Operations on Sets, Assignments of Mathematics

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CHAPTER

1 Operations on Sets

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Animation 1.1: Operations on Sets Source & Credit: elearn.punjab

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1.1 SETS

We know that a set is a collection of well defined distinct objects or symbols. The objects are called its members or elements.

1.1.1 Recognize some important sets and their notations

Set of natural numbers: N = {1, 2, 3,.. .} Set of whole numbers: W = {0, 1,2,.. .} Set of integers: Z = {... , - 3, - 2, - 1, 0, I, 2, 3,.. .} Set of prime numbers: P = {2,3,5,7,11,...} Set of odd numbers: O = {!1, !3, !5,.. .} Set of even numbers: E = {0,!2,!4,...} Set of rational numbers: Q = { pq | p, qdZ q ≠ 0}

1.1.2 Finding subsets of a set

lt is illustrated through the following examples.

Example 1: Write all the subsets of the set {2, 4}

Solution: Following are the subsets of the set {2, 4} f ,{2}, {4}, {2,4}

Example 2: Write all the subsets of the set {3, 5, 7}

Solution: Following are the subsets of the set {3, 5, 7} f , {3}, {5}, {7}, {3,5}, {3, 7}, {5, 7}, {3 5, 7}

Example 3: Write all the subsets of the set X= {a, b, c, d}

Solution: Subsets of X are: f, { a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {c, d}, {a, b, c}, {a, b, d}, {b, c, d}, {a, c, d}, {a, b, c, d}

1.1.3 Definitions

(a) Proper Subset lf A and B are two sets and every element of set A is also an element of set B but at least one element of the set B is not an element of the set A , then the set A is called a proper subset of set B. lt is denoted by AfB and read as set A is a proper subset of the set B. For example, if A={1, 2, 3 } and B={I, 2, 3, 4} then A f B.

Remember that: (i) Every set is a subset of itself. (ii) Empty set is a proper subset of every non-empty set.

(b) Improper Subset lf A and B are two sets and set A is a subset of set B and B is also a subset of set A then A is called an improper subset of set B and B is an improper subset of set A.

Note: (i) All the subsets of a set except the set itself are proper subsets of the set. (ii) Procedure of writing subsets of a given set: First of all write empty set, then singleton sets, (a set containing one element only is called singleton set) then sets having two members and so on. Continue till the number of elements becomes equal to the given set. (iii) Every set is an improper subset of itself. (iv) There is no proper subset of an empty set. (v) There is only one proper subset of a singleton set.

1.1.4 Power Set

A set consisting of all possible subsets of a given set A is called the power set of A and is denoted by P(A).

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Remember that: To find union / intersection of three sets, first we find the union / intersection of any two of them and then the union / intersection of the third set with the resultant set.

Example 1: Verify the associative laws of union (i) Aj(BjC) (ii) (AjB)jC where A = {1,2,3,4}, B = {3,4,5,6,7,8} and C = {6,7,8,9,10}

Solution: (i) Aj(BjC) = {1,2,3,4}j({3,4,5,6,7,a} j {6,7,3,9,10}) = {1, 2, 3, 4} j {3, 4, 5, 6, 7, 8, 9, 10} = {1, 2, 3, 4, 5, 6, 7, 3, 9, 10} ........................... (1) (ii) (A j B)jC = ({1, 2, 3, 4} j {3, 4, 5, 6, 7, 8}) j {6, 7, 8, 9, 10} = {1, 2, 3, 4, 5, 6, 7, 8} j {6, 7, 8, 9, 10} = {I, 2, 3, 4, 5, 6, 7, 8, 9, I0} ............................. (2)

Thus, from (1) and (2), we conclude that Aj(BjC) = (AjB)jC

Example 2: Verify the associative laws of intersection (i) Ak(BkC) and (ii) (AkB)kC for sets given in example 1.

