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MATHEMATICS
SQUARES & CUBES
SQUARES
The square of a number is the product of the number with the number itself. For a given number x, the
square of x is (x × x), denoted by x2.
Perfect squares of square numbers
A natural number is called a perfect square or a square number if it is a square of some natural number.
Examples:
We know that 1 = 12; 4 = 22; 9 = 32; 16 = 42; 25 = 52 and so on.
Thus 1, 4, 9, 16, 25 etc. are perfect squares.
Example 1.
Solution:
Is 196 a perfect square? If so find the number whose square is 196.
Resolving 196 into prime factors, we get
196 = 2 × 2 × 7 × 7.
Thus, 196 can be expressed as a product of pairs of equal factors.
∴ 196 is a perfect square.
Also, 196 = (2 × 7) × (2 × 7) = (14 × 14) = (14)2
Hence, 14 is a number whose square is 196.
Properties of perfect squares
1.
2.
3.
4.
5.
6.
7.
8.
A number ending 2, 3, 7 or 8, is never a perfect square.
A number ending in an odd number of zeros is never a perfect square.
The square of an even number is always even.
The square of an odd number is always odd.
The square of a proper fraction is smaller than the fraction.
For every natural number n, we have
(n + 1)2 – n2 = (n + 1 + n)(n + 1 – n) = {(n + 1) + n}
∴ {(n + 1)2 – n2} = {(n + 1) + n}.
For every natural number n, we have
sum of the first n odd natural numbers = n2.
Three natural numbers m, n, p are said to form a Pythagorean triplet (m, n, p) if m2 + n2 = p2.
Square root
The square root of a number x is that number which when multiplied by itself gives x as the product.
Square root of a perfect square by the prime factorisation method:
When a given number is a perfect square. We find its square root by the following steps:
1.
Resolve the given number into prime factors.
2.
make pairs of similar factors.
3.
take the product of the prime factors, choosing one factor out of every pair.
Square root of a perfect square by the long division method
When numbers are very large, the method of finding their square roots by factorisation becomes lengthy
and difficult. So we use the long division method which is explained in the following steps.
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PDT Courseware– 8th moving to 9TH-S & C-MA- 2
1.
2.
3.
4.
5.
Group the digits in pairs, starting with the digit in the units place. Each pair and the remaining digit
if any, is called a period.
Think of the largest number whose square is equal to or just less than the first period. Take this
number as the divisor and also as the quotient.
Subtract the product of the divisor and the quotient from the first period and bring down the next
period to the right of the remainder. This becomes the new dividend.
Now, the new divisor is obtained by taking two times the quotient and annexing with it a suitable
digit which is also taken as the next digit of the quotient, chosen in such a way that the product of
the new divisor and this digit is equal to or just less than the new dividend.
Repeat step (2), (3) and (4) till all the periods have been taken up. Now the quotient so obtained
is the required square root of the given number.
Square root of numbers in decimal form
Make the number of decimal places even by affixing a zero, if necessary. Now mark periods and find out
the square root by the long division method. Put the decimal point in the square as soon as the integral
part is exhausted.
Illustration 1: Evaluate 42.25 .
Solution:
Using the division method we may find the square root of the given number as show
below.
6 42.25(6.5
36
125 625
625
x
∴
42.25 = 6.5
To find the value of square root correct up to certain places of decimal
If the square root is required correct up to two places of decimal, we shall find it up to 3 places of decimal
and then round it off up to two places of decimal.
Similarly, if the square root is required correct to three places of decimal. We shall find it up to 4 places of
decimal and then round it off up to three places of decimal, and so on.
Illustration 2: Evaluate
Solution:
2 correct up to two places of decimal.
Using the division method we may find the value of
1
2 as shown below.
2.00 00 00(1.414
1
24 100
96
281 400
281
2824 11900
11296
604
∴
2 = 1.414 ⇒
2 = 1.41 (correct upto 2 places of decimal)
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Square roots of fractions
For any positive numbers a and b, we have
(i)
ab = ( a × b )
(ii)
a
=
b
a
b
.
Illustration 3: Evaluate
Solution:
441
.
961
441
441
=
961
961
Now find the square roots of 441 and 961, as shown below.
We have
2 4.41(21
4
41 41
3 9.61(31
9
61 61
41
x
Thus,
⇒
61
x
441 = 21 and
441
=
961
961 = 31
441
21
=
.
31
961
Short cut methods for squaring a number:
1.
Column method for squaring a two digit number
Let the given number have the tens digit = a and the units digit = b. Then, we have to square this number.
Step 1. make three columns, I, II and III, headed by a2, (2 × a × b) and b2 respectively. Write the values of
a2, (2 × a × b) and b2 in columns I, II and III respectively.
