Maharashtra Board Class 9 Maths Chapter 2 Set 2 Algebra Standard Part 1 Real Numbers Solutions

Question 1. Choose the correct alternative answer for the questions given below.
(i) Which one of the following is an irrational number?
(a) \( \sqrt{\frac{16}{25}} \)
(b) \( \sqrt{5} \)
(c) \( \frac{3}{9} \)
(d) \( \sqrt{196} \)
Answer: (b) \( \sqrt{5} \)
In simple words: An irrational number is a number that cannot be written as a simple fraction. Since 5 is not a perfect square, its square root is a non-terminating, non-recurring decimal, making it irrational.

🎯 Exam Tip: To quickly identify irrational square roots, check if the number inside the root is a perfect square. If it is not a perfect square, its square root is always irrational.

Question ii. Which of the following is an irrational number?
(a) 0.17
(b) \( 1.\overline{513} \)
(c) \( 0.27\overline{46} \)
(d) \( 0.101001000... \)
Answer: (d) \( 0.101001000... \)
In simple words: An irrational number has a decimal form that goes on forever without repeating any fixed pattern. Here, the number of zeros keeps increasing, so it never repeats.

🎯 Exam Tip: Look for decimals that do not terminate and do not have a repeating block of digits; these are always irrational.

Question iii. Decimal expansion of which of the following is non-terminating recurring?
(a) \( \frac{2}{5} \)
(b) \( \frac{3}{16} \)
(c) \( \frac{3}{11} \)
(d) \( \frac{137}{25} \)
Answer: (c) \( \frac{3}{11} \)
In simple words: If the denominator of a simplified fraction has prime factors other than 2 or 5, its decimal form will keep repeating forever. Since 11 is not made of 2s or 5s, its decimal is non-terminating and recurring.

🎯 Exam Tip: Quickly factorize the denominator. If you see any prime factor other than 2 or 5, the decimal expansion is non-terminating recurring.

Question iv. Every point on the number line represents which of the following numbers?
(a) Natural numbers
(b) Irrational numbers
(c) Rational numbers
(d) Real numbers
Answer: (d) Real numbers
In simple words: The number line contains all possible numbers, including both fractions and decimals that do not repeat. Together, all these numbers are called real numbers.

🎯 Exam Tip: Remember that the real number line is uniquely and completely filled by real numbers—each point corresponds to exactly one real number.

Question v. The number \( 0.\dot{4} \) in \( \frac{p}{q} \) form is ......
(a) \( \frac{4}{9} \)
(b) \( \frac{40}{9} \)
(c) \( \frac{3.6}{9} \)
(d) \( \frac{36}{9} \)
Answer: (a) \( \frac{4}{9} \)
In simple words: To convert a single repeating digit like \( 0.444... \) into a fraction, we write the repeating digit over 9, which gives us \( \frac{4}{9} \).

🎯 Exam Tip: For any single repeating digit \( 0.\dot{x} \), the fraction form is always \( \frac{x}{9} \). This shortcut saves valuable time in exams.

Question vi. What is \( \sqrt{n} \), if \( n \) is not a perfect square number?
(a) Natural number
(b) Rational number
(c) Irrational number
(d) All of the options
Answer: (c) Irrational number
In simple words: When you take the square root of a number that isn't a perfect square (like 2, 3, or 5), you get a decimal that never ends and never repeats, which is an irrational number.

🎯 Exam Tip: Remember that the square root of any prime number or non-perfect square is always irrational.

Question 1.
(vii) Which of the following is not a surd?
(a) \( \sqrt{7} \)
(b) \( \sqrt[3]{17} \)
(c) \( \sqrt[3]{\sqrt{64}} \)
(d) \( \sqrt{193} \)
Answer: (c) \( \sqrt[3]{\sqrt{64}} \)
In simple words: A surd is a root of a number that cannot be simplified to a whole number or fraction. Since the square root of 64 is 8, and the cube root of 8 is 2, this expression simplifies to a normal whole number and is not a surd.

🎯 Exam Tip: To identify if an expression is not a surd, try to simplify the terms inside the radical step-by-step to see if they result in a rational number.

Question 1.
(viii) What is the order of the surd \( \sqrt[3]{\sqrt{5}} \)?
(a) 3
(b) 2
(c) 6
(d) 5
Answer: (c) 6
In simple words: When you have a root inside another root, you multiply their index numbers together. Here, we multiply the cube root (3) by the square root (2) to get an overall order of 6.

🎯 Exam Tip: Always remember that a standard square root symbol \( \sqrt{x} \) has an implicit index of 2. Multiply this index with any outer root index to find the final order.

Question 1.
(ix) Which one is the conjugate pair of \( 2\sqrt{5} + \sqrt{3} \)?
(a) \( -2\sqrt{5} + \sqrt{3} \)
(b) \( -2\sqrt{5} - \sqrt{3} \)
(c) \( 2\sqrt{3} - \sqrt{5} \)
(d) \( \sqrt{3} + 2\sqrt{5} \)
Answer: (a) \( -2\sqrt{5} + \sqrt{3} \)
In simple words: A conjugate pair is created by changing the sign of one of the terms in the expression. When you multiply a conjugate pair together, the square roots cancel out completely.

🎯 Exam Tip: Look for the option where exactly one sign is inverted. Multiplying the original term by its correct conjugate must always yield a rational number.

Question 1.
(x) The value of \( |12 - (13 + 7) \times 4| \) is ____.
(a) -68
(b) 68
(c) -32
(d) 32
Answer: (b) 68
In simple words: First, add 13 and 7 to get 20, then multiply by 4 to get 80. Subtracting 80 from 12 gives -68, and the absolute value bars turn this negative number into positive 68.

🎯 Exam Tip: Always follow the BODMAS rule inside absolute value bars, and remember that absolute value \( |x| \) can never result in a negative number.

Hints

(ii) Since the decimal expansion is neither terminating nor recurring, \( 0.101001000\dots \) is an irrational number.

