In the previous section we completed the discussion on the basics of irrational numbers. We also saw some solved examples. In this section, we will see their addition and subtraction.
Consider ⊿OAB in fig.16.17(a) below:
• Lengths of both the legs AB and AO is 1 unit. So the hypotenuse OB will be equal to √2 units
• Now, what is the perimeter of ⊿OAB?
• Let us add the sides: Perimeter = OA + AB + BO = 1 + 1 + √2 = 2 + √2
• We have seen that √2 = 1.414213... So, if the unit is meter, we can write:
■ Perimeter of ⊿OAB = 2 + √2 = 2 + 1.41 = 3.41 m (Correct to 1 cm)
■ Perimeter of ⊿OAB = 2 + √2 = 2 + 1.414 = 3.414 m (Correct to 1 mm)
• Now, what is the perimeter of ⊿OBC?
• Let us add the sides: Perimeter = 1 + √2 + √3
• We have seen that √2 = 1.414213... , and √3 = 1.732050... So, if the unit is meter, we can write:
■ Perimeter of ⊿OBC - Perimeter of ⊿OAB = √3 - 1 = 1.732 - 1 = 0.732 (Correct to 1 mm)
The results are same as those obtained earlier.
• Now, what is the perimeter of ⊿OCD?
• Let us add the sides: Perimeter = 1 + √3 + 2 = 3 + √3
• We have seen that √3 = 1.732050... So, if the unit is meter, we can write:
■ Perimeter of ⊿OCD = 3 + √3 = 3 + 1.73 = 4.73 m (Correct to 1 cm)
■ Perimeter of ⊿OCD = 3 + √3 = 3 + 1.732 = 4.732 m (Correct to 1 mm)
• How much is the increase in perimeter from the second triangle? We can write:
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = 4.73 - 4.14 = 0.59 (Correct to 1 cm)
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = 4.732 - 4.146 = 0.586 (Correct to 1 mm)
Let us find the increase using the easy method:
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = (3 + √3) - (1+ √2 + √3) = 3 + √3 - 1 - √2 - √3 = 2 - √2
Now we can convert to decimal form. We get:
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = 2 - √2 = 2 - 1.41 = 0.59 (Correct to 1 cm)
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = 2 - √2 = 2 - 1.414 = 0.586 (Correct to 1 mm)
The results are same as those obtained earlier.
The hypotenuse of a right triangle is 11⁄2 m, and one of it's leg is 1⁄2 m. Calculate the perimeter correct to one cm
Solution:
Given: Hypotenuse = 11⁄2 m = 3⁄2 m and, one leg = 1⁄2 m. Fig.16.18 below shows a rough sketch:
• 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 we get: Other leg = √[(3⁄2)2 - (1⁄2)2] = √[(9⁄4) - (1⁄4)] = √(8⁄4) = √2
■ Thus perimeter = 3⁄2 + 1⁄2 + √2 = 4⁄2 + √2 = 2 + √2
We want the perimeter correct to 1 cm. So we take the value of √2 correct to 1 cm. That is., we take √2 = 1.41 m
Thus perimeter = 2 + √2 = 2 + 1.41 = 3.41 m
Solved example 16.8
The fig.16.19(a) shows an equilateral triangle cut into halves by a line through the vertex.
(i) What is the perimeter of each half?
(ii) what is the difference between the 'perimeter of one half triangle' and the 'perimeter of the whole triangle'?