Step 2. In column III, underline the units digit of b2 and carry over the tens digit of it to column II and add
it to the value of (2 × a × b).
Step 3. In column II, underline the units digit of the number obtained in Step 2 and carry over the tens
digit of it to column I and add it to the value of a2.
Step 4. Underline the number obtained in step 3 in column I.
The underlined digits give the required square number.
Illustration 4: Find the square of
(i)
47
(ii)
86
Solution:
(i) Given number = 47
∴ a = 4 and b = 7
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PDT Courseware– 8th moving to 9TH-S & C-MA- 4
I
II
III
a2
(2 × a × b)
b2
16
+6
22
56
+4
60
49
∴ (47)2 = 2209.
(ii) Given number = 86
∴ a = 8 and b = 6
I
II
III
a2
(2 × a × b)
b2
64
+9
73
96
+3
99
36
∴ (86)2 = 7396.
2.
Diagonal method of squaring a number
Step 1. The given number contains two digits. So draw a square and divide it into 4 subsquares as shown
below. Write down the digits 3 and 9, horizontally and vertically, as shown below.
Step 2. Multiply each digit on the left of the square with digit on the top, one by one. Write the product in
the corresponding subsqure.
If the product is a one digit number. Write it below the diagonal and put 0 above the diagonal.
In case the product is a two digit number, write the tens digit above the diagonal and the units
digit below the diagonal.
Step 3. Starting below the lowest diagonal, sum the digits diagonally.
If the sum is a two digit number, underline the units digit and carry over the tens digit to the next
diagonal.
Step 4. Underline all the digits in the sum above the topmost diagonal.
Step 5. The underline digits give the required square number.
∴ 392 = 1521.
CUBES
CUBE OF A NUMBER
For a given number x, we define, cube of x = x × x × x, denoted by x3.
Example:
(i) 23 = (2 × 2 × 2) = 8. Thus, cube of 2 is 8.
(ii) 33 = (3 × 3 × 3) = 27. Thus, cube of 3 is 27.
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Perfect Cube: A natural number is said to be a perfect cube if it is the cube of some natural number.
13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125 etc.
Thus 1, 8, 27, 64, 125 etc. are perfect cubes.
Example:
Illustration 5: Show that 189 is not a perfect cube.
Resolving 189 into prime factors, we get (3 × 3 × 3) × 7
Making triplets, we find that one triplet is formed i.e.(3 × 3 × 3) and we are left with one
more factor i.e. 7.
Thus, 189 cannot be expressed as a product of triplets.
Hence, 189 is not a perfect cube.
Solved:
CUBE OF NEGATIVE INTEGERS
The cube of a negative integer is always negative.
Example:
(− 1)3 = (− 1) × (− 1) × (− 1) = − 1, (− 2)3 = (− 2) × (− 2) × (− 2) = − 8,
(− 3)3 = (− 3) × (− 3) × (− 3) = − 27, etc.
CUBE OF NEGATIVE INTEGERS
3
a a a a × a × a a3
a
=
We have,   = × × =
.
b b b b × b × b b3
b
3
a3
a
Hence   = 3 .
b
b
PROPERTIES OF CUBE OF NUMBERS
(i)
(ii)
The cube of every even number is even.
The cube of every odd number is odd.
CUBE ROOTS
The cube root of a number x is that number whose cube gives x. We denote the cube root of x by
Example:
(i) Since (2 × 2 × 2) = 8, we have 3 8 = 2 .
(ii)
Since (5 × 5 × 5) = 125, we have 3 125 = 5 .
3
x.
Method of Finding the Cube Root of a Given Number by Factorisation:
Step 1. Express the given number as the product of primes.
Step 2. Make groups in triplets of the same prime.
Step 3. Find the product of primes, choosing one from each triplet.
Step 4. This product is the required cube root of the given numbers.
Illustration 6: Evaluate
Solution:
3
216 .
By prime factorization, we have
216 = 2 × 2 × 2 × 3 × 3 × 3
= (2 × 2 × 2) × (3 × 3 × 3).
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PDT Courseware– 8th moving to 9TH-S & C-MA- 6
∴
3
216 = (2 × 3) = 6.
Cube Root of a Negative Perfect Cube:
Let ‘a’ be a positive integer. Then, (− a) is a negative integer.
We know that (− a)3 = − a3.
3
∴ −a3 = −a .
Thus, cube root of (− a)3 = − (cube root of a3).
Thus, 3 − x = − 3 x .
Cube Root of Product of Integers
We have
3
(
ab =
3
a×3b
Illustration 7: Evaluate
)
3
125 × 64 .