Question 1. Solve the following sub-questions:
(iii) Find the type of decimal expansion for \( \frac{3}{11} \).
(v) Express \( 0.\dot{4} \) as a fraction.
(vii) Determine if \( \sqrt[3]{64} \) is an irrational number.
(viii) Find the order of the surd \( \sqrt[3]{\sqrt{5}} \).
(ix) Write the conjugate of \( 2\sqrt{5} + \sqrt{3} \).
(x) Simplify \( |12 - (13+7) \times 4| \).
Answer:
(iii) Denominator = \( 11 = 1 \times 11 \). Since the denominator has prime factors other than 2 or 5, the decimal expansion of \( \frac{3}{11} \) will be non-terminating recurring.
(v) Let \( x = 0.\dot{4} \)
\( \implies 10x = 4.\dot{4} \)
\( \implies 10x - x = 4.\dot{4} - 0.\dot{4} \)

\( \implies 9x = 4 \)

\( \implies x = \frac{4}{9} \)
(vii) \( \sqrt[3]{64} = 4 \), which is a rational number and not an irrational number.
(viii) \( \sqrt[3]{\sqrt{5}} = \sqrt[3 \times 2]{5} = \sqrt[6]{5} \)

\( \implies \) Order = 6
(ix) The conjugate of \( 2\sqrt{5} + \sqrt{3} \) is \( 2\sqrt{5} - \sqrt{3} \) or \( -2\sqrt{5} + \sqrt{3} \).
(x) \( |12 - (13+7) \times 4| = |12 - 20 \times 4| = |12 - 80| = |-68| = 68 \)
In simple words: These solutions show how to simplify surds, find conjugates, calculate absolute values, and convert recurring decimals into fractions.

🎯 Exam Tip: Remember that the order of a surd \( \sqrt[n]{\sqrt[m]{a}} \) is obtained by multiplying the indices \( n \times m \), and the absolute value of any number is always positive.

Question 2. Write the following numbers in \( \frac{p}{q} \) form.
(i) \( 0.555 \)
(ii) \( 29.\overline{568} \)
(iii) \( 9.315315... \)
(iv) \( 357.417417... \)
(v) \( 30.\overline{219} \)
Answer:
(i) \( 0.555 = \frac{0.555 \times 1000}{1 \times 1000} = \frac{555}{1000} = \frac{5 \times 111}{5 \times 200} = \frac{111}{200} \)
(ii) Let \( x = 29.\overline{568} \) ---(1)
\( x = 29.568568... \)
Multiplying both sides by 1000:
\( 1000x = 29568.568568... \) ---(2)
Subtracting (1) from (2):
\( 1000x - x = 29568.\overline{568} - 29.\overline{568} \)

\( \implies 999x = 29539 \)

\( \implies x = \frac{29539}{999} \)
(iii) Let \( x = 9.315315... = 9.\overline{315} \) ---(1)
Multiplying both sides by 1000:
\( 1000x = 9315.\overline{315} \) ---(2)
Subtracting (1) from (2):
\( 1000x - x = 9315.\overline{315} - 9.\overline{315} \)

\( \implies 999x = 9306 \)

\( \implies x = \frac{9306}{999} = \frac{1034}{111} \)
(iv) Let \( x = 357.417417... = 357.\overline{417} \) ---(1)
Multiplying both sides by 1000:
\( 1000x = 357417.\overline{417} \) ---(2)
Subtracting (1) from (2):
\( 1000x - x = 357417.\overline{417} - 357.\overline{417} \)

\( \implies 999x = 357060 \)

\( \implies x = \frac{357060}{999} = \frac{119020}{333} \)
(v) Let \( x = 30.\overline{219} \) ---(1)
Multiplying both sides by 1000:
\( 1000x = 30219.\overline{219} \) ---(2)
Subtracting (1) from (2):
\( 1000x - x = 30219.\overline{219} - 30.\overline{219} \)

\( \implies 999x = 30189 \)

\( \implies x = \frac{30189}{999} = \frac{10063}{333} \)
Converting these recurring decimals into rational fractions helps in performing precise algebraic calculations.
In simple words: To convert a recurring decimal to a fraction, we multiply it by 10, 100, or 1000 depending on how many digits repeat, subtract the original equation to cancel out the repeating part, and then solve for x.

🎯 Exam Tip: Always count the number of recurring digits under the bar; if there are 3 recurring digits, multiply by 1000 to easily eliminate the decimal part.

Question (ii). Convert \( 29.\overline{568} \) into rational form.
Answer: Let \( x = 29.568568... = 29.\overline{568} \) ...(i)
Since, three numbers i.e. 5, 6 and 8 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
\( 1000x = 29568.568568... \)
\( \implies 1000x = 29568.\overline{568} \) ...(ii)
Subtracting (i) from (ii),
\( 1000x - x = 29568.\overline{568} - 29.\overline{568} \)
\( \implies 999x = 29539 \)
\( \implies x = \frac{29539}{999} \)
\( \implies 29.\overline{568} = \frac{29539}{999} \)
In simple words: To convert a recurring decimal to a fraction, we multiply by 1000 because three digits repeat, subtract the original equation to cancel the repeating part, and then solve for x.

🎯 Exam Tip: Always count the number of repeating digits to decide whether to multiply by 10, 100, or 1000. Since three digits repeat here, we multiply by 1000.

Question (iii). Convert \( 9.315315... \) into rational form.
Answer: Let \( x = 9.315315... = 9.\overline{315} \) ...(i)
Since, three numbers i.e. 3, 1 and 5 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
\( 1000x = 9315.315315... \)
\( \implies 1000x = 9315.\overline{315} \) ...(ii)
Subtracting (i) from (ii),
\( 1000x - x = 9315.\overline{315} - 9.\overline{315} \)
\( \implies 999x = 9306 \)
\( \implies x = \frac{9306}{999} = \frac{9 \times 1034}{9 \times 111} = \frac{1034}{111} \)
\( \implies 9.315315... = \frac{1034}{111} \)
In simple words: Since three digits repeat, we multiply by 1000, subtract the original value to cancel out the repeating decimal, and simplify the resulting fraction.

🎯 Exam Tip: Remember to simplify the final fraction to its lowest terms by finding common factors, which helps in scoring full marks.