Solution: (i) Ak ( Bk C) = { 1, 2, 3, 4 }k ( { 3, 4, 5, 6, 7, 8 }k{ 6, 7, 8, 9, 10 } ) = { 1,2,3,4 }k{ 6, 7,8 } = f ............. (a) (ii) ( A k B )k C = ( { 1, 2, 3, 4 } k { 3, 4, 5, 6, 7, 8 } ) k { 6, 7, 8, 9, 10 } = { 3, 4 } k { 6, 7, 8, 9, 10 } = f ............. (b)

Thus, from (a) and (b), we conclude that Ak(BkC) = (AkB)kC

1.2.2 Verification of Distributive Laws

lf A , B and C are three sets, then Aj ( Bk C) = (A j B) k (A j C) is called the distributive law of union over intersection.

lf A , B and C are three sets, then Ak ( BjC ) = ( AkB ) j ( AkC ) is called the distributive law of intersection over union.

Example: Verify: ( I) Distributive law of union over intersection ( II) Distributive law of intersection over union where A = {1, 2, 3,...,20}, B = {5, 10, 15,...,30} and C = {3, 9, 15, 21,27, 33}.

Solution: ( I )

L.H.S = Aj(BkC) = {1,2,3, ... , 20}j({5, 10 ,15,... ,30}k{3, 9, 15, 21, 27, 33}) = {1, 2, 3, ...,20}j{15} ∴ Aj(BkC) = {1,2,3,...,20} .............(i) Now R.H.S = AjB = {1,2,3,...,20}j{5,10,15,...,30} = {1, 2, 3, ..., 20, 25, 30} and AjC = {1, 2, 3, ...,20}j{3, 9, 15, 21,27, 33} = {1, 2, 3, ..., 20, 21, 27, 33} (AjB)k(AjC) = {1, 2, 3, 4, ... , 20, 25, 30}k{1, 2, 3, ... ,20, 21, 27, 33} ∴ (AjB)k(AjC) = {1,2,3,...,20} .............(ii)

Thus, from (i) and (ii), we conclude that A j (B k C) = (A j B) k (A j C)

(II) L.H.S = Ak(BjC) = {1, 2, 3,...,20}k({5,10,15,.... ,30}j{3, 9, 15, 21, 27, 33}) = {1, 2, 3,... ,20}k{3, 5, 9, 10, 15, 20, 21, 25, 27, 30, 33} ∴ A k (B j C) = ( 3, 5, 9, 10, 15,20} ............. (i) R.H.S = A k B = {1, 2, 3, ..., 20} k {5, 10, 15, , 30} = {5, 10, 15, 20} A k C = {1, 2, 3,...,20} k {3, 9, 15, 21, 27, 33} = {3, 9, 15} (A k B) j (A k c) = {5, 10, 15,20} j {3, 9, 15} ∴ (A k B) j (A k c) = {3, 5, 9, 10, 15, 20} ............ (ii)

Thus, from (i) and (ii), we conclude that Aj(BjC) = (AkB)j(AkC)

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1.2.3 De Morgan’s Laws

lf A and B are the subsets of a universal set U , then (i) (A j B) c^ = Ac k B c^ (ii) (A k B) c^ = Ac j B c

Example: Verify De Morgan’s Laws if: U = {1, 2, 3,...,10}, A = {2, 4, 6} and B = {1,2,3,4,5,6, 7}

Solution: (i) L.H.S = (A j B) c A j B = (2,4,6} j (1,2,3,4,5,6,7} = {1,2, 3, 4, 5, 6, 7} ∴ (A j B) c^ = U - (A j B) = {8,9,10} ............. (i) R.H.S = Ac k B c Ac^ = U - A = {1, 2, 3, ... ,10} - (2, 4, 6} = {1, 3, 5, 7, 8, 9, 10} Bc^ = U - B Bc^ = {1, 2, 3, ...,10} - (1, 2, 3, 4, 5, 6, 7} Bc^ = {8, 9, 10} ∴ Ac k B c^ = {1, 3, 5, 7, 8, 9, 10} k (8,9,10} = {8,9,10} ............. (ii)