We have,
3
125 × 64 = 3 125 × 3 64
= 3 5×5×5 × 3 4×4×4
= (5 × 4) = 20.
Solution:
Cube Root of a rational number
We define
3
a
=
b
3
a
3
b
.
Illustration 8: Evaluate (i)
Solution:
We have,
216
=
(i) 3
2197
(ii)
3
-125
=
512
3
3
216
2197
216
3
2197
3
−125
3
512
=
=
(ii)
3
6×6×6
3
=
6
.
13
3
13 × 13 × 13
3
( −5) × ( −5) × ( −5)
3
8×8×8
-125
.
512
=
−5
.
8
Short-cut Method for Finding the Cube of a Two-digit Number
We have (a + b)3 = a3 + 3a2b + 3ab2 + b3.
Method:
For finding the cube of a two-digit number with the tens digit = a
and the units digit = b, we make four columns, headed by
a3, (3a2 × b), (3a × b2) and b3.
The rest of the procedure is the same as followed in squaring a number by the column
method. We simplify the working as
a2
×a
a3
a2
× 3b
3a2b
b2
× 3a
3ab2
b2
×b
b3
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Cube root of a perfect cube (using pattern)
Look at the following pattern:
1 3 = 1 ∴ 13 − 03 = 1 = 1 + 0 × 6
2 3 = 8 ∴ 23 − 13 = 7 = 1 + 1 × 6
33 = 27 ∴ 33 − 23 = 19 = 1 + 1 × 6 + 2 × 6
…..
…..
……
……
……
……
1× 0
×6
2
2 ×1
= 1+
×6
2
3×2
= 1+
×6
2
……..
……..
= 1+
…..
…..
93 = 729 ∴ 93 − 83 = 217 = 1 + 1 × 6 + 2 × 6 + … + 8 × 6 = 1 +
9×8
×6
2
From above, we get
13 = 1
23 = 1 + 7
33 = 1 + 7 + 19
43 = 1 + 7 + 19 + 37
….. …… …..
……
….. …… …..
……
93 = 1 + 7 + 19 + 37 + …… + 217
∴ 23, 33, 43, …… , 93 are the sums of the first 2, 3, 4, …. 9 numbers out of 1, 7, 19, 37, ….. , 217
2 ×1 

× 6 ,
respectively. The numbers 1, 7, 19, 37, …… , 217 are respectively. The numbers 1,  1 +
2


4×2
 3×2
 

 9×8

× 6 .
× 6  , 1 +
× 6  , ….. ,  1 +
1 +



 

2
2
2
Thus to find the cube root of a perfect cube, we keep on subtracting 1, 7, 19, 37, ….. , 217, …. till we get
the remainder as zero. The number of times the subtraction is carried out, gives the cube root. This is
also called successive subtraction method.
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SOLVED PROBLEMS
Example 1:
Find the square root of 324.
Solution:
By prime factorisation, we get
324 = 2 × 2 × 3 × 3 × 3 × 3
∴ 324 = (2 × 3 × 3) = 18.
Example 2:
Find the square root of 784 by the long division method.
Solution:
Marking periods and using the long division method, we have
2 7 84(28
4
48 384
384
x
∴
784 = 28.
Example 3:
Find the square root 1
Solution:
56
225
=
1
169
169
56
.
169
56
225
225
.
1
=
=
169
169
169
We find the square roots of 225 and 169 separately, as shown below.
∴
1 225(15 1 169(13
1
1
25 125
23 69
69
125
x
x
∴
225 = 15 and
56
225
1
=
=
169
169
15
2
=
=1
.
13
13
⇒
169 = 13
225
169
Example 4:
Find the cube root of (− 1000).
Solution:
We know that 3 −1000 = − 3 1000 .
Resolving 1000 into prime factors, we get
1000 − 2 × 2 × 2 × 5 × 5 × 5
= (2 × 2 × 2( × (5 × 5 × 5).
∴ 3 1000 = (2 × 5) = 10 .
∴
3
−1000 = −
(
3
)
1000 = −10 .
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PDT Courseware– 8th moving to 9TH-S & C-MA- 9
Example 5:
Find the value of (29)3 by the short-cut method.
Solution:
Here, a = 2 and b = 9.
a2
×a
a3
4
×2
8
+ 16
24
a2
× 3b
3a2b
4
× 27
108
+ 55
163
b2
× 3a
3ab2
81
×6
486
+ 72
558
b2
×b
b3
81
×9
729
∴ (29)3 = 24389.
Example 6:
Find the smallest number that must be subtracted from 792 to make it a perfect
cube.