Question (iv). Convert \( 357.417417... \) into rational form.
Answer: Let \( x = 357.417417... = 357.\overline{417} \) ...(i)
Since, three numbers i.e. 4, 1 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
\( 1000x = 357417.417417... \)
\( \implies 1000x = 357417.\overline{417} \) ...(ii)
Subtracting (i) from (ii),
\( 1000x - x = 357417.\overline{417} - 357.\overline{417} \)
\( \implies 999x = 357060 \)
\( \implies x = \frac{357060}{999} = \frac{3 \times 119020}{3 \times 333} = \frac{119020}{333} \)
\( \implies 357.417417... = \frac{119020}{333} \)
In simple words: We multiply by 1000 because three digits repeat, subtract the original equation to eliminate the decimal part, and then simplify the fraction by dividing both numbers by 3.

🎯 Exam Tip: Double-check your subtraction of the whole numbers to avoid simple calculation errors when finding the numerator.

Question 2. (v) Express \( 30.\overline{219} \) in \( \frac{p}{q} \) form.
Answer:
Let \( x = 30.\overline{219} \) ...(i)
\( \therefore x = 30.219219... \)
Since three numbers (2, 1, and 9) are repeating after the decimal point, we multiply both sides by 1000.
\( 1000x = 30219.219219... \)
\( \therefore 1000x = 30219.\overline{219} \) ...(ii)
Subtracting (i) from (ii),
\( 1000x - x = 30219.\overline{219} - 30.\overline{219} \)
\( \therefore 999x = 30189 \)
\( \therefore x = \frac{30189}{999} = \frac{3 \times 10063}{3 \times 333} \)
\( \therefore 30.\overline{219} = \frac{10063}{333} \)
In simple words: To convert a repeating decimal to a fraction, we multiply it by a power of 10 based on how many digits repeat, subtract the original equation to cancel out the repeating part, and then solve for x.

🎯 Exam Tip: Always count the number of repeating digits under the bar to decide whether to multiply by 10, 100, or 1000 to eliminate the decimal part successfully.

Question 3. Write the following numbers in its decimal form.
i. \( \frac{-5}{7} \)
ii. \( \frac{9}{11} \)
iii. \( \sqrt{5} \)
iv. \( \frac{121}{13} \)
v. \( \frac{29}{8} \)
Answer:
i. \( \frac{-5}{7} = -0.\overline{714285} \)
ii. \( \frac{9}{11} = 0.\overline{81} \)
iii. \( \sqrt{5} = 2.2360679... \)
iv. \( \frac{121}{13} = 9.\overline{307692} \)
v. \( \frac{29}{8} = 3.625 \)
The decimal form of a number can be either terminating or non-terminating recurring, depending on its prime factors.
In simple words: To convert a fraction into a decimal, we divide the numerator by the denominator. If the division doesn't end, we put a bar over the repeating digits. For square roots of non-perfect squares like 5, the decimal goes on forever without repeating.

🎯 Exam Tip: When converting fractions to decimals, remember to place a bar over the entire repeating block of digits to show it is recurring.

Question 1. Write the following rational numbers in decimal form:
i. \( \frac{-5}{7} \)
ii. \( \frac{9}{11} \)
Answer:
i. \( \frac{-5}{7} \)
First, perform the long division of \( 5 \) by \( 7 \):
\[ \begin{array}{r} 0.714285... \\ 7 \overline{) 5.000000} \\ \underline{- 0} \phantom{000000} \\ 50 \phantom{00000} \\ \underline{- 49} \phantom{00000} \\ 10 \phantom{0000} \\ \underline{- 7} \phantom{0000} \\ 30 \phantom{000} \\ \underline{- 28} \phantom{000} \\ 20 \phantom{00} \\ \underline{- 14} \phantom{00} \\ 60 \phantom{0} \\ \underline{- 56} \phantom{0} \\ 40 \\ \underline{- 35} \\ 5 \end{array} \]
Since the remainder \( 5 \) repeats, the decimal group \( 714285 \) will keep repeating infinitely. This shows that the rational number has a non-terminating recurring decimal representation.
\( \therefore \frac{-5}{7} = -0.\overline{714285} \)

ii. \( \frac{9}{11} \)
Perform the long division of \( 9 \) by \( 11 \):
\[ \begin{array}{r} 0.81... \\ 11 \overline{) 9.00} \\ \underline{- 0} \phantom{00} \\ 90 \phantom{0} \\ \underline{- 88} \phantom{0} \\ 20 \\ \underline{- 11} \\ 9 \end{array} \]
Since the remainder \( 9 \) repeats, the decimal group \( 81 \) will keep repeating infinitely.
\( \therefore \frac{9}{11} = 0.\overline{81} \)
In simple words: To convert these fractions into decimals, we divide the top number by the bottom number. When the division results in a repeating pattern of remainders, we place a horizontal bar over the repeating digits to show they recur forever.

🎯 Exam Tip: Always remember to carry over the negative sign to your final decimal answer if the original fraction was negative, and ensure the recurring bar covers only the digits that actually repeat.

Question 1. Find the decimal representation of the following:
iii. \( \sqrt{5} \)
iv. \( \frac{121}{13} \)
Answer:

iii. \( \sqrt{5} \)
By using the long division method to find the square root of 5:

\( \therefore \sqrt{5} = 2.2360679... \)

iv. \( \frac{121}{13} \)
By dividing 121 by 13:

\( \therefore \frac{121}{13} = 9.\overline{307692} \)
The decimal representation of this fraction continues indefinitely in a repeating pattern.
In simple words: To find the decimal form of these numbers, we use division. For the square root of 5, the numbers after the decimal point go on forever without repeating, whereas for 121 divided by 13, the block of numbers "307692" repeats over and over.

🎯 Exam Tip: When writing recurring decimals, always place a horizontal bar over the entire repeating block of digits to indicate that the pattern repeats infinitely.