Solution:
1. Let us name the whole triangle as ABC. CD cuts it into two equal halves. This is shown in fig.b
2. Given that ABC is an equilateral triangle. So AB = 2 m, and, AD = BD = 1m
3. Also, both ADC and BDC are right angled
4. Applying Pythagoras theorem to ⊿ADC, we get: CD = √[AC2 - AD2] = √[22 - 12] = √(4 - 1) = √3
5. So perimeter of ⊿ADC = AD + AC + CD = 1 + 2 + √3 = (3 + √3) m
6. Let us take √3 = 1.73 m. Then Perimeter = 3 + √3 = 3 + 1.73 = 4.73 m
7. Perimeter of original triangle ABC = 2 + 2 + 2 = 6 m
8. Difference in perimeter = 6 - (3 + √3) = 6 - 3 - √3 = 3 - √3
9. So difference = 3 - √3 = 3 - 1.73 = 1.27 m
Solved example 16.9
Calculate the perimeter of the triangle shown in fig.16.20 below:
Solution:
1. Let us name the triangle as ABC. This is shown in fig.b
2. Through the vertex B, a perpendicular line BD is drawn to AC. So ∠ADB is 90o
3. Consider ⊿ABD. We know two of it's angles: ∠DAB = 30o and ∠ADB = 90o
4. So ∠ABD = 60o (∵ sum of interior angles of any triangle is 180o)
5. Thus we find that ⊿ABD is a 30, 60, 90 triangle. We have seen the details here. In such a triangle, if we know any one side, the other two sides can be easily calculated.
6. We know the side AB as 2 m. This is the longest side, the hypotenuse, which is opposite the largest angle 90o
7. The longest side is twice the smallest side. Here, the smallest side is BD, as it is opposite the smallest angle 30o
8. So we get: AB = 2BD ⇒ BD = AB/2 = 2/2 = 1 m
9. The medium side is √3 times the smallest side. The medium side is AD, as it is opposite the intermediate angle 60o
10. So we get: AD = √3 × BD = √3 × 1 = √3
11. Thus we get all the sides of ⊿ABD
12. Now consider ⊿BCD. In this triangle, we will get ∠DBC = 105 - 60 = 45
13. We will also get ∠DCB = 180 -(30 + 105) = 180 - 135 = 45
14. So, in ⊿BCD, ∠B = ∠C = 45. Base angles are equal. So it is an isosceles triangle
15. The sides opposite to the base angles will be equal. That is., BD = CD
16. But in (8), we obtained BD as 1 m. So, CD is also equal to 1 m
17. Now we have the two components of the side AC of the original triangle ABC:
18. AC = AD + CD = {√3 [from 10] + 1 [from 16]} = 1 + √3
19. Now come back to ⊿BCD. It is a right triangle, and it's two legs are now known:
BD = 1 m, and, CD = 1 m [from 16]
20. Applying Pythagoras theorem, we get: BC = √(12 + 12) = √(1 + 1) = √2
21. Thus we get all the sides of the original triangle ABC
• AB = 2 m [given] • BC = √2 [from 20] • AC = (1 + √3) [from 18]
22. Thus perimeter = 2 + √2 + 1 + √3 = 3 + √2 + √3 = 3 + 1.41 + 1.73 = 6.14 m (correct to 1 cm)
Solved example 16.10
We have seen how we can draw a series of right triangles as in the fig.16.21(a) below:
(i) What are the lengths of the sides of the tenth triangle?
(ii) How much more is the perimeter of the tenth triangle, than the perimeter of the ninth triangle?
(iii) How do we write in algebra, the difference in perimeter of the nth triangle, and that of the triangle just before it?
Solution:
We have seen this series before. We did some basic calculations on this series, and worked upto the fourth triangle. See fig.16.17 at the beginning of this section.
Let us write a summary based on those calculations:
1. 1st triangle:
Both legs are equal to 1. Hypotenuse = √2
2. 2nd triangle:
short leg = 1. Long leg = √2. Hypotenuse = √3
3. 3rd triangle:
short leg = 1. Long leg = √3. Hypotenuse = √4
4. 4th triangle:
short leg = 1. Long leg = √4. Hypotenuse = √5
- - -
- - -
So a pattern has emerged.
