. We can use the equation:

■ Area after the removal of equilateral triangle from a sector of radius 'r' and central angle 60^o

= [(^π⁄₆) -  (^√3⁄₄)]r²  cm². Details here.

10. In our problem, r = 2 cm. So we get:

Area = [(^π⁄₆) -  (^√3⁄₄)]2² =  [(^π⁄₆) -  (^√3⁄₄)]×4

11. Four such areas are present. So the total curved area = [(^π⁄₆) -  (^√3⁄₄)]×4×4 = [(^π⁄₆) -  (^√3⁄₄)]×16

12. In addition, we have two equilateral triangles of side 2 cm. We can find the area of any equilateral triangle, if we know the side. It is given by: Area = (^√3⁄₄)s². Details here.
13. In our problem, s = 2 cm. So the area of two triangles 
= 2×(^√3⁄₄)2² = 2√3 cm². 
14. Required area = (11) + (13) 
= {[(^π⁄₆) -  (^√3⁄₄)]×16} + {2√3} = {^8π⁄₃ - 4√3} + {2√3} = {^8π⁄₃ - 2√3}

Solved example 21.24
The fig.21.42(a) shows three circles drawn with their centres on each vertex of an equilateral triangle and passing through the other two vertices.

Fig.21.42

Find the area common to the three circles
Solution:
1. We are given fig.a. Let us add a few more details:
2. The equilateral triangle is named as ΔABC. Circles are drawn centred on each vertex. The three circles have a common area.
3. This common area consists of:
• Three green coloured regions in fig.b 
• The equilateral triangle ABC
4. The three green regions are equal. So we need to find the area of only one.

5. Consider the sector ABC in the circle centred at A. It's boundaries are:

The minor arc BC, and the two radial lines AC and AB

6. The equilateral triangle ABC can be separated out from this sector. The remaining portion will be a curved region.

7. We know how to calculate the area of this curved region. We can use the equation:

■ Area after the removal of equilateral triangle from a sector of radius 'r' and central angle 60^o

= [(^π⁄₆) -  (^√3⁄₄)]r²  cm². Details here.

8. In our problem, r = 2 cm. So we get:

Area = [(^π⁄₆) -  (^√3⁄₄)]2² =  [(^π⁄₆) -  (^√3⁄₄)]×4

9. Three such areas are present. So the total curved area = [(^π⁄₆) -  (^√3⁄₄)]×4×3 = [(^π⁄₆) -  (^√3⁄₄)]×12
10. We can find the area of any equilateral triangle, if we know the side. It is given by: Area = (^√3⁄₄)s². Details here.
11. In our problem, s = 2 cm. So the area of ΔABC
= (^√3⁄₄)2² = √3 cm². 
12. Required area = (9) + (11) 
= {[(^π⁄₆) -  (^√3⁄₄)]×12} + {√3} = {2π - 3√3} + {√3} = {2π - 2√3} = 2(π-√3) cm²

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We have completed the discussion on Area of sectors. In the next section we will see some solved examples in general from this chapter.

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Chapter 21.8 - Area of Sector

In the previous section we completed the discussion on Length of arcs. In this section we will see Area of sectors.

In fig.21.35 below, AB is an arc. We can see two radii: OA and OB. They are drawn from the two ends of the arc towards the centre of the circle. 

Fig.21.35

■ The area enclosed by an arc, and two radii is called a sector. It's symbol is ⌔. So, in fig.21.35, OAB is a sector. Let us find the method to calculate the area of a sector.

• Look at fig.21.36 below. The circle with radius ‘r’ is divided into four equal parts.