Question v. Express \( \frac{29}{8} \) in decimal form.
Answer:
Perform the long division of 29 by 8:
\( \begin{array}{r} 3.625 \\ 8 \overline{) 29.000} \\ \underline{-24} \\ 50 \\ \underline{-48} \\ 20 \\ \underline{-16} \\ 40 \\ \underline{-40} \\ 0 \end{array} \)

\( \implies \frac{29}{8} = 3.625 \)
In simple words: To convert a fraction into a decimal, we divide the top number by the bottom number. Since 29 divided by 8 leaves no remainder after three decimal places, it gives us the exact terminating decimal 3.625.

🎯 Exam Tip: Always write down the zero placeholders after the decimal point in long division to keep your subtraction columns perfectly aligned.

Question 4. Show that \( 5 + \sqrt{7} \) is an irrational number. [3 Marks]
Answer:
Let us assume that \( 5 + \sqrt{7} \) is a rational number. So, we can find co-prime integers 'a' and 'b' (\( b \neq 0 \)) such that:
\( 5 + \sqrt{7} = \frac{a}{b} \)

\( \implies \sqrt{7} = \frac{a}{b} - 5 \)
Since 'a' and 'b' are integers, \( \frac{a}{b} - 5 \) is a rational number and so \( \sqrt{7} \) must also be a rational number.
But this contradicts the fact that \( \sqrt{7} \) is an irrational number.
Our assumption that \( 5 + \sqrt{7} \) is a rational number is wrong.

\( \implies 5 + \sqrt{7} \) is an irrational number.
In simple words: We start by pretending that the number is rational. But when we rearrange the equation, it claims that an irrational number equals a rational one, which is impossible and proves our assumption was wrong.

🎯 Exam Tip: Clearly state the contradiction between the rational right-hand side and the known irrationality of the square root term to secure full marks.

Question 5. Write the following surds in simplest form.
(i) \( \frac{3}{4}\sqrt{8} \)
(ii) \( -\frac{5}{9}\sqrt{45} \)
Answer:
(i) \( \frac{3}{4}\sqrt{8} \)
\( \frac{3}{4}\sqrt{8} = \frac{3}{4}\sqrt{4 \times 2} \)

\( \implies \frac{3}{4} \times 2\sqrt{2} \)

\( \implies \frac{3}{2}\sqrt{2} \)

(ii) \( -\frac{5}{9}\sqrt{45} \)
\( -\frac{5}{9}\sqrt{45} = -\frac{5}{9}\sqrt{9 \times 5} \)

\( \implies -\frac{5}{9} \times 3\sqrt{5} \)

\( \implies -\frac{5}{3}\sqrt{5} \)
In simple words: To simplify a surd, find the largest perfect square number that divides the number inside the root, take its square root outside, and multiply it by the fraction.

🎯 Exam Tip: Always look for perfect square factors like 4, 9, 16, or 25 inside the radical to simplify surds to their lowest terms.

Question 5. Simplify the following surds:
(i) \( \frac{3}{4}\sqrt{8} \)
(ii) \( -\frac{5}{9}\sqrt{45} \)
Answer:
(i) \( \frac{3}{4}\sqrt{8} = \frac{3}{4} \times \sqrt{4 \times 2} \)
\( = \frac{3}{4} \times 2\sqrt{2} \)
\( \therefore \frac{3}{4}\sqrt{8} = \frac{3}{2}\sqrt{2} \)

(ii) \( -\frac{5}{9}\sqrt{45} = -\frac{5}{9} \times \sqrt{9 \times 5} \)
\( = -\frac{5}{9} \times 3\sqrt{5} \)
\( \therefore -\frac{5}{9}\sqrt{45} = -\frac{5}{3}\sqrt{5} \)
This process helps us work with smaller, more manageable numbers under the radical.
In simple words: To simplify a surd, we find the largest perfect square factor of the number inside the square root, take its square root outside, and simplify the remaining fraction.

🎯 Exam Tip: Always look for the largest perfect square factor (like 4, 9, 16, 25) under the radical sign to simplify surds quickly and avoid calculation errors.

Question 6. Write the simplest form of rationalising factor for the given surds.
(i) \( \sqrt{32} \)
(ii) \( \sqrt{50} \)
(iii) \( \sqrt{27} \)
(iv) \( \frac{3}{5}\sqrt{10} \)
(v) \( 3\sqrt{72} \)
(vi) \( 4\sqrt{11} \)
Answer:
(i) \( \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \)
Now, \( 4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8 \), which is a rational number.
\( \therefore \sqrt{2} \) is the simplest form of the rationalising factor of \( \sqrt{32} \).

(ii) \( \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \)
Now, \( 5\sqrt{2} \times \sqrt{2} = 5 \times 2 = 10 \), which is a rational number.
\( \dots \sqrt{2} \) is the simplest form of the rationalising factor of \( \sqrt{50} \).

(iii) \( \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \)
Now, \( 3\sqrt{3} \times \sqrt{3} = 3 \times 3 = 9 \), which is a rational number.
\( \therefore \sqrt{3} \) is the simplest form of the rationalising factor of \( \sqrt{27} \).

(iv) \( \frac{3}{5}\sqrt{10} \)
Now, \( \frac{3}{5}\sqrt{10} \times \sqrt{10} = \frac{3}{5} \times 10 = 3 \times 2 = 6 \), which is a rational number.
\( \therefore \sqrt{10} \) is the simplest form of the rationalising factor of \( \frac{3}{5}\sqrt{10} \).

(v) \( 3\sqrt{72} = 3\sqrt{36 \times 2} = 3 \times 6\sqrt{2} = 18\sqrt{2} \)
Now, \( 18\sqrt{2} \times \sqrt{2} = 18 \times 2 = 36 \), which is a rational number.
\( \therefore \sqrt{2} \) is the simplest form of the rationalising factor of \( 3\sqrt{72} \).

(vi) \( 4\sqrt{11} \)
Now, \( 4\sqrt{11} \times \sqrt{11} = 4 \times 11 = 44 \), which is a rational number.
\( \therefore \sqrt{11} \) is the simplest form of the rationalising factor of \( 4\sqrt{11} \).
This shows that multiplying an irrational surd by its rationalising factor always results in a clean integer or fraction.
In simple words: A rationalising factor is a number we multiply a surd by to turn it into a normal, rational number. To find the simplest one, first simplify the surd and then find the smallest square root that clears the radical.

