Chapter 22 - Real Numbers

In the previous section we completed the discussion on Circles, arcs and sectors. In a previous chapter, we have seen different types of numbers. (see fig.16.1). In this chapter, we will see how and where 'Real numbers' fit into the list of numbers.

First let us recall how numbers are represented on a number line. In fig.22.1(a), a horizontal line segment is drawn between points A and B. We will fix this distance AB as 1 unit.

Fig.22.1

• So, if we take two line segments, each equal to AB, and put them end to end, we will get a distance of 2 units. This distance is the distance from zero to 2 on the number line shown in fig.d

• If we take 3 such segments and put them end to end, we will get a distance of 3 units. This distance is the distance from zero to 3 on the number line shown in fig.d. In this way, we can find the position of any number on the number line.

Now what about fractions? Consider fig.b. The segment AB is divided into 2 equal parts at C. So AC = BC = ¹⁄₂ unit. 
• If we take two line segments, one equal to AB, and the other equal to AC, and put them end to end, we will get a distance 1¹⁄₂ = ³⁄₂ units. If we measure out this end to end distance (from zero) on the number line, we can mark ³⁄₂. It is shown in fig.d

Another example: In fig.c, the segment AB is divided into 3 equal parts at D and E. So AD = DE = BE = ¹⁄₃ unit. 
• If we take three line segments, two of them equal to AB, and the third equal to AD, and put them end to end, we will get a distance 2¹⁄₃ = ⁷⁄₃ units.  If we measure out this end to end distance (from zero) on the number line, we can mark ⁷⁄₃. It is shown in fig.d

In the above examples, we marked points using a geometrical method. That is., we measured the distance from the 'standard 1 unit', or it's fractions, and then marked it on the number line. Can we do the marking with out actual measuring? Let us analyse:

• ³⁄₂ does not give us any problems because in decimal form, it is 1.5. It is a terminating decimal. 
• But ⁷⁄₃ is a different case. We discussed about such numbers here. It is a recurring decimal. It is written as ⁷⁄₃ = 2.333... Using this decimal form, we can represent ⁷⁄₃ on the number line by the following procedure: 

2.333... has infinite number of decimal places. The digit '3' repeats forever. If more number of decimal places are taken, we will get more accuracy. If we want such a greater accuracy, we will have to 'zoom in' on the region between 2 and 3. Because 2.333... lies in between 2 and 3. In the fig.22.2 below, ‘zoom level 1’ shows the portion between 2 and 3, at an enlarged scale.

Fig.22.2

In this zoom level, we have enough space to clearly mark ten subdivisions between 2 and 3. So the third subdivision will be 2.3 and the fourth subdivision will be 2.4. They are specially marked in red because 2.333... lies between them

Next, we zoom in on the region between this 2.3 and 2.4. This is shown as 'zoom level 2'. In this zoom level, we have enough space to clearly mark ten subdivisions between 2.3 and 2.4. So the third subdivision will be 2.33 and the fourth subdivision will be 2.34. They are specially marked in red because 2.333... lies between them

Next, we zoom in on the region between this 2.33 and 2.34. This is shown as 'zoom level 3'. In this zoom level, we have enough space to clearly mark ten subdivisions between 2.33 and 2.34. So the third subdivision will be 2.333 and the fourth subdivision will be 2.334. They are specially marked in red because 2.333... lies between them

So, if we take the left red subdivision in zoom level 3, it will represent 2.333. It is fairly accurate. We know that 0.003 is a small quantity. But we are able to mark it by 'zooming in'. It is like zooming in on Google Maps. At higher zoom levels, we are able to see smaller details. But in our case of number line, there is no limit. We can mark any quantity, however small it may be. All we need to do, is to zoom in to the required level. 
So we can conclude that:
• Any natural number can be represented on a number line
• Any fraction can be represented on a number line

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Now we try to do the reverse: We are given a number line with a mark  on it. We want to represent that mark as a number or a fraction. 
Consider the mark P in fig.22.3(a). It is exactly at 4

