In the previous section we saw the details about √3. We also saw some solved examples. In this section, we will see more solved examples.
Solved example 16.4
Any odd number can be written as the difference of two perfect squares. Using this property, draw the lines of lengths (i) √7 cm and (ii) √11 cm
Solution:
Let us first write some examples which demonstrate the property.
• 3 = 22 - 12 = 4 - 1 • 5 = 32 - 22 = 9 - 4 • 7 = 42 - 32 = 16 - 9 • 9 = 52 - 42 = 25 - 16
In the above series, we have obtained 7. We can continue like this until we get 11. But that will not be necessary. There is a definite formula to find the squares: If 'n' is the number,
• the first square will be the square of (n+1) ⁄2
• the second square will be the square of (n-1) ⁄2
• So for 11, the first square will be the square of (11+1) ⁄2 = 12⁄2 = 6
• the second square will be the square of (11-1) ⁄2 = 10⁄2 = 5
• Thus we get 11 = 62 - 52 = 36 - 25
• So we can solve the problem using the following two information:
♦ 7 = 42 - 32 ♦ 11 = 62 - 52
• Consider 7 = 42 - 32 . Taking square root on both sides, we get: √7 = √(42 - 32)
• It is clear that √7 is one leg of a right triangle, whose other leg is 3, and hypotenuse is 4
The above derivation can be explained as follows:
• Using Pythagoras theorem, hypotenuse2 = (leg1)2 + (leg2)2 ⇒ (leg1)2 = hypotenuse2 - (leg2)2
• Taking square root on both sides we get: leg1 = √[hypotenuse2 - (leg2)2]
• So if we construct a right triangle with hypotenuse 4 cm, and one leg 3 cm, the other leg will be equal to √7 cm. This is shown in the fig.16.13(a) below. We have learned about such construction here.
• In the fig.16.13(a), AB represents √7 cm
• Similarly, if we construct a right triangle with hypotenuse 6 cm, and one leg 5 cm, the other leg will be equal to √11 cm. This is shown in the fig.16.13(b). In the fig., XY represents √11 cm
Solved example 16.5
Explain two different methods for drawing a line of length √13 cm
Solution:
Method 1: We can use the same method that we saw in the previous example.
• For 13, the first square will be the square of (13+1) ⁄2 = 14⁄2 = 7
• the second square will be the square of (13-1) ⁄2 = 12⁄2 = 6
• Thus we get 13 = 72 - 62 = 49 - 36
The triangle is shown in fig.16.14(a) below. In the fig., PQ represents √13
The second method is shown in fig.b. In this method, we draw a series of right triangles, each above the previous. The hypotenuse OE of the last triangle is equal to √13. The steps are as follows:
1. Draw ⊿OAB, with OA = 3 cm, and AB = 1 cm
2. The the hypotenuse OB = √(OA2 + AB2) = √(32 + 12) = √(9 + 1) = √10
3. Draw ⊿OBC with OB as base and BC = 1 cm
4. The the hypotenuse OC = √(OB2 + BC2) = √[(√10)2 + 12] = √[10 + 1] = √11
5. Draw ⊿OCD with OC as base and CD = 1 cm
6. The the hypotenuse OD = √(OC2 + CD2) = √[(√11)2 + 12] = √[11 + 1] = √12
7. Draw ⊿ODE with OD as base and DE = 1 cm
8. The the hypotenuse OE = √(OD2 + DE2) = √[(√12)2 + 12] = √[12 + 1] = √13
PQ in fig.(a), and OE in fig.(b) represents the same length √13. This can be checked using divider, as shown in the animation in fig.16.16 below:
Solved example 16.16
Find 3 fractions, larger than √2 and smaller than √3
Solution:
1. Consider the equation: √2 × √2 = 2
2. Similarly we have: √3 × √3 = 3
3. 1.5 × 1.5 = 2.25
4. Comparing (1) and (3), we get 1.5 > √2 (∵ 2.25>2)
5. Comparing (2) and (3), we get 1.5 < √3 (∵ 2.25<3)
6. So we can write: √2 < 1.5 < √3. That is., 1.5 is a quantity, that lies in between √2 and √3
7. Similarly, 1.6 also lies between √2 and √3. (∵ 1.6 ×1.6 = 2.56, and this 2.56 lies in between (√2 × √2) and (√3 × √3)
8. Similarly, 1.7 also lies between √2 and √3. (∵ 1.7 ×1.7 = 2.89, and this 2.89 lies in between (√2 × √2) and (√3 × √3)
9. So we can write: √2 < 1.5 < 1.6 < 1.7 < √3. That is., the quantities 1.5, 1.6 and 1.7 lies in between √2 and √3.
The result in (9) can be pictorially represented as:
√2 × √2 = 2
1.5 × 1.5 = 2.25
1.6 ×1.6 = 2.56
1.7 ×1.7 = 2.89
√3 × √3 = 3
10. Now we write 1.5, 1.6 and 1.7 as fractions:
•1.5 = 15⁄10 ; •1.6 = 16⁄10 ; and •1.7 = 17⁄10
11. Thus, the required three fractions are 15⁄10, 16⁄10 and 17⁄10
So we have seen the basics about irrational numbers. In the next section, we will see their addition and subtraction.