🎯 Exam Tip: To find the simplest rationalising factor, always simplify the surd first. The remaining irrational part under the square root is your simplest rationalising factor.

Question 6. Find the simplest form of the rationalising factor of the following:
(v) \( 3\sqrt{72} \)
(vi) \( 4\sqrt{11} \)
Answer:
(v) \( 3\sqrt{72} = 3\sqrt{36 \times 2} = 3 \times 6\sqrt{2} = 18\sqrt{2} \)
Now, \( 18\sqrt{2} \times \sqrt{2} = 18 \times 2 = 36 \), which is a rational number. This shows that multiplying by the simplest square root term successfully eliminates the radical.
\( \therefore \sqrt{2} \) is the simplest form of the rationalising factor of \( 3\sqrt{72} \).

(vi) \( 4\sqrt{11} \)
\( 4\sqrt{11} \times \sqrt{11} = 4 \times 11 = 44 \), which is a rational number.
\( \dots \sqrt{11} \) is the simplest form of the rationalising factor of \( 4\sqrt{11} \).
In simple words: To find the simplest rationalising factor, simplify the surd first and then find the smallest square root term that multiplies with it to make it a whole number.

🎯 Exam Tip: Always simplify the surd to its lowest form first before finding the rationalising factor to avoid using larger, non-simplified roots.

Question 7. Simplify.
(i) \( \frac{4}{7}\sqrt{147} + \frac{3}{8}\sqrt{192} - \frac{1}{5}\sqrt{75} \)
(ii) \( 5\sqrt{3} + 2\sqrt{27} + \frac{1}{\sqrt{3}} \)
(iii) \( \sqrt{216} - 5\sqrt{6} + \sqrt{294} - \frac{3}{\sqrt{6}} \)
(iv) \( 4\sqrt{12} - \sqrt{75} - 7\sqrt{48} \)
(v) \( 2\sqrt{48} - \sqrt{75} - \frac{1}{\sqrt{3}} \)
Answer:
(i) \( \frac{4}{7}\sqrt{147} + \frac{3}{8}\sqrt{192} - \frac{1}{5}\sqrt{75} \)
\( = \frac{4}{7}\sqrt{49 \times 3} + \frac{3}{8}\sqrt{64 \times 3} - \frac{1}{5}\sqrt{25 \times 3} \)
\( = \frac{4}{7} \times 7\sqrt{3} + \frac{3}{8} \times 8\sqrt{3} - \frac{1}{5} \times 5\sqrt{3} \)
\( = 4\sqrt{3} + 3\sqrt{3} - \sqrt{3} \)
\( = (4 + 3 - 1)\sqrt{3} \)
\( = 6\sqrt{3} \)
\( \therefore \frac{4}{7}\sqrt{147} + \frac{3}{8}\sqrt{192} - \frac{1}{5}\sqrt{75} = 6\sqrt{3} \)

(ii) \( 5\sqrt{3} + 2\sqrt{27} + \frac{1}{\sqrt{3}} \)
\( = 5\sqrt{3} + 2\sqrt{9 \times 3} + \frac{1 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
\( = 5\sqrt{3} + 2 \times 3\sqrt{3} + \frac{\sqrt{3}}{3} \)
\( = 5\sqrt{3} + 6\sqrt{3} + \frac{\sqrt{3}}{3} \)
\( = 11\sqrt{3} + \frac{\sqrt{3}}{3} \)
\( = \left(11 + \frac{1}{3}\right)\sqrt{3} \)
\( = \frac{34}{3}\sqrt{3} \)
\( \therefore 5\sqrt{3} + 2\sqrt{27} + \frac{1}{\sqrt{3}} = \frac{34}{3}\sqrt{3} \)

(iii) \( \sqrt{216} - 5\sqrt{6} + \sqrt{294} - \frac{3}{\sqrt{6}} \)
\( = \sqrt{36 \times 6} - 5\sqrt{6} + \sqrt{49 \times 6} - \frac{3 \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}} \)
\( = 6\sqrt{6} - 5\sqrt{6} + 7\sqrt{6} - \frac{3\sqrt{6}}{6} \)
\( = (6 - 5 + 7)\sqrt{6} - \frac{\sqrt{6}}{2} \)
\( = 8\sqrt{6} - \frac{1}{2}\sqrt{6} \)
\( = \left(8 - \frac{1}{2}\right)\sqrt{6} \)
\( = \frac{15}{2}\sqrt{6} \)
\( \therefore \sqrt{216} - 5\sqrt{6} + \sqrt{294} - \frac{3}{\sqrt{6}} = \frac{15}{2}\sqrt{6} \)

(iv) \( 4\sqrt{12} - \sqrt{75} - 7\sqrt{48} \)
\( = 4\sqrt{4 \times 3} - \sqrt{25 \times 3} - 7\sqrt{16 \times 3} \)
\( = 4 \times 2\sqrt{3} - 5\sqrt{3} - 7 \times 4\sqrt{3} \)
\( = 8\sqrt{3} - 5\sqrt{3} - 28\sqrt{3} \)
\( = (8 - 5 - 28)\sqrt{3} \)
\( = -25\sqrt{3} \)
\( \therefore 4\sqrt{12} - \sqrt{75} - 7\sqrt{48} = -25\sqrt{3} \)

(v) \( 2\sqrt{48} - \sqrt{75} - \frac{1}{\sqrt{3}} \)
\( = 2\sqrt{16 \times 3} - \sqrt{25 \times 3} - \frac{1 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \)
\( = 2 \times 4\sqrt{3} - 5\sqrt{3} - \frac{\sqrt{3}}{3} \)
\( = 8\sqrt{3} - 5\sqrt{3} - \frac{\sqrt{3}}{3} \)
\( = 3\sqrt{3} - \frac{\sqrt{3}}{3} \)
\( = \left(3 - \frac{1}{3}\right)\sqrt{3} \)
\( = \frac{8}{3}\sqrt{3} \)
\( \therefore 2\sqrt{48} - \sqrt{75} - \frac{1}{\sqrt{3}} = \frac{8}{3}\sqrt{3} \)
This step-by-step simplification ensures that all terms are expressed with the same radicand before performing arithmetic operations.
In simple words: To simplify these expressions, convert each surd into its simplest form with matching square root terms, rationalise any denominators, and then add or subtract their coefficients.