Fig.21.36

So P, Q, R and S are the quadrant points. Let us see four cases:

Case 1:

1. In fig.a, the area of sector OPQ is one fourth of the total area of the 'circle with radius r cm'.

2. We know that the total area = πr². So area of the sector OPQ = (¹⁄₄ × πr²)

3. The central angle of this sector is 90^o

4. So we can write: 90^o → (¹⁄₄ × πr²) 

5. So 1^o → [(¹⁄₄ × πr²) ÷ 90]

6. [(¹⁄₄ × πr²) ÷ 90] = ^πr2⁄_(4×90) = ^πr2⁄₃₆₀

7. So 1^o → ^πr2⁄₃₆₀. That means, If the central angle of a sector with radius 'r' is 1^o, then it's area will be equal to ^πr2⁄₃₆₀ cm²  

8. Another form of writing this is: In a circle of radius 'r' cm, every 1^o will give a sector of area ^πr2⁄₃₆₀ cm².

Case 2:

1. In fig.b, the area of sector OPQR is one half of the total area of the 'circle with radius r cm'.

2. We know that the total area = πr². So area of the sector OPQR = (¹⁄₂ × πr²)

3. The central angle of this sector is 180^o

4. So we can write: 180^o → (¹⁄₂ × πr²) 

5. So 1^o → [(¹⁄₂ × πr²) ÷ 180]

6. [(¹⁄₂ × πr²) ÷ 180] = ^πr2⁄_(2×180) = ^πr2⁄₃₆₀

7. So 1^o → ^πr2⁄₃₆₀. That means, If the central angle of a sector with radius 'r' is 1^o, then it's area will be equal to ^πr2⁄₃₆₀ cm²  

8. Another form of writing this is: In a circle of radius 'r' cm, every 1^o will give a sector of area ^πr2⁄₃₆₀ cm².

Case 3:

1. In fig.c, the area of sector OPQRS is three fourth of the total area of the 'circle with radius r cm'.

2. We know that the total area = πr². So area of the sector OPQRS = (³⁄₄ × πr²)

3. The central angle of this sector is 270^o

4. So we can write: 270^o → (¹⁄₂ × πr²) 

5. So 1^o → [(³⁄₄ × πr²) ÷ 270]

6. [(³⁄₄ × πr²) ÷ 270] = ^3πr2⁄_(4×270) = ^πr2⁄₃₆₀

7. So 1^o → ^πr2⁄₃₆₀. That means, If the central angle of a sector with radius 'r' is 1^o, then it's area will be equal to ^πr2⁄₃₆₀ cm²  

8. Another form of writing this is: In a circle of radius 'r' cm, every 1^o will give a sector of area ^πr2⁄₃₆₀ cm².

Case 4:

1. In fig.d, the area of sector OPQRSP is one full of the total area of the 'circle with radius r cm'.

2. We know that the total area = πr². So area of the sector OPQRSP = πr²

3. The central angle of this sector is 360^o

4. So we can write: 360^o → πr² 

5. So 1^o → [(πr²) ÷ 360]

6. [(πr²) ÷ 360] = ^πr2⁄₃₆₀

7. So 1^o → ^πr2⁄₃₆₀. That means, If the central angle of a sector with radius 'r' is 1^o, then it's area will be equal to ^πr2⁄₃₆₀ cm²  

8. Another form of writing this is: In a circle of radius 'r' cm, every 1^o will give a sector of area ^πr2⁄₃₆₀ cm².
■ In all the four cases, we get the same result. We can write it in the form of a theorem:

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Theorem 21.3
• The radius of a circle is 'r' cm
• Every 1^o central angle in this circle will give a sector of area (^πr2⁄₃₆₀) cm².
This is shown in the fig.21.37 below:

Fig.21.37

• So the area of a sector with radius 'r' cm, and central angle x^o will be equal to
(^πr2⁄₃₆₀) × x = (^πr2x⁄₃₆₀)cm²

---

A sample calculation:
■ Area of a sector of central angle 120^o in a circle of radius 3 cm:
• Given: r = 3 cm, and central angle x = 120^o
• Area of the sector = (^π×32⁄₃₆₀) × 120 = (^9π⁄₃) = 3π cm²