Solved example 16.4
Any odd number can be written as the difference of two perfect squares. Using this property, draw the lines of lengths (i) √7 cm and (ii) √11 cm
Solution:
Let us first write some examples which demonstrate the property.
• 3 = 22 - 12 = 4 - 1 • 5 = 32 - 22 = 9 - 4 • 7 = 42 - 32 = 16 - 9 • 9 = 52 - 42 = 25 - 16
In the above series, we have obtained 7. We can continue like this until we get 11. But that will not be necessary. There is a definite formula to find the squares: If 'n' is the number,
• the first square will be the square of (n+1) ⁄2
• the second square will be the square of (n-1) ⁄2
• So for 11, the first square will be the square of (11+1) ⁄2 = 12⁄2 = 6
• the second square will be the square of (11-1) ⁄2 = 10⁄2 = 5
• Thus we get 11 = 62 - 52 = 36 - 25
• So we can solve the problem using the following two information:
♦ 7 = 42 - 32 ♦ 11 = 62 - 52
• Consider 7 = 42 - 32 . Taking square root on both sides, we get: √7 = √(42 - 32)
• It is clear that √7 is one leg of a right triangle, whose other leg is 3, and hypotenuse is 4
The above derivation can be explained as follows:
• Using Pythagoras theorem, hypotenuse2 = (leg1)2 + (leg2)2 ⇒ (leg1)2 = hypotenuse2 - (leg2)2
• Taking square root on both sides we get: leg1 = √[hypotenuse2 - (leg2)2]
• So if we construct a right triangle with hypotenuse 4 cm, and one leg 3 cm, the other leg will be equal to √7 cm. This is shown in the fig.16.13(a) below. We have learned about such construction here.
 |
| Fig.16.13 |
• Similarly, if we construct a right triangle with hypotenuse 6 cm, and one leg 5 cm, the other leg will be equal to √11 cm. This is shown in the fig.16.13(b). In the fig., XY represents √11 cm
Solved example 16.5
Explain two different methods for drawing a line of length √13 cm
Solution:
Method 1: We can use the same method that we saw in the previous example.
• For 13, the first square will be the square of (13+1) ⁄2 = 14⁄2 = 7
• the second square will be the square of (13-1) ⁄2 = 12⁄2 = 6
• Thus we get 13 = 72 - 62 = 49 - 36
The triangle is shown in fig.16.14(a) below. In the fig., PQ represents √13
 |
| Fig.16.15 |
1. Draw ⊿OAB, with OA = 3 cm, and AB = 1 cm
2. The the hypotenuse OB = √(OA2 + AB2) = √(32 + 12) = √(9 + 1) = √10
3. Draw ⊿OBC with OB as base and BC = 1 cm
4. The the hypotenuse OC = √(OB2 + BC2) = √[(√10)2 + 12] = √[10 + 1] = √11
5. Draw ⊿OCD with OC as base and CD = 1 cm
6. The the hypotenuse OD = √(OC2 + CD2) = √[(√11)2 + 12] = √[11 + 1] = √12
7. Draw ⊿ODE with OD as base and DE = 1 cm
8. The the hypotenuse OE = √(OD2 + DE2) = √[(√12)2 + 12] = √[12 + 1] = √13
PQ in fig.(a), and OE in fig.(b) represents the same length √13. This can be checked using divider, as shown in the animation in fig.16.16 below:
 |
| Fig.16.16 |
Find 3 fractions, larger than √2 and smaller than √3
Solution:
1. Consider the equation: √2 × √2 = 2
2. Similarly we have: √3 × √3 = 3
3. 1.5 × 1.5 = 2.25
4. Comparing (1) and (3), we get 1.5 > √2 (∵ 2.25>2)
5. Comparing (2) and (3), we get 1.5 < √3 (∵ 2.25<3)
6. So we can write: √2 < 1.5 < √3. That is., 1.5 is a quantity, that lies in between √2 and √3
7. Similarly, 1.6 also lies between √2 and √3. (∵ 1.6 ×1.6 = 2.56, and this 2.56 lies in between (√2 × √2) and (√3 × √3)
8. Similarly, 1.7 also lies between √2 and √3. (∵ 1.7 ×1.7 = 2.89, and this 2.89 lies in between (√2 × √2) and (√3 × √3)
9. So we can write: √2 < 1.5 < 1.6 < 1.7 < √3. That is., the quantities 1.5, 1.6 and 1.7 lies in between √2 and √3.
The result in (9) can be pictorially represented as:
√2 × √2 = 2
1.5 × 1.5 = 2.25
1.6 ×1.6 = 2.56
1.7 ×1.7 = 2.89
√3 × √3 = 3
10. Now we write 1.5, 1.6 and 1.7 as fractions:
•1.5 = 15⁄10 ; •1.6 = 16⁄10 ; and •1.7 = 17⁄10
11. Thus, the required three fractions are 15⁄10, 16⁄10 and 17⁄10
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