🎯 Exam Tip: When simplifying surds with fractions, always rationalise the denominator first to make it easier to combine like terms.

Question. Simplify the following expressions:
(ii) \( 5\sqrt{3} + 2\sqrt{27} + \frac{1}{\sqrt{3}} \)
(iii) \( \sqrt{216} - 5\sqrt{6} + \sqrt{294} - \frac{3}{\sqrt{6}} \)
Answer:
(ii) \( 5\sqrt{3} + 2\sqrt{27} + \frac{1}{\sqrt{3}} \)
\( = 5\sqrt{3} + 2\sqrt{9 \times 3} + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \) (Rationalizing the denominator of the third term helps simplify the expression)
\( = 5\sqrt{3} + 2 \times 3\sqrt{3} + \frac{\sqrt{3}}{3} \)
\( = 5\sqrt{3} + 6\sqrt{3} + \frac{\sqrt{3}}{3} \)
\( = \left(5 + 6 + \frac{1}{3}\right)\sqrt{3} \)
\( = \left(11 + \frac{1}{3}\right)\sqrt{3} \)
\( = \left(\frac{33 + 1}{3}\right)\sqrt{3} \)
\( = \frac{34}{3}\sqrt{3} \)
\( \therefore 5\sqrt{3} + 2\sqrt{27} + \frac{1}{\sqrt{3}} = \frac{34}{3}\sqrt{3} \)

(iii) \( \sqrt{216} - 5\sqrt{6} + \sqrt{294} - \frac{3}{\sqrt{6}} \)
\( = \sqrt{36 \times 6} - 5\sqrt{6} + \sqrt{49 \times 6} - \frac{3}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} \) (We find perfect square factors for each radicand to simplify them)
\( = 6\sqrt{6} - 5\sqrt{6} + 7\sqrt{6} - \frac{3\sqrt{6}}{6} \)
\( = 6\sqrt{6} - 5\sqrt{6} + 7\sqrt{6} - \frac{1}{2}\sqrt{6} \)
\( = \left(6 - 5 + 7 - \frac{1}{2}\right)\sqrt{6} \)
\( = \left(8 - \frac{1}{2}\right)\sqrt{6} \)
\( = \left(\frac{16 - 1}{2}\right)\sqrt{6} \)
\( = \frac{15}{2}\sqrt{6} \)
\( \therefore \sqrt{216} - 5\sqrt{6} + \sqrt{294} - \frac{3}{\sqrt{6}} = \frac{15}{2}\sqrt{6} \)
In simple words: To simplify these expressions, we first convert all surds into their simplest form by finding perfect square factors. Then, we rationalize any denominators and combine the like terms together just like normal numbers.

🎯 Exam Tip: Always rationalize the denominator first and find the largest perfect square factor of the numbers inside the square root to simplify surds quickly.

Question 7. Simplify:
(iv) \( 4\sqrt{12} - \sqrt{75} - 7\sqrt{48} \)
(v) \( 2\sqrt{48} - \sqrt{75} - \frac{1}{\sqrt{3}} \)
Answer:
(iv) \( 4\sqrt{12} - \sqrt{75} - 7\sqrt{48} \)
\( = 4\sqrt{4 \times 3} - \sqrt{25 \times 3} - 7\sqrt{16 \times 3} \)
\( = 4 \times 2\sqrt{3} - 5\sqrt{3} - 7 \times 4\sqrt{3} \)
\( = 8\sqrt{3} - 5\sqrt{3} - 28\sqrt{3} \)
\( = (8 - 5 - 28)\sqrt{3} \)
\( = (-25)\sqrt{3} \)
\( \therefore 4\sqrt{12} - \sqrt{75} - 7\sqrt{48} = -25\sqrt{3} \)

(v) \( 2\sqrt{48} - \sqrt{75} - \frac{1}{\sqrt{3}} \)
\( = 2\sqrt{16 \times 3} - \sqrt{25 \times 3} - \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \)
\( = 2 \times 4\sqrt{3} - 5\sqrt{3} - \frac{1}{3}\sqrt{3} \)
\( = 8\sqrt{3} - 5\sqrt{3} - \frac{1}{3}\sqrt{3} \)
\( = \left(8 - 5 - \frac{1}{3}\right)\sqrt{3} \)
\( = \left(3 - \frac{1}{3}\right)\sqrt{3} \)
\( = \left(\frac{9 - 1}{3}\right)\sqrt{3} \)
\( = \frac{8}{3}\sqrt{3} \)
\( \therefore 2\sqrt{48} - \sqrt{75} - \frac{1}{\sqrt{3}} = \frac{8}{3}\sqrt{3} \)
In simple words: To simplify these expressions, we break down the numbers inside the square roots into factors that are perfect squares, pull them out, and then add or subtract the remaining like terms.

🎯 Exam Tip: Always look for the largest perfect square factor (like 4, 16, 25) when simplifying surds to make the calculation faster and avoid errors.