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Now we will see a special case: Sectors with central angle 60^o.
Consider the minor arc AB in fig.21.38(a) below:

Fig.21.38

• The points A and B are joined by a straight line. Then we get a separate area:
• The area enclosed between the minor arc AB and the straight line AB. This area is shown in blue colour. We want to calculate this area. We can proceed as follows:
1. Consider the whole sector OAB (fig.b). It's area can be easily calculated:
(^πr2⁄₃₆₀) × 60 = (^πr2⁄₆) cm²
2. Now join A and B by a straight line (fig.c). Now we have a triangle OAB. What is the peculiarity of this triangle?
• OA = OB = r. So it is an isosceles triangle
3. In isosceles triangles, base angles are equal. So ∠A = ∠B 
• Since they are equal, we will put ∠A = ∠B = x^o
4. Sum of the three interior angles in any triangle is 180^o. So we can write:
60 + x + x = 180 ⇒ 60 +2x = 180 ⇒ 2x = 120 ⇒ x = 60^o
5. So we get ∠A = ∠B = 60^o. So all the three angles are 60^o. It is an equilateral triangle
6. We can find the area of any equilateral triangle, if we know the side. Details here.
So area of ΔOAB = (^√3⁄₄)r².
7. Area of the blue region = Area of ⌔ OAB - Area of ΔOAB = [(^πr2⁄₆) - (^√3⁄₄)r²] 
= [(^π⁄₆) -  (^√3⁄₄)]r²  cm².

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In the next section we will see some solved examples.

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Chapter 21.7 - Length of Arc - Solved examples

In the previous section we have seen the central angle of an arc for each 1 cm length. In this section we will see some solved examples.

Solved example 21.16

In a circle, the length of an arc is 3π cm. The central angle of this arc is 40^o. What is the perimeter of the circle? What is it's radius?
Solution:

1. Both values of the arc is given to us:
• Length of arc = 3π cm
• Central angle of arc = 40^o
2. We need an equation which gives the relation between the two. We can use theorem 21.1
According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
3. So for 40^o, the length of arc will be (^πr⁄₁₈₀) × 40 = (^πr⁄_4.5) cm
4. We can equate it to the given length. So we get: (^πr⁄_4.5) = 3π cm ⇒ r = 3 × 4.5 = 13.5 cm
5. Perimeter = 2πr = 2 × 13.5 × π =  27π cm

Solved example 21.17
In a circle, the length of an arc is 4 cm. It's central angle is 25^o
(i) In the same circle, what is the length of an arc whose central angle is 75^o?
(ii) In a circle of radius one and a half times the radius of this circle, what is the length of an arc whose central angle is 75^o?
Solution:
1. Both values of the arc is given to us:
• Length of arc = 4 cm
• Central angle of arc = 25^o
2. We need an equation which gives the relation between the two. We can use theorem 21.1
According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
3. So for 25^o, the length of arc will be (^πr⁄₁₈₀) × 25 = (^πr⁄_7.2) cm
4. We can equate it to the given length. So we get: (^πr⁄_7.2) = 4 cm ⇒ πr = 4 × 7.2 = 28.8 cm
• We need not divide 28.8 by π to get the actual radius because, we will be using 'πr' as a whole in our calculations.
Part (i)
1. We use theorem 21.1 again:
For 75^o, the length of arc =  (^πr⁄₁₈₀) × 75 = (^28.8⁄₁₈₀) × 75 = 12 cm
Part (ii)
1. The radius is one and a half times. So new πr = 28.8 ×1.5 = 43.2 cm
For 75^o, the length of arc =  (^πr⁄₁₈₀) × 75 = (^43.2⁄₁₈₀) × 75 = 18 cm