Fig.22.3

So we write: The mark P represents number 4. 
Consider mark Q. It is exactly at 6.5 
So we write: The mark Q represents number 6.5, which is equal to ¹³⁄₂
Consider mark R. We have seen that any recurring decimal can be represented on a number line. If it is given that R is 5.333... , can we represent it as a fraction? 
The answer is yes. The procedure is as follows:
• Let x = 5.333...
• 10x = 10 × 5.333... = 53.333... = 48 + 5.333... = 48 + x (∵ x = 2.333..)
⇒ 10x = 48 + x ⇒ 9x = 48 ⇒ x = ⁴⁸⁄₉ = 5³⁄₉ = 5¹⁄₃ = ¹⁶⁄₃

So, even if the given mark on the number line is a recurring decimal, we can convert it into a fraction. Let us see a few more examples:
■ Express 1.272727... as a fraction:
1. We can write: 1.272727... = 1.2̅7
2. A line is drawn above 2 and 7. That means the block '27' repeats for ever
3. Let x = 1.272727... Since two digits are repeating, we will multiply by 100
• 100x = 100 × 1.272727... = 127.272727... = 126 + 1.272727... = 126 + x (∵ x = 1.272727..)
⇒ 100x = 126 + x ⇒ 99x = 126 ⇒ x = ¹²⁶⁄₉₉ = 1³⁄₁₁

■ Express 0.2353535... as a fraction:
1. We can write: 0.2353535... = 0.23̅5
2. A line is drawn above 3 and 5. That means the block '35' repeats for ever
3. Let x = 0.2353535... Since two digits are repeating, we will multiply by 100
• 100x = 100 × 0.2353535... = 23.535353... =  23.3 + 0.2353535... = 23.3 + x (∵ x = 0.2353535...)
⇒ 100x = 23.3 + x ⇒ 99x = 23.3 ⇒ x = ²³³⁄₉₉₀

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■ Based on the above discussion, we can write the following:
1. Any natural number can be represented on a number line
2. Any fraction can be represented on a number line
    ♦ The fractions  like ¹⁄₂, ¹⁄₅, ³⁄₅ etc., that give terminating decimals
    ♦ The fractions like ¹⁄₃, ¹⁄₇, ³⁄₅ etc., that give non-terminating, recurring decimals
■ Reverse of the above is also possible: 
If a point is marked on a number line, it can be represented by any one of the two categories given below:
1. A natural number
2. A fraction
    ♦ The fraction may be one  like ¹⁄₂, ¹⁄₅, ³⁄₅ etc., that give terminating decimals
    ♦ The fraction may be one like ¹⁄₃, ¹⁄₇, ³⁄₅ etc., that give non-terminating, recurring decimals

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■ Any number that can be expressed as a fraction ^p⁄_q where q is not equal to zero is called a rational number. 

• Simple natural numbers like 2, 5 etc., can be written as ²⁄₁ and ⁵⁄₁. So they are rational numbers. 
• Decimals like 0.2, 0.35, 0.5 etc., can be easily written in the form of fractions (with non-zero denominators). So they are rational numbers.
• Non-terminating recurring decimals like 0.333..., 1.2̅7, 0.23̅5, etc., can be written in the form of fractions (with non-zero denominators). So they are rational numbers.

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In this section we saw the details about rational numbers. In the next section we will see irrational numbers.

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Chapter 16.3 - Square root of 7, 11 and 13

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

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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]

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• 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

• 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

Fig.16.15

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:

Fig.16.16

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.

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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

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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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So we have seen the basics about irrational numbers. In the next section, we will see their addition and subtraction.

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Chapter 16.2 - Square root of 3

In the previous section we saw the details about √2. In this section, we will see √3. First, we will try to get a pictorial representation of √3. Consider the fig.16.7 below:

Fig.16.7

• ⊿ABC is a right triangle, right angled at A. The perpendicular sides AB and AC are 1 unit each. We have seen in the previous section that, the hypotenuse BC will be equal to √2. Now, we are going to add some extra geometrical figures, based on this ⊿ABC:
• ⊿BDC is drawn adjacent to ⊿ABC
• ⊿BDC is right angled at B
• One perpendicular side BC of ⊿BDC is the same hypotenuse of ⊿ABC. So BC = √2
• The other perpendicular side BD is equal to 1 unit
• Then, what is the hypotenuse CD of ⊿BDC?
• Applying Pythagoras theorem to ⊿BDC, we get: 
CD² = BD² + BC² ⇒  CD² = 1² + (√2)² ⇒ CD² = 1 + 2 ⇒ CD² = 3. ∴ CD = √3
• So CD is a pictorial representation of √3 
• But there is more. With base CD, draw a square CDEF (coloured in green). All sides of this square will be equal to CD
• That is., all sides of the square CDEF will be equal to √3
• What is the area of a square with side √3?
• Obviously, the area will be equal to (√3×√3) = (√3)²  = 3
• So we can write: √3 is the side of a square whose area is 3 sq.units
Let us write a summary:
■ Length of the side of a square, whose area is 3 sq.units, is equal to √3
■ Hypotenuse of a right triangle, whose perpendicular sides are '1 unit' and '√2 unit', is equal to √3
■ Diagonal of a rectangle, whose adjacent sides are 1 unit and √2 unit, is equal to √3 

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Like √2, √3 is also an irrational number. The calculation of it's approximate value is shown in fig.16.8 below:

Fig.16.8

Based on the above, we can write:
• √3 ≈ 1.7 (correct to 1 decimal place) 
• √3 ≈ 1.73 (correct to 2 decimal places) 
• √3 ≈ 1.732 (correct to 3 decimal places) and so on... 
The symbol '≈' is read as 'almost equal to'.
• In this way, we can calculate the approximate values of irrational numbers like √5, √7 etc., 

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We will now see some solved examples

Solved example 16.1
In the fig.16.9(a) below, the square on the hypotenuse of the top most right triangle is drawn.

Fig.16.9

Calculate the length of it's side, and it's area.
Solution:
• In the given fig.16.9(a), the lower most triangle has both it's perpendicular sides 1 m length.
• Also, bases of all the triangles are 1m, and all triangles are right angled. 
• In this problem, it is better to give names (as shown in fig.b) to the triangles and squares. Then we will be able to write the steps easily.   

1. Consider the lower most ⊿OAB. Applying Pythagoras theorem, we get: 
2. OB² = OA² + AB² ⇒  OB² = 1² + 1² ⇒ OB² = 1 + 1 ⇒ OB² = 2. ∴ OB = √2
3. Consider the next upper ⊿OBC. Applying Pythagoras theorem, we get:
4. OC² = OB² + BC² ⇒  OC² = (√2)² + 1² ⇒ OC² = 2 + 1 ⇒ OC² = 3. ∴ OC = √3
5. Consider the next upper ⊿OCD. Applying Pythagoras theorem, we get:
6. OD² = OC² + CD² ⇒  OD² = (√3)² + 1² ⇒ OD² = 3 + 1 ⇒ OD² = 4. ∴ OD = √4
7. Consider the next upper ⊿ODE. Applying Pythagoras theorem, we get:
8. OE² = OD² + DE² ⇒  OE² = (√4)² + 1² ⇒ OE² = 4 + 1 ⇒ OE² = 5. ∴ OE = √5
■ So the length of the side of the square EFGO is √5
■ ∴ Area = (√5×√5) = (√5)²  = 5 sq.m
Solved example 16.2
In the fig.16.10(a) below, a square is drawn on the altitude of an equilateral triangle. The side of the triangle is 2 m

