In the previous section we saw theorem 27.4 and it's converse. We also saw a solved example. In this section we will see a few more solved examples.
Solved example 27.3
How do we draw a triangle with angles 40o, 60o and 80o within a circle of 2.5 cm radius?
Solution:
1. First step is to draw a rough sketch. This is shown in fig.27.19(a) below.
• Let A, B and C be the three vertices. All of them lie on a circle
2. Join A and B to the centre O
3. Separate out the minor arc AB. This is shown in fig(b)
4. Now we have an arc AB which subtends ∠ACB on the alternate arc.
5. ∠ACB = 80o. So, using the converse of theorem 27.4, we get: ∠AOB = 2 × 80 = 160. This is marked in fig(b)
6. Again separate out the minor arc BC. This is shown in fig(c)
7. Now we have an arc BC which subtends ∠BAC on the alternate arc.
8. ∠BAC = 40o. So, using the converse of theorem 27.4, we get: ∠BOC = 2 × 40 = 80. This is marked in fig(c)
9. So now we have the angles between the radial lines from centre O. We can do the actual construction.
10. At any convenient point O, draw a circle of radius 2.5 cm
11. Draw a radial line OA in any convenient direction. Draw OB such that ∠OAB is 160o. This is shown in fig.27.20 below:
11. Next draw OC such that ∠BOC is 80o.
12. Thus the three vertices are obtained. Draw AB, BC and CA to get the required ΔABC.
Solved example 27.4
In the fig 27.21(a) below, what percentage of the total circumference is the length of the arc ADB
Solution:
1. Separate out the arc ADB as shown in fig.27.21(b)
2. This arc ADB subtends ∠ACB at point B on the alternate arc
3. ∠ACB = 60o. So, using the converse of theorem 27.4, we get: ∠AOB = 2 × 60 = 120. This is marked in fig(b)
4. So we have the central angle of ADB. If we assume the radius of the circle to be 'r' cm, we can find the length of arc ADB. We can use theorem 21.1
5. So length of arc ADB = πr⁄180 ×120 = 2πr⁄3 . .
6. The whole circumference = 2πr
7. Now we take the ratio:
Length of arc ADB⁄Whole circumference = 2πr⁄3 ÷ 2πr = 2πr⁄3 × 1⁄2πr = 1⁄3 = 33.33%
8. So length of the arc ADB is 33.33% of the whole circumference
Solved example 27.5
What is the radius of the circle shown in fig.27.22(a) below
Solution:
1. Separate out the arc ABC. This is shown in fig(b)
2. Consider ⊿ABC. Vertices A and C are on the circle. Angle at B is given as 90o.
• So by theorem 27.1, Arc ABC will be a semicircle.
• That means, AC is the diameter. So midpoint of AC will be the centre O of the circle
• If we find the length of OA or OC, we will have the radius
• So our aim is to find OA or OC. For that, the following steps can be used:
3. Angle at A = [180 - (90+45)] = [180 -90 -45] = 45o.
4. So the base angles of ⊿ABC are equal. It is an isosceles triangle
Thus we get: AB = BC = 3 cm
5. So we have the two legs of ⊿ABC. Applying Pythagoras theorem, we get:
AC2 = AB2 + BC2 ⇒ AC2 = 32 + 32 ⇒ AC2 = 2 × 32 ⇒ AC2 = 2 × 9 ⇒ AC = 3√2
6. So OA = OC = AC⁄2 = 3√2⁄2 = 1.5√2
Solved example 27.6
What is the area of the circle shown in fig.27.23(a) below
Solution:
1. We can use a standard result that we derived in the previous section. let us write it's steps:
Solved example 27.3
How do we draw a triangle with angles 40o, 60o and 80o within a circle of 2.5 cm radius?
Solution:
1. First step is to draw a rough sketch. This is shown in fig.27.19(a) below.
• Let A, B and C be the three vertices. All of them lie on a circle
 |
| Fig.27.19 |
3. Separate out the minor arc AB. This is shown in fig(b)
4. Now we have an arc AB which subtends ∠ACB on the alternate arc.
5. ∠ACB = 80o. So, using the converse of theorem 27.4, we get: ∠AOB = 2 × 80 = 160. This is marked in fig(b)
6. Again separate out the minor arc BC. This is shown in fig(c)
7. Now we have an arc BC which subtends ∠BAC on the alternate arc.
8. ∠BAC = 40o. So, using the converse of theorem 27.4, we get: ∠BOC = 2 × 40 = 80. This is marked in fig(c)
9. So now we have the angles between the radial lines from centre O. We can do the actual construction.
10. At any convenient point O, draw a circle of radius 2.5 cm
11. Draw a radial line OA in any convenient direction. Draw OB such that ∠OAB is 160o. This is shown in fig.27.20 below:
 |
| Fig.27.20 |
11. Next draw OC such that ∠BOC is 80o.
12. Thus the three vertices are obtained. Draw AB, BC and CA to get the required ΔABC.
Solved example 27.4
In the fig 27.21(a) below, what percentage of the total circumference is the length of the arc ADB
 |
| Fig.27.21 |
1. Separate out the arc ADB as shown in fig.27.21(b)
2. This arc ADB subtends ∠ACB at point B on the alternate arc
3. ∠ACB = 60o. So, using the converse of theorem 27.4, we get: ∠AOB = 2 × 60 = 120. This is marked in fig(b)
4. So we have the central angle of ADB. If we assume the radius of the circle to be 'r' cm, we can find the length of arc ADB. We can use theorem 21.1
5. So length of arc ADB = πr⁄180 ×120 = 2πr⁄3 . .
6. The whole circumference = 2πr
7. Now we take the ratio:
Length of arc ADB⁄Whole circumference = 2πr⁄3 ÷ 2πr = 2πr⁄3 × 1⁄2πr = 1⁄3 = 33.33%
8. So length of the arc ADB is 33.33% of the whole circumference
Solved example 27.5
What is the radius of the circle shown in fig.27.22(a) below
 |
| Fig.27.22 |
1. Separate out the arc ABC. This is shown in fig(b)
2. Consider ⊿ABC. Vertices A and C are on the circle. Angle at B is given as 90o.
• So by theorem 27.1, Arc ABC will be a semicircle.