• The short leg is always 1 • The long leg is square root of 'number of the triangle' • The hypotenuse is square root of 'number of the triangle + 1'. Based on this, we can write:
9. 9th triangle:
short leg = 1. Long leg = √9. Hypotenuse = √10
10. 10th triangle:
short leg = 1. Long leg = √10. Hypotenuse = √11
(ii) Perimeter of the 9th triangle = 1 + √9 + √10
Difference in perimeter = (1 + √10 + √11) - (1 + √9 + √10) = 1 + √10 + √11 -1 - √9 - √10 = √11 - √9
(iii) Now we can write an algebraic form:
• Perimeter of the nth triangle = 1 + √n + √(n+1)
• Perimeter of the (n-1)th triangle = 1 + √(n-1) + √n
Difference = (1 + √n + √(n+1)) - (1 + √(n-1) + √n) = 1 + √n + √(n+1) - 1 - √(n-1) - √n = √(n+1) - √(n-1)
Solved example 16.11
The legs of a right triangle are √2 cm and √3 cm long.
(i) What is the length of the hypotenuse? (ii) What is the difference between the hypotenuse and the 'sum of the legs'?
Solution:
Given: Lengths of legs = √2 cm and √3 cm. Fig.16.22 below shows a rough sketch:
1. ABC is a right triangle. So applying Pythagoras theorem, we get:
2. BC2 = AB2 + AC2 = (√3)2 + (√2)2 = 3 + 2 = 5 ⇒Hypotenuse = BC = √5 cm = 2.23 cm
3. Sum of legs = √2 + √3 = 1.41 + 1.73 = 3.14 cm
4. Difference = 3.14 - 2.23 = 0.91 cm
So we have completed the discussion on addition and subtraction. In the next section, we will see multiplication.
Consider ⊿OAB in fig.16.17(a) below:
 |
| Fig.16.17 |
• Now, what is the perimeter of ⊿OAB?
• Let us add the sides: Perimeter = OA + AB + BO = 1 + 1 + √2 = 2 + √2
• We have seen that √2 = 1.414213... So, if the unit is meter, we can write:
■ Perimeter of ⊿OAB = 2 + √2 = 2 + 1.41 = 3.41 m (Correct to 1 cm)
■ Perimeter of ⊿OAB = 2 + √2 = 2 + 1.414 = 3.414 m (Correct to 1 mm)
• Now consider fig.16.17(b). A new ⊿OBC is drawn on top of ⊿OAB. Length of BC = 1
• We have seen this combination before. We know that OC = √3• Now, what is the perimeter of ⊿OBC?
• Let us add the sides: Perimeter = 1 + √2 + √3
• We have seen that √2 = 1.414213... , and √3 = 1.732050... So, if the unit is meter, we can write:
■ Perimeter of ⊿OBC = 1 + √2 + √3 = 1 + 1.41 + 1.73 = 4.14 m (Correct to 1 cm)
■ Perimeter of ⊿OBC = 1 + √2 + √3 = 1 + 1.414 + 1.732 = 4.146 m (Correct to 1 mm)
• The new ⊿OBC is larger than the first ⊿OAB
• So the perimeter of ⊿OBC will be greater than that of ⊿OAB
• How much is the increase in perimeter? We can write:
■ Perimeter of ⊿OBC - Perimeter of ⊿OAB = 4.14 - 3.41 = 0.73 (Correct to 1 cm)
■ Perimeter of ⊿OBC - Perimeter of ⊿OAB = 4.146 - 3.414 = 0.732 (Correct to 1 mm)
There is an easier method to find the increase. That is., to do the calculations with out changing to decimal form. So we write:
■ Perimeter of ⊿OBC - Perimeter of ⊿OAB = (1 + √2 + √3) - (2 + √2) = 1 + √2 + √3 - 2 - √2 = √3 -1
Now we can convert to decimal form. We get:
■ Perimeter of ⊿OBC - Perimeter of ⊿OAB = √3 - 1 = 1.73 - 1 = 0.73 (Correct to 1 cm)■ Perimeter of ⊿OBC - Perimeter of ⊿OAB = √3 - 1 = 1.732 - 1 = 0.732 (Correct to 1 mm)
The results are same as those obtained earlier.
• Now consider fig.16.17(c). One more triangle is drawn on top of ⊿OBC . Length of CD = 1
• We have seen this combination also before. We know that OD = √4 = 2• Now, what is the perimeter of ⊿OCD?