Thus, from (i) and (ii) we have (AjB) c^ =A c kB c

(ii) L.H.S = (A k B) c A k B = {2, 4, 6} k {1, 2, 3, 4, 5, 6, 7} = {2,4,6} (A k B)c^ = U - (A k B) = {1,2,3,..., 10} - {2,4, 6} = {1,3,5,7,8,9,10} .......... (iii) R.H.S = Ac^ = U - A = {1,2,3,..., 10} - {2,4, 6} = {1,3,5,7,8,9,10} and Bc^ = U - B = {1,2,3,..., 10} - {1,2,3,4,5,6, 7}

∴ Ac j B c^ = {1,3,5, 7,8,9,10} j {8,9,10} = {1,3,5,7,8,9,10} ............ (iv) Thus, from (iii) and (iv) , we have (A k B)c^ = Ac j B c

EXERCISE 1.

1. Verify: (a) A j B =B j A and (b) A k B =B k A, when (i) A = {1,2,3,......10}, B = {7,8,9,10,11,12} (ii) A = {1,2,3,......15}, B = {6,8,10,...., 20} 2. Verify: (a) X j (Y j Z ) = (X j Y ) j Z and (b) X k (Y k Z) = (X k Y) k Z, when (i) X = {a, b, c, d}, Y = {b, d, c,f} and Z = {c,f g, h} (ii) X = {1,2,3,..., 10}, Y = {2,4, 6, 7,8} and Z = {5, 6, 7,8} (iii) X = { - 1, 0,2,4,5}, Y = {1,2,3,4, 7} and Z = {4, 6,8,10} (iii) X = {1,2,3, ..., 14}, Y = {6,8, 10,..., 20} and Z = {1,3,5, 7} 3. Show that: if A = {a,b,c},B = {b,d,f} and C = {a,f,c}, A j( B k C ) = ( A j B )k( A j C ) 4. Show that: if A = {0}, B = {0,1} and C = { }, Aj(BkC) = (AjB) k (AjC) 5. Verify De Morgan’s Laws if: U = N, A = f , and B = P

1.3 VENN DIAGRAM

Operations on Sets Through Venn-diagram A universal set is represented in the form of a rectangle, its subsets are represented in the form of closed figures inside the rectangle Adjoining figure is the representation for A 5 U through Venn-diagram.

August De Morgan (1806 - 1871), a formulated British mathematician De Morgan's (^) laws.who

JohnEnglish Venn mathematician (1834-1923), whoan introduced Venn diagrams.

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Distributive Laws (a) Distributive Law of Intersection over Union Ak(BjC) = (AkB)j(AkC) Let A = {1,3,5, 7,9,10}, B = {2,4, 6,8,9,10} and C = {2,3,5, 7,11,13} L.H.S = Ak(BjC) BjC = {2,4, 6,8,9,10} j {2,3,5, 7,11,13} = {2,3,4,5, 6, 7,8,9,10,11,13} Ak(BjC) = {1,3,5, 7,9,10} k{2,3,4, 5, 6, 7, 8,9,10,11,13} = {3,5, 7,9,10} Horizontal lines represent BjC and vertical lines Ak(BjC). Thus, slanting lines represent Ak(BjC).

R.H.S = (AkB)j(AkC) AkB = {1,3,5,7,9,10}k{2,4,6,8,9,10} = {9,10} AkC = {l,3,5,7,9,10}k{2,3,5,7,11,13} = (3,5,7} (AkB)j(AkC) = {9,10}k{3, 5,7} = {3,5,7,9,10} Horizontal lines represent A k B , vertical lines represent A k C and slanting lines represent (A k B) j (A k C ). From fig. (ix) and (x), it is clear that A k (B j C) = (A k B) j (A k C) Hence distributive law of intersection over union holds.