• That means, AC is the diameter. So midpoint of AC will be the centre O of the circle
• If we find the length of OA or OC, we will have the radius
• So our aim is to find OA or OC. For that, the following steps can be used:
3. Angle at A = [180 - (90+45)] = [180 -90 -45] = 45o.
4. So the base angles of ⊿ABC are equal. It is an isosceles triangle
Thus we get: AB = BC = 3 cm
5. So we have the two legs of ⊿ABC. Applying Pythagoras theorem, we get:
AC2 = AB2 + BC2 ⇒ AC2 = 32 + 32 ⇒ AC2 = 2 × 32 ⇒ AC2 = 2 × 9 ⇒ AC = 3√2
6. So OA = OC = AC⁄2 = 3√2⁄2 = 1.5√2
Solved example 27.6
What is the area of the circle shown in fig.27.23(a) below
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| Fig.27.23 |
1. We can use a standard result that we derived in the previous section. let us write it's steps:
(i) Consider any rectangle in which, all the four corners lie on a circle
(ii) Then both the diagonals of that rectangle will be diameters of the circle
(iii) So the centre of the circle will bisect both the diagonals.
(iii) So the centre of the circle will bisect both the diagonals.
2. So the diagonal AC will be a diameter of the circle
3. AC will split the rectangle into two right angled triangles: ⊿ABC and ⊿ADC
4. Consider ⊿ADC. Hypotenuse is not given.
• But one leg is 3 cm and the other leg is 4 cm. So it is a 3,4,5 triangle
• That is., hypotenuse of ⊿ADC is 5 cm
5. The centre of the circle bisects the diagonal.
So radius of the circle = AC⁄2 = 5⁄2 = 2.5
6. So area of the circle = πr2 = π × 2.52 = 6.25π.
Solved example 27.7
How do we draw a triangle with two of the angles 40o and 120o within a circle of 3 cm radius?
Solution:
1. First step is to draw a rough sketch. This is shown in fig.27.24(a) below.
• Let A, B and C be the three vertices. All of them lie on a circle
• The third angle can be calculated as: [180 - (120+40)] = [180 -120-40] = 20o. But we don't really need it
2. Join A and B to the centre O
3. Separate out the minor arc AB. This is shown in fig(b)
4. Now we have an arc AB which subtends ∠ACB on the alternate arc.
5. ∠ACB = 120o. So, using the converse of theorem 27.4, we get: ∠AOB = 2 × 120 = 240. This is marked in fig(b)
6. Again separate out the minor arc BC. This is shown in fig(c)
7. Now we have an arc BC which subtends ∠BAC on the alternate arc.
8. ∠BAC = 40o. So, using the converse of theorem 27.4, we get: ∠BOC = 2 × 40 = 80. This is marked in fig(c)
9. So now we have the angles between the radial lines from centre O. We can do the actual construction.
10. At any convenient point O, draw a circle of radius 3 cm
11. Draw a radial line OA in any convenient direction. Draw OB such that ∠OAB is 240o. This is shown in fig.27.25 below:
11. Next draw OC such that ∠BOC is 80o.
12. Thus the three vertices are obtained. Draw AB, BC and CA to get the required ΔABC.
3. AC will split the rectangle into two right angled triangles: ⊿ABC and ⊿ADC
4. Consider ⊿ADC. Hypotenuse is not given.
• But one leg is 3 cm and the other leg is 4 cm. So it is a 3,4,5 triangle
• That is., hypotenuse of ⊿ADC is 5 cm
5. The centre of the circle bisects the diagonal.
So radius of the circle = AC⁄2 = 5⁄2 = 2.5
6. So area of the circle = πr2 = π × 2.52 = 6.25π.
Solved example 27.7
How do we draw a triangle with two of the angles 40o and 120o within a circle of 3 cm radius?
Solution:
1. First step is to draw a rough sketch. This is shown in fig.27.24(a) below.
• Let A, B and C be the three vertices. All of them lie on a circle
• The third angle can be calculated as: [180 - (120+40)] = [180 -120-40] = 20o. But we don't really need it
 |
| Fig.27.24 |
3. Separate out the minor arc AB. This is shown in fig(b)
4. Now we have an arc AB which subtends ∠ACB on the alternate arc.
5. ∠ACB = 120o. So, using the converse of theorem 27.4, we get: ∠AOB = 2 × 120 = 240. This is marked in fig(b)
6. Again separate out the minor arc BC. This is shown in fig(c)
7. Now we have an arc BC which subtends ∠BAC on the alternate arc.
8. ∠BAC = 40o. So, using the converse of theorem 27.4, we get: ∠BOC = 2 × 40 = 80. This is marked in fig(c)
9. So now we have the angles between the radial lines from centre O. We can do the actual construction.
10. At any convenient point O, draw a circle of radius 3 cm
11. Draw a radial line OA in any convenient direction. Draw OB such that ∠OAB is 240o. This is shown in fig.27.25 below:
 |
| Fig.27.25 |
12. Thus the three vertices are obtained. Draw AB, BC and CA to get the required ΔABC.
In the next section, we will see a few more solved examples.
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