• Let us add the sides: Perimeter = 1 + √3 + 2 = 3 + √3
• We have seen that √3 = 1.732050... So, if the unit is meter, we can write:
■ Perimeter of ⊿OCD = 3 + √3 = 3 + 1.73 = 4.73 m (Correct to 1 cm)
■ Perimeter of ⊿OCD = 3 + √3 = 3 + 1.732 = 4.732 m (Correct to 1 mm)
• How much is the increase in perimeter from the second triangle? We can write:
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = 4.73 - 4.14 = 0.59 (Correct to 1 cm)
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = 4.732 - 4.146 = 0.586 (Correct to 1 mm)
Let us find the increase using the easy method:
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = (3 + √3) - (1+ √2 + √3) = 3 + √3 - 1 - √2 - √3 = 2 - √2
Now we can convert to decimal form. We get:
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = 2 - √2 = 2 - 1.41 = 0.59 (Correct to 1 cm)
■ Perimeter of ⊿OCD - Perimeter of ⊿OBC = 2 - √2 = 2 - 1.414 = 0.586 (Correct to 1 mm)
The results are same as those obtained earlier.
We will see some solved examples
Solved example 16.7The hypotenuse of a right triangle is 11⁄2 m, and one of it's leg is 1⁄2 m. Calculate the perimeter correct to one cm
Solution:
Given: Hypotenuse = 11⁄2 m = 3⁄2 m and, one leg = 1⁄2 m. Fig.16.18 below shows a rough sketch:
 |
| Fig.16.18 |
• 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 we get: Other leg = √[(3⁄2)2 - (1⁄2)2] = √[(9⁄4) - (1⁄4)] = √(8⁄4) = √2
■ Thus perimeter = 3⁄2 + 1⁄2 + √2 = 4⁄2 + √2 = 2 + √2
We want the perimeter correct to 1 cm. So we take the value of √2 correct to 1 cm. That is., we take √2 = 1.41 m
Thus perimeter = 2 + √2 = 2 + 1.41 = 3.41 m
Solved example 16.8
The fig.16.19(a) shows an equilateral triangle cut into halves by a line through the vertex.
 |
| Fig.16.19 |
(ii) what is the difference between the 'perimeter of one half triangle' and the 'perimeter of the whole triangle'?
Solution:
1. Let us name the whole triangle as ABC. CD cuts it into two equal halves. This is shown in fig.b
2. Given that ABC is an equilateral triangle. So AB = 2 m, and, AD = BD = 1m
3. Also, both ADC and BDC are right angled
4. Applying Pythagoras theorem to ⊿ADC, we get: CD = √[AC2 - AD2] = √[22 - 12] = √(4 - 1) = √3
5. So perimeter of ⊿ADC = AD + AC + CD = 1 + 2 + √3 = (3 + √3) m
6. Let us take √3 = 1.73 m. Then Perimeter = 3 + √3 = 3 + 1.73 = 4.73 m
7. Perimeter of original triangle ABC = 2 + 2 + 2 = 6 m
8. Difference in perimeter = 6 - (3 + √3) = 6 - 3 - √3 = 3 - √3
9. So difference = 3 - √3 = 3 - 1.73 = 1.27 m
Solved example 16.9
Calculate the perimeter of the triangle shown in fig.16.20 below:
 |
| Fig.16.20 |
1. Let us name the triangle as ABC. This is shown in fig.b
2. Through the vertex B, a perpendicular line BD is drawn to AC. So ∠ADB is 90o
3. Consider ⊿ABD. We know two of it's angles: ∠DAB = 30o and ∠ADB = 90o
4. So ∠ABD = 60o (∵ sum of interior angles of any triangle is 180o)
5. Thus we find that ⊿ABD is a 30, 60, 90 triangle. We have seen the details here. In such a triangle, if we know any one side, the other two sides can be easily calculated.