Question 8. Rationalize the denominator.
(i) \( \frac{1}{\sqrt{5}} \)
(ii) \( \frac{2}{3\sqrt{7}} \)
(iii) \( \frac{1}{\sqrt{3}-\sqrt{2}} \)
(iv) \( \frac{1}{3\sqrt{5}+2\sqrt{2}} \)
(v) \( \frac{12}{4\sqrt{3}-\sqrt{2}} \)
Answer:
(i) \( \frac{1}{\sqrt{5}} \)
Multiply the numerator and denominator by \( \sqrt{5} \):
\( \frac{1 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} = \frac{\sqrt{5}}{5} \)

(ii) \( \frac{2}{3\sqrt{7}} \)
Multiply the numerator and denominator by \( \sqrt{7} \):
\( \frac{2 \times \sqrt{7}}{3\sqrt{7} \times \sqrt{7}} = \frac{2\sqrt{7}}{3 \times 7} = \frac{2\sqrt{7}}{21} \)

(iii) \( \frac{1}{\sqrt{3}-\sqrt{2}} \)
Multiply the numerator and denominator by the conjugate \( \sqrt{3}+\sqrt{2} \):
\( \frac{1 \times (\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} = \frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3}+\sqrt{2}}{3-2} = \sqrt{3}+\sqrt{2} \)

(iv) \( \frac{1}{3\sqrt{5}+2\sqrt{2}} \)
Multiply the numerator and denominator by the conjugate \( 3\sqrt{5}-2\sqrt{2} \):
\( \frac{1 \times (3\sqrt{5}-2\sqrt{2})}{(3\sqrt{5}+2\sqrt{2})(3\sqrt{5}-2\sqrt{2})} = \frac{3\sqrt{5}-2\sqrt{2}}{(3\sqrt{5})^2 - (2\sqrt{2})^2} = \frac{3\sqrt{5}-2\sqrt{2}}{45 - 8} = \frac{3\sqrt{5}-2\sqrt{2}}{37} \)

(v) \( \frac{12}{4\sqrt{3}-\sqrt{2}} \)
Multiply the numerator and denominator by the conjugate \( 4\sqrt{3}+\sqrt{2} \):
\( \frac{12 \times (4\sqrt{3}+\sqrt{2})}{(4\sqrt{3}-\sqrt{2})(4\sqrt{3}+\sqrt{2})} = \frac{12(4\sqrt{3}+\sqrt{2})}{(4\sqrt{3})^2 - (\sqrt{2})^2} = \frac{12(4\sqrt{3}+\sqrt{2})}{48 - 2} = \frac{12(4\sqrt{3}+\sqrt{2})}{46} = \frac{6(4\sqrt{3}+\sqrt{2})}{23} = \frac{24\sqrt{3}+6\sqrt{2}}{23} \)
In simple words: To rationalize a denominator, we multiply the top and bottom of the fraction by the square root term (or its conjugate partner) so that the square root on the bottom disappears.

🎯 Exam Tip: When rationalizing binomial denominators, always use the conjugate identity \( (a-b)(a+b) = a^2 - b^2 \) to easily eliminate the square roots from the denominator.

Question 1. Rationalize the denominator of the following:
(i) \( \frac{1}{\sqrt{5}} \)
(ii) \( \frac{2}{3\sqrt{7}} \)
(iii) \( \frac{1}{\sqrt{3}-\sqrt{2}} \)
Answer: Rationalizing the denominator helps simplify expressions by removing radical signs from the bottom of a fraction.
(i) \( \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \) ... [Multiplying the numerator and denominator by \( \sqrt{5} \)]
\( = \frac{1 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}} \)
\( \therefore \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \)

(ii) \( \frac{2}{3\sqrt{7}} = \frac{2}{3\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} \) ... [Multiplying the numerator and denominator by \( \sqrt{7} \)]
\( = \frac{2 \times \sqrt{7}}{3\sqrt{7} \times \sqrt{7}} = \frac{2\sqrt{7}}{3 \times 7} \)
\( \therefore \frac{2}{3\sqrt{7}} = \frac{2\sqrt{7}}{21} \)

(iii) \( \frac{1}{\sqrt{3}-\sqrt{2}} = \frac{1}{(\sqrt{3}-\sqrt{2})} \times \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})} \) ... [Multiplying the numerator and denominator by \( (\sqrt{3}+\sqrt{2}) \)]
\( = \frac{1 \times (\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})} \)
\( = \frac{\sqrt{3}+\sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} \)
\( = \frac{\sqrt{3}+\sqrt{2}}{3 - 2} \)
\( \therefore \frac{1}{\sqrt{3}-\sqrt{2}} = \sqrt{3}+\sqrt{2} \)
In simple words: To rationalize a denominator, we multiply both the top and bottom of the fraction by a term that eliminates the square root from the bottom. This makes the fraction much easier to work with in calculations.

🎯 Exam Tip: When rationalizing a binomial denominator like \( \sqrt{a} - \sqrt{b} \), always multiply by its conjugate \( \sqrt{a} + \sqrt{b} \) to use the identity \( (x-y)(x+y) = x^2 - y^2 \).

Question 1. Rationalize the denominator of \( \frac{1}{\sqrt{3} - \sqrt{2}} \)
Answer: To rationalize the denominator of \( \frac{1}{\sqrt{3} - \sqrt{2}} \), we multiply the numerator and denominator by its conjugate partner \( \sqrt{3} + \sqrt{2} \). This helps to eliminate the radical signs from the denominator.
\( \frac{1}{\sqrt{3} - \sqrt{2}} = \frac{1 \times (\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} \)
\( = \frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} \)
[Since \( (a+b)(a-b) = a^2 - b^2 \)]
\( = \frac{\sqrt{3} + \sqrt{2}}{3-2} \)
\( = \frac{\sqrt{3} + \sqrt{2}}{1} \)

\( \implies \frac{1}{\sqrt{3} - \sqrt{2}} = \sqrt{3} + \sqrt{2} \)
In simple words: To get rid of the square roots in the bottom of the fraction, we multiply both top and bottom by the same numbers but with a plus sign instead of a minus sign. This simplifies the bottom to just 1.

🎯 Exam Tip: Always write the algebraic identity \( (a-b)(a+b) = a^2 - b^2 \) in brackets next to the step where you use it to secure full marks.