(b) Distributive Law of Union over Intersection Aj(BkC) = (AjB)k(AjC) A = {1,3,5,7,9,10}, B = {2,4,6,8,9,10} and C = {2,3,5, 7,11,13} L.H.S = Aj( BkC) BkC = {2,4, 6, 8,10}k {2,3, 5, 7,11,13} = {2} Horizontal lines represent BkC, vertical lines represent Aj(BkC ). Thus, slanting lines represent Aj(BkC ). Aj(BkC) = {1, 3, 5, 7, 9,10,}j{2} = {1,2, 3, 5, 7, 9,10} R.H.S = (AjB) k (Aj C) AjB = {1,3, 5, 7,9,10} j {2,4, 6, 8,9,10} = {1,2,3,4, 5, 6, 7, 8,9,10} AjC = {1,3,5, 7,9,10} j {2,3,5, 7,11,13} = {1,2,3,5, 7,9,10,11,13} (AjB)k(AjC) = {1,2,3,4,5, 6, 7,8,9,10} k {1,2,3,5, 7,9,10,11,13} ={1,2,3,5,7,9,10}

Horizontal lines represent AjB , vertical lines represent AjC and slanting lines represent (AjB)k(AjC). From fig. (xi) and (xii), it is clear that Aj(BkC) = (AjB)k(AjC). Hence, distributive law of union over intersection holds.

EXERCISE 1.

1. Verify the commutative law of union and intersection of the following sets through Venn diagrams. (i) A = {3,5, 7,9,11,13} (ii) The sets N and Z B = {5,9,13,17,21,25} (iii) C = { x | x d N/ 8 7 x 7 18} (iv) The sets is and O D = { y | y dN/ 9 7 y 7 19} 2. Copy the following figures and shade according to the operation mentioned below each: 3. For the given sets, verify the following laws through venn diagram. (i) Associative law of Union of sets. (ii) Associative law of Intersection of sets. (iii) Distributive law of Union over intersection of sets. (iv) Distributive law of Intersection over Union of sets. (a) A = {2,4, 6,8,10,12}, B = {1,3,5,7,9,11} and C = {3, 6,9,12,15} (b) A = { x | x dZ/ 8 7 x 7 25}, B = {y|ydZ/ - 2<y<6} and C = {z|zdz/z 7 8}

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4. Copy the following Venn diagrams and shade according to the operation, given below each diagram.

REVIEW EXERCISE 1

1. Four options are given against each statement. Encircle the correct one. 2. Write the short answers of the following questions. i. Define a set. ii. What is the difference between whole numbers and natural numbers? iii. Define the proper and improper subsets. iv. Define a power set. v. Define De Morgan’s Laws. 3. Write all subsets of the following sets. i. A = {e ,f, g} and B = {1,3,5} ii. Write the power set of {a, b, c} iii. Verify De Morgan’s Laws if U = {a, b, c, d, e}, A = {9, 6} and B = {a,b,c}

SUMMARY

• Set is defined as “a collection of well defined distinct objects”. These objects are called elements or members of a set.
• A set A is a subset of a set B if every element in set A is also an element in set B.
• The empty set is a subset of all sets.
  • If A is a subset of B and A is not equal to B (i.e. there exists at least one element of B not contained in A ), then A is a proper subset of B , denoted by AfB.
  • If A is a subset of B and A is equal to B (i.e. every element of B is also the element of A ), then A is an improper subset of B , denoted by, A = B.
  • Intersection of two sets AkB , is a set which consist of only the common elements of both A and B.
  • Union of two sets AjB , is a set which consists of elements of both A and B with common elements represented only once. - A and B are any two sets, then i. AjB = BjA (Commutative law over union) ii. AkB = BkA (Commutative law over intersection)
  • Let A, B and C be any three sets, then i. Aj(BjC) = (AjB)jC (Associative law of union of sets) ii. Ak(BkC) = (AkB)kC (Associative law of intersection of sets)
  • Let A, B and C be any three sets, then distributive laws are given below i. Aj(BkC)=(AjB)k(AjC) (distributive law of union over intersection) ii. Ak(BjC) = (AkB)j(AkC) (distributive law of intersection over union)
  • Let A, B and C be any three sets then according to the De Morgan laws. i. (AjBc^ = AckB c ii. (AkB) c^ = AcjB c
  • A Venn diagram is a pictorial representation of sets and operations performed on sets.

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2.1 IRRATIONAL NUMBERS

2.1.1 Definition of an Irrational Number

The numbers which cannot be written in the form pq where p , qdz and q ≠ 0, are called irrational numbers. We know that there is no such rational number whose square is 2.