Solution:
792 − 1 = 791 ; 791 − 7 = 784 ; 784 − 19 = 765 ; 765 − 37 = 728 ; 728 − 61 = 667 ;
667 − 91 = 576 ; 576 − 127 = 449 ; 449 − 169 = 280 ; 280 − 217 = 63.
The next number to be subtracted from 63 is 271 which is greater than 63. Hence 63 is to
be subtracted from 792.
Example 7:
What least number should be subtracted from 5634 so that the resulting number
becomes a perfect square ?
Solution:
The remainder 9 shows that if the square root will be 75 and the resulting number will be
a perfect square.
∴ 9 is to be subtracted.
7
75
5634
49
734
725
9
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PDT Courseware– 8th moving to 9TH-S & C-MA- 10
ASSIGNMENT PROBLEMS
Subjective
1.
The area of a square field is 60025 m2. A man cyles along its boundary at 18 km/hr. In how much
time will he return to the starting point?
2.
Evaluate the following
(i)
.4225
(ii)
(iii)
10609
(iv)
3
-512
343
15876
3.
Without adding, find the sum (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23).
4.
Evaluate (218)2 − (217)2.
5.
Find the least square number which is exactly divisible by each of the numbers 8, 12, 15 and 20?
6.
Find the value of (71)3 by short cut method.
7.
What least number should be multiplied with 4464 to make it a perfect square?
8.
Find the smallest number which should divide 24696 so that it be comes a perfect cube. Find the
cube root of the quotient.
9.
If
10.
Find the largest perfect square number of three digits divisible by 10, 15 and 25?
1296
x
=
, then find x ?
x
2.25
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Objective
Multiple choice questions (singe option correct):
1.
The least numbers which must be subtracted from the number 2361 so as to leave a perfect
square
(A) 37
(B) 47
(C) 57
(D) 67
2.
The square roots of the number 7.29 is
(A) 3.7
(C) 5.7
(B) 2.7
(D) 6.7
The square root of 0.008281 is
(A) 0.091
(C) 0.089
(B) 0.071
(D) 0.099
3.
4.
The square root of
16641
is
4489
129
67
127
(C)
65
127
67
129
(D)
65
(A)
5.
(B)
The square root of 56
569
is
1225
263
35
265
(C)
35
263
33
266
(D)
35
The cubes of the number 35 is
(A) 42375
(C) 42377
(B) 42376
(D) 42875
(B)
(A)
6.
7.
Which of the following numbers are not perfect cubes?
(A) 64
(B) 216
(C) 243
(D) 1728
8.
The cube roots of − 226981 is
(A) −53
(C) −69
(B) −61
(D) −57
The cube roots of number 389017 is
(A) 75
(C) 73
(B) 72
(D) 79
9.
10.
The cube roots of
(A) 7/5
(C) 3/5
343
is
125
(B) 5/7
(D) 5/9
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PDT Courseware– 8th moving to 9TH-S & C-MA- 12
11.
12.
The square root of 0.4 is
(A) 0.2
(C) 0.63
The value of
(B) 0.02
(D) none of these
2
will be approximately
3
(A) 0.67
(C) 0.82
(B) 8.2
(D) 6.7
13.
By what smallest number 13500 be multiplied so that the product becomes a perfect cube ?
(A) 2
(B) 3
(C) 5
(D) none of these
14.
The square of x is an odd number, and then x can be
(A) 2248
(B) 1392
(C) 2223
(D) 28
15.
To get a perfect square we should divide 3698 by
(A) 2
(B) 3
(C) 4
(D) 5
16.
The least number of 4 digits which is perfect square is
(A) 1000
(B) 1004
(C) 1024
(D) none of these
17.
The unit digit of the square of 12796 is
(A) 5
(C) 7
(B) 6
(D) 8
The number of perfect squares upto 200 is
(A) 12
(C) 14
(B) 13
(D) 15
The square of 1001 is
(A) 10101
(C) 10200201
(B) 1002001
(D) None of these
18.
19.
20.
The product of two numbers is 3042. If one number is 18 times the other, the smallest number is
(A) 13
(B) 14
(C) 15
(D) 16
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PDT Courseware– 8th moving to 9TH-S & C-MA- 13
ANSWERS
TO
ASSIGNMENT
Subjective
1.
196 sec.
2.
(i)
3.
144
4.
435
5.
3600
6.
357911
7.
31
8.
14
9.
9
10.
900
.65
(ii)
−8/7
(iii)
103
(iv)
126
2.
6.
10.
14.
18.
B
D
A
C
C
3.
7.
11.
15.
19.
A
C
C
A
B
4.
8.
12.
16.
20.
A
B
C
C
A
Objective
1.
5.
9.
13.
17.
C
A
C
A
B
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