Solved example 21.18
From a bangle of radius 3 cm, a piece is to be cut out to make a ring of radius ¹⁄₂ cm. 
(i) What is the central angle of the piece to be cut out?
(ii) The remaining part of the bangle was bent to make a smaller bangle. What is it's radius?
Solution:
1. When a piece is cut out from a bangle, it would be an arc. What is the length of this arc?
Ans: Enough to make a ring of radius ¹⁄₂ cm 
2. So, if r is the radius of the ring, it's perimeter 2πr must be equal to the length of the arc cut out
3. But radius of the ring is given as ¹⁄₂ cm. So length of cut out arc = 2π × ¹⁄₂ = π cm
4. So we have the length of arc. From that, we need to find the central angle. We can use theorem 21.2.
5. For every 1 cm of an arc on a circle of radius r, the central angle will be (¹⁸⁰⁄_πr)^o
6. So for π cm, the angle will be (¹⁸⁰⁄_πr) × π = (¹⁸⁰⁄_r)^o .
7. But r is given as 3 cm. So angle = (¹⁸⁰⁄₃) = 60^o. This is the answer for part (i)
8. π cm is cut out. So remaining arc length = 2πr - π = (2r-1)π = (2×3 - 1)π = 5π
9. So the smaller bangle is made using an arc of length 5π. That means, perimeter of the smaller bangle = 5π
10. Let r₁ be the radius of the smaller bangle. Then it's perimeter = 2πr₁
11. So we get: 2πr₁= 5π ⇒ r₁ = 2.5 cm. This is the answer for part (ii)

Solved example 21.19

In fig.21.33(a) below, parts of a circle centred at each vertex of an equilateral triangle, and passing through the other two vertices is shown. 

Fig.21.33

What is the perimeter of this fig.?

Solution:

1. In fig.21.33(b), more details are added. ABC is the equilateral triangle. 

2. A circle is drawn centred at each vertex. What is the radius of those circles?

Ans: Any of the above three circles centred on a vertex, passes through the other two vertices. For example, in the fig.b, the circle centred at A passes through B and C.

Also, it is an equilateral triangle. So radius of any circle is the side of the triangle, and is equal to 4 cm.

3. An arc is taken out between the 'other two vertices', from each circle. 

4. For an equilateral triangle, all three angles are 60^o

5. So we have 3 equal arcs

• Each of them have a radius of 4 cm. (∵ they are part of a circle with 4 cm radius)

• Each of them have central angle 60^o.  

6. We want the length of these arcs. We can use  theorem 21.2.

According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
7. So for 60^o, the length of arc will be (^π×4⁄₁₈₀) × 60 = (⁴⁄₃)π cm
8. Thus the total perimeter = 3×(⁴⁄₃)π = 4π cm

Solved example 21.20

Parts of a circle are drawn, centred at each vertex of a regular octagon, and a fig. is cut out as in fig.21.34(b) below. Calculate the perimeter of fig.b

Fig.21.34

Solution:
1. The measure of each interior angle of a regular polygon is (^[180(n-2)]⁄_n). Where n is the number of sides.
2. We are given a regular octagon. It has 8 sides. So each interior angle is equal to:
(^[180(8-2)]⁄₈) = (^180×6⁄₈) = 135^o.
3. We want to find the perimeter in fig.b. It consists of equal arcs. How are these arcs formed?
We get the answer from fig.c
Equal circles are centred at each vertex. Then the portions outside the octagon are removed.
Radius of each circle is 1 cm

4. So we have 8 equal arcs

• Each of them have a radius of 1 cm. (∵ they are part of a circle with 1 cm radius)

• Each of them have central angle 135^o.  

5. We want the length of these arcs. We can use  theorem 21.2.

According to the theorem, For every 1^o central angle, the length of arc will be (^πr⁄₁₈₀) cm 
6. So for 135^o, the length of arc will be (^π×1⁄₁₈₀) × 135 = (³⁄₄)π cm
7. Thus the total perimeter = 8×(³⁄₄)π = 6π cm

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In the next section we will see Area of Sector.

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