Therefore, the square root of 2 is not a rational number. Similarly (^2) , 2, 5 7 and 2 3 are not rational numbers. These are called irrational numbers. The set of irrational numbers is denoted by Q ′^. It can also be defined as a number whose decimal representation is non-terminating and non-recurring is called an irrational number.

2.1.2 Recognition of Rational and Irrational Number

We have already learnt about rational numbers and irrational numbers. Now we recognize these numbers with the help of the following examples.

Example 1: Which of the following numbers are rational numbers?

(^2) , 9, 7 , 16 , 6 , 5, 7, 25 3 9 25 11

Solution:

The numbers 2 , 9, 7 , 16 , 6 , 3 9 25 11

• (^) and 25 are rational numbers because all of

these numbers can be expressed in the form of pq , where p, qd Z and q ≠ 0.

Example 2: Which of the following numbers are irrational numbers? 2, 1.7320505, 4, 2.236068, 16, 17, 19, 25, 37 Solution: The numbers 2, 1.7320505, 2.236068, 17, 19 and 37 are irrational numbers because all of these cannot be written in the form of pq , where p , qd Z and q ≠ 0

2.1.3 Real Numbers

Now we define the set of Real Numbers as: “The union of the set of rational numbers Q and the set of irrational numbers Q ′^ is called the set of Real Numbers and is denoted by R. i.e ., R = Q j Q ′

2.1.4 Demonstration of Non-Terminating / Non Repeating

decimals

- Terminating decimal fractions The decimal fraction in which the number of digits after the decimal point is finite or, while converting a rational number to the decimal fraction the division process ends, then it is called a terminating decimal fraction. These fractions can easily be converted in the form p q of rational numbers where^ p, qdZ^ and^ q^ ≠^ 0, as 0.25, 3.125 and 0.0625 etc. are also the examples of terminating decimal fractions. Look at the following examples:

Example 1: Convert common fraction 9 4

to decimal.

Solution:

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Example 2: Convert common fraction 1 9

to decimal.

Solution:

∴

9 =^ 0.1111....^ (terminating/repeating)

Non Terminating decimal fractions The decimal fraction in which the number of digits after the decimal point is infinite or while converting a rational number into the decimal fraction, the division process does not end, then it is called a non-terminating/non-repeating decimal fraction. It can be explained through the following example:

Example 3: Convert common fraction 9 7

to decimal.

Solution:

= 1.285714......

We have seen that in Example 1 the decimal 2.25 has terminated/ended after 2 digits and in Example 2, the decimal 0.1111 non terminating but repeating. Whereas in Example 3, the decimal fraction 1.285714 ... does not end but it goes on forever. The (...... ) or the line over decimals indicates that decimal are non-terminating. It may also be noted that none of the digits is being repeated. So, this type of decimal fraction is known as non-terminating and non-repeating decimal.

Note that: The decimals which are non-terminating and non-repeating are called irrational numbers.

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In the above pattern we notice that: (i) Each row starts and ends by digit 1. (ii) The digits increase upto the number whose square is required and then decrease. (iii) The number of digits in each row increases by 2. (iv) The difference of any two consecutive squares is an odd number. (v) The number of digits in a particular row is the addition of the number and the previous consecutive numbers whose squares are to be found.

Consider another pattern of squares of natural numbers.

12 = 1 = 1 22 = 1 + 3 = 4 3 2 = 1 + 3 + 5 = 9 42 = 1 + 3 + 5 + 7 = 16 5 2 = 1 + 3 + 5 + 7 + 9 = 25 6 2 = 1 + 3 + 5 + 7 + 9 + 11 = 36 72 = 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 82 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 9 2 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81 10 2 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

We observed the pattern and note that: (i) The summation is an ascending order. (ii) The square of each number is written as the sum of odd numbers only. (iii) Each row of the pattern starts from an odd number 1. (iv) The number of odd numbers in each row is equal to the number whose square is to be found. (v) The sum of each row is equal to the required square. (vi) The last odd number in each row is one less than the double of the given number.