Fig.16.10

(i) What is the altitude of the triangle?
(ii) What is the area of the square?
(iii) What are the lengths of the other two sides of the triangle shown in fig.16.10(b)?
Solution:
• We are given fig.16.10(a). It is better to give names (as shown in fig.c) to the triangle and square. Then we will be able to write the steps easily.
1. Given ABC is an equilateral triangle. So AB=BC=CA=2 m.
2. CD is the altitude. So AD=BD=1 m
3. Altitude CD is perpendicular to AB. So triangles ADC and BDC are right angled at D
4. Applying Pythagoras theorem to ⊿ADC, we get:
5. AC² = AD² + CD² ⇒ CD² = AC² - AD²  ⇒ CD² = 2² - 1² ⇒ CD² = 4 - 1 ⇒ CD² = 3. ∴ CD = √3
■ Thus we get: Altitude of the triangle = √3 m   
6. Given that CDEF is a square. In (5), we calculated it’s one side CD as √3 m
■ So area of CDEF = (√3×√3) = (√3)²  = 3 sq.m
7. We are given a triangle PQR, with two angles and it's included side
8. Sum of the interior angles of any triangle is 180^o
9. So we can write: ∠RPQ + ∠PQR + ∠PRQ = 180 ⇒ 60+30+∠PRQ =180 ⇒ 90+∠PRQ =180 ⇒ ∠PRQ = 180-90 =90
10. Thus we find that, PQR is a right angled triangle.
11. We are required to find the sides PR and QR
12. We cannot apply Pythagoras theorem because, only one side PQ is known. We must use some other method
13. Consider our first equilateral triangle ABC. It is shown separately in fig.d.
14. As it is an equilateral triangle, ∠CAB = ∠ABC = ∠BCA = 60^o
15. Now consider ⊿ADC. The ∠CAD is same as ∠CAB.
16. But from (14), ∠CAB = 60^o. So we get: ∠CAD = 60^o
17. Now apply the sum of interior angle property to ⊿ADC
18. We can write: ∠CAD + ∠ADC + ∠DCA = 180 ⇒ 60+90+∠DCA =180 ⇒ 150 +∠DCA =180 ⇒ ∠DCA = 180-150 =30^o
19. In fig.b, we have a triangle PQR, with two angles 30^o and 60^o, with an included side 2m
20. In fig.d, we have a triangle ADC, with two angles 30^o and 60^o, with an included side 2m
21. It is a case of ASA congruence. The two triangles are exactly the same. We can obtain the ‘missing details in ⊿PQR’ from ⊿ABC. For that, first we must write the correspondence:
22. From the figs.(b) and (d), it is obvious that P↔A, Q↔C and R↔D. Because,
      ♦ ∠P and ∠A have the same measure (60^o) 
      ♦ ∠Q and ∠C have the same measure (30^o)
      ♦ ∠R and ∠D have the same measure (90^o)
23. Now we write the corresponding of the sides.
    ♦ We have P↔A and Q↔C. So PQ↔AC
    ♦ We have Q↔C and R↔D. So QR↔CD
    ♦ We have P↔A and R↔D. So PR↔AD
■ Now we can write the missing details in ⊿PQR: We want the lengths PR and QR.
    ♦ From (23), we have PR↔AD. So PR = AD = 1 cm 
    ♦ From (23), we have QR↔CD. So QR = CD = √3 cm [from (5)]

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What we saw in the above problem, is a special triangle. It has one angle 30^o , and another 60^o. The third angle will then obviously be 90^o. Let us randomly make a few such triangles. They are shown in the fig.16.11 below:

Fig.16.11

Fig.16.11.a shows the exact ⊿ADC which we saw in fig.16.10.c above. It has two angles 30^o and 60^o, and the included side 2 m. The other three are drawn randomly. But with the specified angles. How can we draw random triangles with the specified angles? 

There is an easy method:

Consider for example ⊿XYZ in fig.16.11.b. It can be drawn by the following procedure:
1. Draw a line XY of any convenient length, in any convenient direction. It is the first line of our triangle. It's length and direction does not matter for the construction.
2. But the second and third lines should be drawn in specified directions. That is., at the specified angles:
    ♦ Draw the second line from X, at 30^o to XY. Here the length does not matter for the construction. But the angle does.
    ♦ Draw the third line at Y, at 60^o to XY. Here also, the length does not matter for the construction, but the angle does.
3. The second and third lines will meet at Z. The triangle is complete. We can see that, in the construction of this triangle, we are not aware of the lengths of sides. We completed the triangle using angles only.
4. Once the triangle is complete, we must measure the sides and write them near the corresponding sides. We must also write the angles.
In this way, three random triangles ⊿XYZ, ⊿MNO, and ⊿UVW are constructed in the fig.16.11. All of them are right angled triangles
5.All of them have:
• One smallest side. This is the side opposite to the smallest angle which is the 30^o angle. We will call it 'small side'
• One medium side. This is the side opposite to the intermediate angle which is the 60^o angle. We will call it 'medium side' 
• One largest side. This is the side opposite to the largest angle which is the 90^o  angle. We will call it 'large side' 
6. Now, in each triangle, consider the ratio:
small side : medium side : large side
7. In fig.a we will get 1:√3:2 [remember that √3 ≈ 1.73, which lies in between 1 and 2. So √3 is indeed the medium side] 
8. In ⊿XYZ of fig.b, we will get 1.81 : 3.14 : 3.62. 
• Is there a common factor? Let us divide by the smallest side:
• 1.81/1.81 = 1,   3.14/1.81 = 1.73,   3.62/1.81 = 2
• Consider the second value: 3.14/1.81 = 1.73. But this 1.73 is an approximate value of √3. So we can write:
■ small side : medium side : large side
=  1.81 : 3.14 : 3.62 ⇒ 1 : 1.73 : 2 ⇒ 1 : √3 : 2
9. In ⊿MNO of fig.c, we will get 1.45 : 2.51 : 2.90 
• Is there a common factor? Let us divide by the smallest side:
• 1.45/1.45 = 1,   2.51/1.45 = 1.73,   2.90/1.45 = 2
• Consider the second value: 2.51/1.45 = 1.73. But this 1.73 is an approximate value of √3. So we can write:
■ small side : medium side : large side
=  1.45 : 2.51 : 2.90 ⇒ 1 : 1.73 : 2 ⇒ 1 : √3 : 2
10. In ⊿UVW of fig.d, we will get 2.19 : 3.79 : 4.38 
• Is there a common factor? Let us divide by the smallest side:
• 2.19/2.19 = 1,   3.79/2.19 = 1.73,   4.38/2.19 = 2
• Consider the second value: 3.79/2.19 = 1.73. But this 1.73 is an approximate value of √3. So we can write:
■ small side : medium side : large side
=  2.19 : 3.79 : 4.38 ⇒ 1 : 1.73 : 2 ⇒ 1 : √3 : 2

So we saw 3 random 30^o, 60^o , 90^o triangles. In all of them we got the same ratio. In fact, in any such triangle that we consider, we will get the same ratio. So we can write:
■ If the angles of a triangle are 30^o, 60^o and 90^o, then the ratio
small side : medium side : large side = 1 : √3 : 2

Another way of saying this is:
■ If the angles of a triangle are 30^o, 60^o and 90^o, then:
• The medium side will always be √3 times the smallest side
• The largest side will always be 2 times the smallest side

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• 30^o, 60^o, 90^o triangles are encountered many times in science and engineering topics. 
• One of the 'Set squares' in the geometry box have 30^o, 60^o, 90^o angles. 
we will now see a solved example.
Solved example 16.3
Fig.16.12 below shows a triangle ABC. Find the missing lengths

Fig.16.12

Solution:
In the problem, AB is given as 2.47. We have to find AC and BC
1. AB is the side opposite to the intermediate angle 60^o. So AB is the medium side. Thus AC is the smallest side (opposite to 30^o), and BC is the largest side (opposite to 90^o)
2. We know that, the medium side is √3 times the smallest side. 
3. So AB = √3 × AC ⇒ AC = AB/√3 ⇒ AC = 2.47/1.73 = 1.43
4. Thus we get: Smallest side = AC = 1.43
5. We know that, the largest side is 2 times the smallest side.
6. So BC = 2 × AC = 2 × 1.43 = 2.86
7. Thus we get: largest side = BC = 2.86 
8. Check: Applying Pythagoras theorem, we get: BC² = AB² + AC²

⇒ 2.86² = 2.47² + 1.43² ⇒  8.18 = 6.1 + 2.045 ⇒ 8.18 = 8.145
Both sides are approximately equal.

• Let us repeat the steps from (3) using a more exact value of √3:

3. AB = √3 × AC ⇒ AC = AB/√3 ⇒ AC = 2.47/1.7320508 = 1.426055
4. Thus we get: Smallest side = AC = 1.426055
5. We know that, the largest side is 2 times the smallest side.
6. So BC = 2 × AC = 2 × 1.426055 = 2.85211
7. Thus we get: largest side = BC = 2.85211 
8. Check: Applying Pythagoras theorem, we get: BC² = AB² + AC²

⇒ 2.85211² = 2.47² + 1.426055² ⇒  8.1345 = 6.1009 + 2.0336 ⇒ 8.1345 = 8.1345
So, when we increase the number of decimal places, the accuracy increases, and we get exactly equal quantities on both sides.

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In the next section, we will see more solved examples.

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