Question 2. Rationalize the denominator:
(iv) \( \frac{1}{3\sqrt{5} + 2\sqrt{2}} \)
Answer: To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is \( 3\sqrt{5} - 2\sqrt{2} \). This process helps us eliminate irrational numbers from the bottom of the fraction.
\( \frac{1}{3\sqrt{5} + 2\sqrt{2}} = \frac{1}{3\sqrt{5} + 2\sqrt{2}} \times \frac{3\sqrt{5} - 2\sqrt{2}}{3\sqrt{5} - 2\sqrt{2}} \)
[Multiplying the numerator and denominator by \( (3\sqrt{5} - 2\sqrt{2}) \)]
\( = \frac{1 \times (3\sqrt{5} - 2\sqrt{2})}{(3\sqrt{5} + 2\sqrt{2})(3\sqrt{5} - 2\sqrt{2})} \)
\( = \frac{3\sqrt{5} - 2\sqrt{2}}{(3\sqrt{5})^2 - (2\sqrt{2})^2} \)
[Since \( (a+b)(a-b) = a^2 - b^2 \)]
\( = \frac{3\sqrt{5} - 2\sqrt{2}}{(9 \times 5) - (4 \times 2)} \)
\( = \frac{3\sqrt{5} - 2\sqrt{2}}{45 - 8} \)

\( \implies \frac{1}{3\sqrt{5} + 2\sqrt{2}} = \frac{3\sqrt{5} - 2\sqrt{2}}{37} \)
In simple words: We multiply the top and bottom by the conjugate partner \( 3\sqrt{5} - 2\sqrt{2} \) to eliminate the square roots from the denominator, leaving us with a clean whole number 37 at the bottom.

🎯 Exam Tip: Be very careful when squaring terms like \( (3\sqrt{5})^2 \). Remember to square both the whole number and the root: \( 3^2 \times 5 = 9 \times 5 = 45 \).

Question iv. Rationalize the denominator: \( \frac{1}{3\sqrt{5} + 2\sqrt{2}} \)
Answer:
To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is \( (3\sqrt{5} - 2\sqrt{2}) \). This process of removing radicals from the denominator is essential for simplifying algebraic expressions.
\( \frac{1}{3\sqrt{5} + 2\sqrt{2}} = \frac{1 \times (3\sqrt{5} - 2\sqrt{2})}{(3\sqrt{5} + 2\sqrt{2})(3\sqrt{5} - 2\sqrt{2})} \)
\( = \frac{3\sqrt{5} - 2\sqrt{2}}{(3\sqrt{5})^2 - (2\sqrt{2})^2} \)
\( = \frac{3\sqrt{5} - 2\sqrt{2}}{45 - 8} \)

\( \implies \frac{1}{3\sqrt{5} + 2\sqrt{2}} = \frac{3\sqrt{5} - 2\sqrt{2}}{37} \)
In simple words: To rationalize the denominator, we multiply both the top and bottom by the opposite sign partner of the bottom part, which helps clear out the square roots from the bottom.

🎯 Exam Tip: Always remember that the conjugate of \( a + \sqrt{b} \) is \( a - \sqrt{b} \). Multiplying them together eliminates the square root using the identity \( (a+b)(a-b) = a^2 - b^2 \).

Question v. Rationalize the denominator: \( \frac{12}{4\sqrt{3} - \sqrt{2}} \)
Answer:
To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator, which is \( (4\sqrt{3} + \sqrt{2}) \). Rationalizing denominators makes it much easier to add or subtract fractions containing square roots.
\( \frac{12}{4\sqrt{3} - \sqrt{2}} = \frac{12}{(4\sqrt{3} - \sqrt{2})} \times \frac{(4\sqrt{3} + \sqrt{2})}{(4\sqrt{3} + \sqrt{2})} \)
[Multiplying the numerator and denominator by \( (4\sqrt{3} + \sqrt{2}) \)]
\( = \frac{12(4\sqrt{3} + \sqrt{2})}{(4\sqrt{3} - \sqrt{2})(4\sqrt{3} + \sqrt{2})} \)
\( = \frac{12(4\sqrt{3} + \sqrt{2})}{(4\sqrt{3})^2 - (\sqrt{2})^2} \)
[\( \because (a + b)(a - b) = a^2 - b^2 \)]
\( = \frac{12(4\sqrt{3} + \sqrt{2})}{(16 \times 3) - 2} \)
\( = \frac{12(4\sqrt{3} + \sqrt{2})}{48 - 2} \)
\( = \frac{12(4\sqrt{3} + \sqrt{2})}{46} \)

\( \implies \frac{12}{4\sqrt{3} - \sqrt{2}} = \frac{6(4\sqrt{3} + \sqrt{2})}{23} \)
In simple words: We multiply the top and bottom by \( 4\sqrt{3} + \sqrt{2} \) to remove the square roots from the bottom, then simplify the fraction by dividing both 12 and 46 by 2.

🎯 Exam Tip: Don't forget to simplify the final fraction by dividing the numerator and denominator by their greatest common divisor to get full marks.

Question 1. Draw three or four circles of different radii on a card board. Cut these circles. Take a thread and measure the length of circumference and diameter of each of the circles. Note down the readings in the given table. (Textbook pg.no.23 )
Answer:
The completed table with the measured readings is shown below. This hands-on activity helps visually demonstrate the constant relationship between a circle's boundary and its width.

In simple words: When you measure any circle, its diameter is always twice its radius, and dividing its outer boundary (circumference) by its width (diameter) always gives a constant value of about 3.1.

🎯 Exam Tip: Remember that the ratio of circumference to diameter (\( \frac{c}{d} \)) is always constant and is represented by the Greek letter \( \pi \) (pi), which is approximately equal to 3.14 or \( \frac{22}{7} \).

Question 2. To find the approximate value of \( \pi \), take the wire of length 11 cm, 22 cm and 33 cm each. Make a circle from the wire. Measure the diameter and complete the following table. Verify that the ratio of circumference to the diameter of a circle is approximately \( \frac{22}{7} \).

Answer:
By measuring the diameter of each circle formed by the wires, we get the following completed table:

Measuring these physical wires carefully helps us visually grasp how the ratio remains constant regardless of the circle's size.
\( \therefore \) The ratio of circumference to the diameter of each circle is approximately \( \frac{22}{7} \).
In simple words: No matter how big or small a circle is, if you divide the distance around it by the width across its center, you will always get a value very close to \( \frac{22}{7} \) (which is about 3.14).

🎯 Exam Tip: Always remember that the ratio of a circle's circumference to its diameter is a constant value denoted by \( \pi \), which is approximately \( \frac{22}{7} \) or 3.14.