EXERCISE 2.

1. Find the perfect square of the following numbers. (i) 7 (ii) 11 (iii) 19

(iv) 25 (v) 37 (vi) 75

2. Write the summation patterns for the following squares. (i) 6 2 (ii) 72 (iii) 42

(iv) 5 2 (v) 32 (vi) 82

2.3 SQUARE ROOT

2.3.1 Finding the square root of (a) a natural number (b) a common fraction

(c) a decimal given in perfect square form, by prime factorization and

division method

The square root of a positive number is that positive number whose square is the given number. The symbol used for square root is.

(a) Finding square root of a natural number.

- By Prime Factorization Method First of all find prime factors, then make pairs of these factors. Choose one prime number from each pair and then find the product of all those prime factors, which will be the square root of the given number.

Example 1: Find the square root of 225

Solution: 225 = 3 x 3 x 5 x 5

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225 3 3 5 5 = 3 5 = 15

= × × × ×

Example 2: Find the square roots of 576

Solution:

Example 3: Find the square roots of 1600

Solution:

- By Division Method: To find the square root of natural numbers by division method, we will proceed as under: (i) Make pairs of digits from right to left. If the number of digits is even, we have complete pairs. If the number of digits is odd, the last digit on extreme left will remain single. (ii) Look for the numbers whole square is equal to or less than the number of extreme left, which may be a single digit or a pair. This number will be divisor as well as quotient.

(iii) Subtract the product. Bring down the next pair to the right of the remainder. (iv) Double the quotient and write as divisor as ten’s digit. (v) Look for the number whose square will be equal or less than the dividend. Write that number with the right side of the quotient as well as with divisor at unit place.

Example 1: Find the square root of 625

Solution:

∴ 625 = 25

Example 2: Find the square root of 1024

Solution:

∴ 1024 = 32

Example 3: Find the square root of 15129

× × × × × × ×

× × × × × × ×

× × ×

× × × × × × ×

× × × × × × ×

× × ×

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Example 2: Find the square root of 967 121 Solution:

EXERCISE 2.

1. Find the square root of the following fractions by prime factorization.

(i)^4964 `

(ii)^121625 (iii)^196441

(iv) 11336 (v)^676729 (vi) (^12 )

2. Find the square root of the following fractions by division method.

(i)^144225 (ii)^169256 (iii)^784841

(iv)^1024 1225 (v) 5 41 1 64 2 (vi) 967 121 (c) Finding square root of a decimal

- By Prime Factorization We convert the decimal to common fraction and then find square root.

Example 1: Find the square root of decimal 0.

Solution:

Example 2: Find the square root of decimal 2.

Solution:

967 1156 121 121 Now 967 1156 121 121 1156 121 34 11 3 11 9 67 31 121 11

∴ =

Now 64 64 (^100 ) 2 2 2 2 2 2 2 2 5 5 2 2 2 2 2 2 2 2 5 5 2 2 2 8 2 5 10

0.64 0.

= ×^ ×^ ×^ ×^ ×

× × ×

= ×^ ×^ ×^ ×^ ×

× × ×

= ×^ × =

×

= = ×^ ×^ ×

× × ×

× × ×

× × ×

= × =

×

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- By Division Method: For using this method the following steps will be taken. (i) Make pairs of digits on the left side of the decimal point from right to left. (ii) Make pairs of digits on the right side of the decimal point from left to right. (iii) Place the decimal point in the quotient while bringing down the pair after the decimal point. (iv) While bringing down two pairs at a time, place a zero in the quotient. This method is illustrated with the following examples.

Example 1: Find the square root of 180.

Solution:

∴ 180.9025 =13. Example 2: Find the square root of 0.

Solution:

∴ 0.053361 =0.

Example 3: Find the square root of decimal 152.

Solution:

∴ 152.7696 =12.

EXERCISE 2.

1. Find the square root of the following decimals by prime factorization. (i) 1.21 (ii) 0.64 (iii) 7.

(iv) 1.44 (v) 1.69 (vi) 12.