6. We know the side AB as 2 m. This is the longest side, the hypotenuse, which is opposite the largest angle 90o
7. The longest side is twice the smallest side. Here, the smallest side is BD, as it is opposite the smallest angle 30o
8. So we get: AB = 2BD ⇒ BD = AB/2 = 2/2 = 1 m
9. The medium side is √3 times the smallest side. The medium side is AD, as it is opposite the intermediate angle 60o
10. So we get: AD = √3 × BD = √3 × 1 = √3
11. Thus we get all the sides of ⊿ABD
12. Now consider ⊿BCD. In this triangle, we will get ∠DBC = 105 - 60 = 45
13. We will also get ∠DCB = 180 -(30 + 105) = 180 - 135 = 45
14. So, in ⊿BCD, ∠B = ∠C = 45. Base angles are equal. So it is an isosceles triangle
15. The sides opposite to the base angles will be equal. That is., BD = CD
16. But in (8), we obtained BD as 1 m. So, CD is also equal to 1 m
17. Now we have the two components of the side AC of the original triangle ABC:
18. AC = AD + CD = {√3 [from 10] + 1 [from 16]} = 1 + √3
19. Now come back to ⊿BCD. It is a right triangle, and it's two legs are now known:
BD = 1 m, and, CD = 1 m [from 16]
20. Applying Pythagoras theorem, we get: BC = √(12 + 12) = √(1 + 1) = √2
21. Thus we get all the sides of the original triangle ABC
• AB = 2 m [given] • BC = √2 [from 20] • AC = (1 + √3) [from 18]
22. Thus perimeter = 2 + √2 + 1 + √3 = 3 + √2 + √3 = 3 + 1.41 + 1.73 = 6.14 m (correct to 1 cm)
Solved example 16.10
We have seen how we can draw a series of right triangles as in the fig.16.21(a) below:
 |
| Fig.16.21 |
(ii) How much more is the perimeter of the tenth triangle, than the perimeter of the ninth triangle?
(iii) How do we write in algebra, the difference in perimeter of the nth triangle, and that of the triangle just before it?
Solution:
We have seen this series before. We did some basic calculations on this series, and worked upto the fourth triangle. See fig.16.17 at the beginning of this section.
Let us write a summary based on those calculations:
1. 1st triangle:
Both legs are equal to 1. Hypotenuse = √2
2. 2nd triangle:
short leg = 1. Long leg = √2. Hypotenuse = √3
3. 3rd triangle:
short leg = 1. Long leg = √3. Hypotenuse = √4
4. 4th triangle:
short leg = 1. Long leg = √4. Hypotenuse = √5
- - -
- - -
So a pattern has emerged.
• The short leg is always 1 • The long leg is square root of 'number of the triangle' • The hypotenuse is square root of 'number of the triangle + 1'. Based on this, we can write:
9. 9th triangle:
short leg = 1. Long leg = √9. Hypotenuse = √10
10. 10th triangle:
short leg = 1. Long leg = √10. Hypotenuse = √11
Now we can write the answers:
(i) Perimeter of the 10th triangle = 1 + √10 + √11(ii) Perimeter of the 9th triangle = 1 + √9 + √10
Difference in perimeter = (1 + √10 + √11) - (1 + √9 + √10) = 1 + √10 + √11 -1 - √9 - √10 = √11 - √9
(iii) Now we can write an algebraic form:
• Perimeter of the nth triangle = 1 + √n + √(n+1)
• Perimeter of the (n-1)th triangle = 1 + √(n-1) + √n
Difference = (1 + √n + √(n+1)) - (1 + √(n-1) + √n) = 1 + √n + √(n+1) - 1 - √(n-1) - √n = √(n+1) - √(n-1)
Solved example 16.11
The legs of a right triangle are √2 cm and √3 cm long.
(i) What is the length of the hypotenuse? (ii) What is the difference between the hypotenuse and the 'sum of the legs'?
Solution:
Given: Lengths of legs = √2 cm and √3 cm. Fig.16.22 below shows a rough sketch:
 |
| Fig.16.22 |
2. BC2 = AB2 + AC2 = (√3)2 + (√2)2 = 3 + 2 = 5 ⇒Hypotenuse = BC = √5 cm = 2.23 cm
3. Sum of legs = √2 + √3 = 1.41 + 1.73 = 3.14 cm
4. Difference = 3.14 - 2.23 = 0.91 cm
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