2. Find the square root of the following decimals by division method. (i) 0.3249 (ii) 0.5184 (iii) 10.

(iv) 20.5209 (v) 648.7209 (vi) 2981.

(vii) 7613.609536 (viii) 0.00868624 (ix) 2374.

1 x 21 = 21 2 x 22 = 44 3 x 23 = 69 4 x 24 = 96 1 x 261 = 261 2 x 262 = 524 3 x 263 = 789 4 x 264 = 1056 5 x 265 = 1325 1 x 2621 = 2681 2 x 2682 = 5364 3 x 2683 = 8049 4 x 2684 = 10736 5 x 2685 = 13425

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2. Find the square root of the following upto two decimal places.

(i) 3.6 (ii) 6.4 (iii) 28.

(iv) 63.34 (v) 816.081 (vi) 36.

2.3.3 Use the Rule to Determine the Number of Digits in the Square Root of

a Perfect Square

Rule: Let n be the number of digits in the perfect square then its square root contains:

(i) 2

n (^) digits if n is even

(ii)^1 2

n + (^) digits if n is odd

Now we apply the above rule for finding the number of digits in the square root of a perfect square with the help of following examples:

Example 1: Find the number of digits in the square root of 49729

Solution: Number of digits of the given number = 5 n = 5 is odd, so mentioned above rule (ii) will be applied

∴ Thus the number of digits in the square root will be 1 5 1 2 2

=^ n^ +^ = +^6 2

∴ To check the answer, we proceed as under

∴ 49729 = 223 The square root 223 has 3 digits

Example 2: Find the number of digits in the square root of 10329796

Solution: Number of digits ( n ) = 8 Now n = 8 is even, so part (i) of the rule will be applied

∴ The number of digits in the square root 8 4 2 2

=^ n = = Now, we can verify it

∴ 10329796 = 3214 The square root 3214 has 4 digits

EXERCISE 2.

1. Find the number of digits in the square root of the following perfect square (i) 63504 (ii) 66564 (iii) 50625 (iv) 837225 (v) 839056 (vi) 1054729 (vii) 1577536 (viii) 2119936 (ix) 3283344 (x) 614656 (xi) 7778521 (xii) 12880921

2.3.4 Real Life Problems Involving Square Root

Example 1: 1225 students stand in rows in such a way that the number of rows is equal to the number of students in a row. How many students are there in each row?

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Solution: Since the number of students in a row is the same as the number of rows, square root of 1225 will be found.

Thus, the number of students in each row = 35

Example 2: A rectangular field has an area of 18432 square meters. Its width is half as long as its length. Find its perimeter.

Solution: Since the width of the field is half as long as its length, this rectangle can be divided into two square regions.

∴ The area of each square region = 18432 2

= 9216 m^2 To find the length of its side, we will find the square root of 9216.

∴ The width of each side (width) 96 meters. So the length of the rectangle = 96 x 2 = 192 meters. Thus the perimeter = 2(192 + 96) = 2(288) = 576 meters.

Example 3: Find the least number which, when subtracted from 58780, the answer is a complete square.

Solution: To find which number is subtracted from the given number, we find the square root of 58780 and the remainder will be the required number.

Remaining Number = Given number - Remainder = 58780 - 216 = 58564 Thus, if 216 is subtracted from 58780, the remaining number 58564 will be a complete square.

EXERCISE 2.

1. The area of a square field is 14400 sq. meter. Find the length of the side of the square.
2. The area of a square field is 422500 sq. meter. How much string is required for fixing along the sides as a fence?
3. A gardener wants to plant 122500 trees in his field in such a way that the number of trees in a row is equal to the number of rows. How many trees will he plant in each row?
4. The area of a rectangular field is 10092 sq. meter. Its length is three times as long as its width. Find its perimeter.
5. The area of a circular region is 616 sq. decimeter. Find its radius.  π^ ≅^227 
6. A rectangular field has an area 28800 sq. meter. Its length is twice as long as its width. What is the length of its sides?
7. Find that least number which, when subtracted from 109087, the answer is a complete square.